# The Chebyshev Set Problem in Riesz Space

Authors: Shengwei Wu; Jiarui Zhao; Yanyan Xu; Guanggui Chen; Na Cheng
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, we mainly study the best approximation theory in Riesz space, which is not constructed by the norm, but only rely on the order structure. Based on the order structure, we propose the concept of the order best approximation in Riesz space and discuss some problems related to the order best approximation, including some sufficient and necessary conditions for satisfying the order best approximation set. Finally, we consider the order best approximation projection and its related properties.


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## Body

## 1. Introduction

The order structures occurred in a natural manner in many fields, such as spaces of real continuous functions, spaces of measurable functions and so on. In 1928, Riesz [1] first proposed the concept of Riesz space and emphasized that we must consider the order structure, while studying the algebraic and topological properties of normed spaces. Therefore, the research on partially ordered vector space has caused great interest of researchers and developed greatly in recent years. In 1936, Kantorovich [2] established the axiomatic basis of lattice structure in Riesz space and studied some properties of lattice structure in Riesz space. In the mid-1930s, Birkhoff’s work [3–7] started the general development of lattice theory. In a series of wonderful articles [8–11], they presented the importance of lattice theory which can provide a unified framework for the hitherto unrelated development of many mathematical disciplines. In particular, numerous scholars combined functional analysis with lattice theory to transform specific analysis problems into pure algebra, topology, and order structure problems, which made the application of concepts, theorems, and methods more extensive and in-depth. For example, Schaefer [12] studied the basic theories of vector lattices and Banach Lattices, which laid the foundation for operator theory. After that, Aliprantis and Burkinshaw [13] had further exploration and promotion of operator theory in Riesz space. Luxemburg [14] pointed out that the order structure was another crucial structure in real analysis in addition to the basic arithmetic structure which allowed the comparison of the quantities represented by inequalities and studies many other complex characteristics. Luxemburg and Zaanen [15] and Zaanen [16] systematically summarized some basic theories of Riesz Space, making the research of Riesz Space more systematic and efficient. Many analytical properties such as various kinds of convergences in terms of the partial ordering were studied in [17–21].The idea of approximation has always been an important idea in various fields of mathematics. Based on algebraic structures, the classical approximation theory studies the approximation characteristics of various function classes in normed linear space [21–23]. More and more scholars also try to combine the order structures with the approximation theory and obtain some approximation characteristics in Banach Lattices. In 1992, Kurc [24] discussed some monotonic properties of Luxemburg norm in Musielak-Orlicz space and gave some applications to the best approximation. Then, Hudzik and Kurc [25] and Chen et al. [26] continued Kurc’s work and discussed the more general best approximation problems in Banach Lattices by using some monotonicities. Therefore, the purpose of this paper is to continue the above research by combining the order structure with the best approximation problems to study the best approximation theory in Riesz space.This paper is organized as follows. In Section2, we recall some notations of the best approximation in normed linear spaces and Riesz space. In Section 3, we consider the problem of the best approximation in Riesz space which is the main content and innovation of this paper. In Section 4, we further study the order best approximation projection and some related properties such as the boundedness and continuity.


## 2. Preliminary

First, we introduce the definition of the classical best approximation in normed linear spaces.Definition 1. See [27].
LetE be a normed linear space and S be a nonempty subset of E. Given f∈E, the best approximation of S to f is defined as
(1)ef,S=infu∈Sf−u.The definition of the classical best approximation in normed linear spaces is constructed by the norm. In this paper, we would study the best approximation theory which is not constructed by the norm, but only rely on the order structure. We consider our problem by recalling some definitions in Riesz space. For details of Riesz Space, refer to [28].Definition 2. See [28].
The relation “≤” in E is called a partial ordering if
(1)
a≤a for every a∈E(2)
∀a,b,c∈E, a≤b, and b≤c imply a≤c(3)
∀a,b∈E, a≤b, and b≤a imply a=b
The setE with a partial ordering is called a partially ordered set, denoted as E,≤. With out of generality, we denote E,≤ as E.Suppose thatE is a partially ordered set, andY is a nonempty subset of E. If x and y are points in E such that x≤y or y≤x, we say that x and y are comparable. The point x0∈E satisfies y≤x0 for all y∈Y, and then x0 is called an upper bound of Y. The subset Y is now said to be bounded above. If x0 is an upper bound of Y such that x0≤x′ for any other upper bound x′ of Y, then x0 is called a supremum of Y, denoted by supY. Similarly, the definitions of lower bound, bounded below, and infimum are analogous, denoted the infimum of Y as infY. If there are only two elements x and y in Y, then supY (resp., infY) can be denoted by x∨y (resp., x∧y). The partially ordered set E is called Dedekind complete, if every nonempty subset of E that is bounded above (bounded below) has a supremum (infimum). In particular, the partially ordered set E is said to be lattice if every subset consisting of two points has a supremum and an infimum.Definition 3. See [28].
LetE be a linear space over a field F (ℝ or ℂ) with a partially order “≤”. Then, E is called the Riesz space if for all f,g,h∈E and α∈ℝ+,
(2)f≤g⇒f+h≤g+h,θ≤f⇒θ≤αf.(1)
E is a latticewhere θ is the zero element in E. Unless otherwise specified, θ is used to represent the zero element in the Riesz space.
Assume thatE is a Riesz space, and f,g∈E, we denote the following notations as
(3)f+≔f∨θ,f−≔−f∨θ,f≔f∨−f,andE+≔f:θ≤f∈E.
Iff∧g=θ, then f is called disjoint with g and denoted as f⊥g. D is a nonempty subset of E, and then we can define the disjoint complement of D by
(4)Dd=f∈E:f⊥g,∀g∈D.Definition 4. See [28].
LetE be a Riesz space, a,b∈E, and a≤b. Then, the order interval is defined as
(5)a,b≔f∈Ea≤f≤b.Definition 5. See [28].
LetE be a Riesz space. The subset S is called order bounded if there exist an order interval a,b such that x∈a,b, for all x∈S.Definition 6.
LetE be a Riesz space. Then, E is called a linearly ordered space if for any x and y in E, and then x and y are comparable.Definition 7. See [28].
LetE be a Riesz space.
(1)
The linear subspaceV of E is called a Riesz subspace of E if for all members f and g of V, and the elements f∨g∈V and f∧g∈V hold(2)
The Riesz subspaceS of E is called an ideal in E if for any f∈S and g≤f, and then g∈S(3)
The idealB in E is called a band if, whenever a subset of B possesses a supremum in E, this supremum is a member of BDefinition 8. See [28].
LetE be a Riesz space, and fn be a sequence in E. Then,
(1)
fn is said to be increasing decreasing, if f1≤f2≤⋯f1≥f2≥⋯.This will denoted by fn↑fn↓. If fn↑ and supfn=f, we say fn↑f. Similarly, if fn↓ and inffn=f, we say fn↓f(2)
Thefn is said to converge in order to f if there exists a sequence pn and pn↓θ, such that f−fn≤pn for all n, we call f the order limit of fn. We shall denote this by fn⟶of or limn⟶∞fn=f(3)
A subsetV of E is said to be an order closed set if for any sequence fn in V and fn⟶of, and then f∈V
Some properties and conclusions of the order convergence have been introduced in the following lemmas, which are crucial to our subsequent research on the order best approximation.Lemma 9. See [28].
(1)
Iffn⟶of, fn⟶og, then f=g(2)
Iffn⟶of, gn⟶og, then for any α,β∈R, we have αfn+βgn⟶oαf+βg. Therefore, fn∨gn⟶of∨g, fn∧gn⟶of∧g, and fn⟶of(3)
Iffn⟶of and fn≥g, then f≥g(4)
Iffn⟶of, then for any subsequence fnk:k=1,2,⋯, we have fnk⟶of.


## 3. Order Best Approximation in Riesz Space

In this section, we consider the problem of the best approximation in Riesz space which is the main content and innovation of this paper. Based on the concept of the best approximation in normed linear spaces and the order structure characteristics in Riesz spaces, we propose the concept of the order best approximation in Riesz spaces and further obtain some related properties.Definition 10.
LetE be a Dedekind complete Riesz space, ∅≠S⊂E, f∈E.The quantity
(6)ef,S≔infu∈Sf−uis called the order best approximation of f by elements in S. If ef,S=f−u0 for some u0∈S, then u0 is said to be the order best approximation element of f by elements in S. Denoted by LSf, the order best approximation element set of f, which contains all the order best approximation elements of f in S.From the definition of the order best approximation set, the following proposition can be obtained.Proposition 11.
IfS is a nonempty subset of Riesz space E, x∈E, and z∈LSx, then z∈LSy for all y∈x∧z,x∨z.Proof.
We first provey−z=x−z−x−y, for any y∈x∧z,x∨z.
In fact, we have(7)x−z−x−yx∧z−z+x∨z−z−x∧z−y∧z+x∨z−y∨z=z−x∧z+x∨z−z−y∧z−x∧z+x∨z−y∨z=y∨z−y∧z=y−z.
Then for anyk∈S, by Definition 10, there is
(8)y−z=x−z−x−y≤x−k−x−y≤y−k.
So,y−z≤infk∈Sy−k. Because z∈LSx, then y−z=infk∈Sy−k, z∈LSy.
Obviously, according to the definition of the order best approximation setLSf, there are following three cases: LSf=∅, LSf≠∅, and LSf1; LSf=1 (where LSf is the cardinality of the set LSf). Based on above cases of the order best approximation set LSf, we introduce the following definitions.Definition 12.
LetS is a nonempty subset of Riesz space E.
(1)
S is said to be an order proximinal set if LSf≠∅ for every f∈E(2)
S is said to be an order uniqueness set if LSf≠∅ or LSf=1 for every f∈E(3)
S is said to be an order Cheybeshev set if LSf=1 for every f∈EIn [27], according to the classical definition of the best approximation set, a linear subspace must be an order proximinal set. However, in this paper, a linear subspace in Riesz space is not an order proximinal set, and the following example illustrates this result.Example 13.
LetX≠∅ and μ be the nonnegative countable addable measure. I is the finite or countable union of the set X with finite measure, and MI,μ is the real vector space formed by all the measurable functions defined on I,σ is a set with a measure zero on I. For all f,g∈MI,μ,f≈g, f and g are not equal only on the set of measure zero. Obviously, “≈” is an equivalence relationship. Denote by L0I,μ, the space formed by equivalence classes in MI,μ. Define the partial ordering “≤“on L0I,μ with f≤g, if fx≤gx for ∀x∈I/σ. Then, L0I,μ is Dedekind complete (ref. [28]).Now, we start to illustrate that the linear space ofL0I,μ is not an order proximinal set. For all f∈L0I,μ, we consider the linear space L01I,μ=spanf. Define the function g as follows:
(9)g=3f,x∈Ω,f,x∈I\Ω,where Ω is a set with a measure of nonzero on X. The order best approximation of g in linear space L01X,μ can be estimated as
(10)0≤eg,L01X,μ≤g−3f∧g−f=0.Here, 0 represents zero transformation. Thus, we have(11)eg,L01X,μ=0⇒g−kf=0⇒g=kf∈L01X,μ.This contradicts the construction ofg. Therefore, the linear space in L0X,μ is not necessarily an order proximinal set.In the following parts of this section, we discuss some main properties of the order proximinal set, the order uniqueness set, and the order Cheybeshev set. We will henceforth assume, unless otherwise stated, that the Riesz space is Dedekind complete.Theorem 14.
The order proximinal set is an order closed lattice.Proof.
LetE be a Riesz space and S is an order proximinal set in E. First, we prove S is an order closed set. Otherwise, according to the definition of order closed set, there exists fn⊂S such that fn⟶of and f∉S. Thus, there is a sequence pn and pn↓θ such that fn−f≤pn for all n. Then,
(12)infnfn−f=θandef,S≤infu∈Sf−u≤infnf−fn=θ.Hence,ef,S=θ. Since S is an order proximinal set in E, there exists an order best approximation element u0∈S such that ef,S=f−u0=θ, which implies f=u0∈S. In this way, contradictions are arise, and we can obtain that S is an order closed set.Now, we prove the order proximinal setS is also a lattice. Otherwise, we assume that S is not a lattice. By the properties of lattice, we can obtain that there are two elements x,y∈S, but their upper or lower bounds do not exist in S. Without losing generality, we might as well assume that x∨y∈E, but x∨y∉S. Since S is an order proximinal set, there exists u∗∈S which is the best approximation element of x∨y with respect to the set S. However, for x∨y∈E, we have
(13)x∨y−x∧x∨y−y=θ∨y−x∧x−y∨θ=x−y−∧x−y+=θ.Thus, we obtainex∨y,S=infu∈S∣x∨y−u∣=x∨y−u∗=θ, which implies that x∨y∈S. This creates a contradiction. Then the conclusion can be proved.Remark 15.
Theorem14 shows that the order proximinal set is an order closed lattice in Riesz space, but the converse is not all true. The following example will show that a set is an order closed lattice but not an order proximinal set.Example 16.
Let us consider the spaceℝ2 with the coordinate order, i.e. x=x1,x2, y=y1,y2, and x≤y if and only if x1≤y1,x2≤y2. It is obvious that ℝ2 is Dedekind complete. Assume that S=x=x1,x2x1=x2,1≤x1≤2. Obviously, S is an order closed lattice. However, we find that S is not an order proximinal set. In fact, for x0=1,2∈ℝ2, obviously, x0∉S, there exists x′=1,1,x′′=2,2∈S, such that
(14)x0−x′∧x0−x″=0,0.Therefore,Ex,S=0,0.So, ifS is an order proximinal set, then ∃x∗∈S, such that x0−x∗=0,0, then x0=x∗∈S which shows a contradiction. Therefore, it can be concluded that the set may not be an order proximinal set although it satisfies the properties of order closed lattice.What kind of order closed lattice must be an order proximinal set? We would discuss it from the following two aspects. On the one hand, the order interval must be an order proximinal set (Theorem18); on the other hand, we would consider it in a linearly ordered space (Theorem 20). In order to prove Theorem 18, we recall some lemmas.Lemma 17.
See [28] (minimal decomposition theorem).
LetE be a Riesz space. If f=u−v with u and v in E+, then there exists f+≤u and f−≤v, such that f=f+−f−.Theorem 18.
Any order closed interval in Riesz space is an order proximinal set.Proof.
LetE be a Dedekind complete Riesz space and S=a,b be an order closed interval of E. Then for each x∈E, the set x−g∀g∈S is equivalent to the order interval M=x−b,x−a. Thus, we obtain infg∈Sx−g=infm∈Mm. For any m∈M, we have
(15)x−b+≤m+≤x−a+,x−a−≤m−≤x−b−.
Takingu=x−b+−x−a−, it is truly that u∈x−b,x−a. From Lemma 17, there is a minimum positive decomposition
(16)u+≤x−b+,u−≤x−a−,such that u=u+−u− and u+∧u−=θ. Thus,
(17)u≤x−b++x−a−≤m++m−=m,which shows that u is a lower bound of the set m∀m∈M. Taking into account u∈x−b,x−a, we obtain u=infm∈Mm. So, for each x∈E, there exists an u∈x−b,x−a such that
(18)infg∈Sx−g=infm∈Mm=u.
And foru∈x−b,x−a, there exists a g0∈S such that u=x−g0. Then,
(19)infg∈S∣x−g∣=x−g0,which shows S is an order proximinal set by definition.In real number space, we have the Bolzano-Weierstrass theorem. In linearly ordered spaces, we find that there is a similar conclusion as follows.Lemma 19.
For every order bounded sequence in linearly ordered spaces, there exists an order convergent subsequence.Proof.
Assuming thatxk is an order bounded sequence in the strictly ordered space E, then there exists an order interval a,b such that xk⊆a,b. Taking c=1/2a+b∈a,b, we can find that all points of xk are included in the order interval a,c and c,b. Because any two elements in E can be compared, there is at least one interval of a,c and c,b which contains infinite points of xk and denoted as a1,b1. Obviously, b1−a1=1/2b−a holds. Continuing the above process, we can get an ordered interval sequence an,bn satisfying
(20)a,b⊃a1,b1⊃a2,b2⊃⋯⊃an,bn⊃⋯,bn−an=12nb−a,∀n.(1)
an,bn contains infinite number of points of xk
SinceE is Dedekind complete, we have an≤bn,n=1,2,⋯, which means an has a supremum denoted as ξ. Choosing a subsequence kn of k and xkn∈an,bn, then
(21)xkn−ξ≤bn−an=12nb−a.
With the Dedekind completeness ofE, we obtain 1/2nb−a⟶oθ. Therefore, xkn is order convergent to ξ.Theorem 20.
Every order closed set in a linearly ordered space is an order proximinal set.Proof.
LetE be a linearly ordered space and S be an order closed set in E. For all x∈E, there exists a sequence pn⊂E, and pn↓θ, such that ex,S≤ex,S+pn. Since ex,S=infu∈Sx−u, there exists a sequence un⊂S such that x−un≤ex,S+pn. In fact, for all u∈S and ex,S+pn≤x−u holds. Then, ex,S+pn≤infu∈Sx−u, which shows the contradiction. Hence, we have
(22)x−un≤ex,S+pn⇒un≤x+ex,S+pn≤x∣ex,S+p1,∀n∈ℕ.
Then,un is an order bounded sequence. According to Lemma 9 (4), there exists an order convergent subsequence unk of un and denotes the order limit by ξ. Since the sequence x−un converges in order to ex,S, we obtain that the subsequence x−unk also order converges to ex,S by Lemma 19. Thus, by Lemma 9 (44), we have
(23)unk⟶oξ⇒x−unk⟶ox−ξ⇒x−unk⟶ox−ξ.
Hence,ex,S≤∣x−ξ∣=limk⟶∞x−unk=ex,S holds. Therefore, we obtain ex,S=∣x−ξ∣, and it follows that ξ∈LSx.
With the arbitrariness ofx, we could easily conclude that S is an order proximinal set.
In this position, we discuss the properties of the order uniqueness sets.Lemma 21. See [28].
Iff⊥g,then it has the following properties:
(1)
f⊥g, for any α∈R, we have αf⊥g(2)
f⊥g, and then f+g=f−g=f+g=f−gTheorem 22.
A convex set in Riesz space is an order uniqueness set.Proof.
LetE be a Dedekind complete Riesz space, and S be a convex set in E. Assume that the conclusion is not true, then there exists a f∈E, such that f1, f2∈LSf, f1≠f2, i.e., ef,S=f−f1=f−f2. Since S is a convex set in E, then f1+f2/2∈S. From the definition of the order best approximation, we obtain
(24)f−f1≤f−f1+f22≤f−f12+f−f22=ef,S.
Then,f−f1+f2/2=ef,S. Therefore, we have
(25)2f−f1−f2∧f2−f+f−f1=f−f1−f−f2=θ.
So,2f−f1∧f2−f1=θ. Hence, by Lemma 21, there is
(26)f−f1⊥f2−f1⇒f−f2=f−f1−f2−f1=f−f1+f2−f1.
Therefore, we obtainf2−f1=θ⇒f1=f2, and the conclusion is proved.Remark 23.
Theorem22 also shows that for any point of Riesz space, if there exists an order best approximation element in some convex set, then the order best approximation element must be unique.
Considering that the order interval and the linear subspace of Riesz space are convex sets, we can obtain some corollaries.Corollary 24.
Every order interval of Riesz space is an order uniqueness set.Corollary 25.
Every linear subspace of Riesz space is an order uniqueness set.Theorem 26.
IfS is an order uniqueness set and the elements of S are disjoint with each other, then S must be a single point set.Proof.
LetE be a Riesz space and S be an order uniqueness set with the elements being disjoint with each other in E. Assume that S is not a single point set, then there exists a,b∈S and a≠b, such that a⊥b, and at least one of a,b is not equal to θ. Let us set a≠θ, and we have a+b/2∉S (in fact, if a+b/2∈S, then by Lemma 21, there is a+b/2∧a=a/2+b/2∧a=a/2≠θ,, i.e., a+b/2 and a are not disjoint). Thus, we could discuss it from the following two cases.
(1)
IfS=a,b, then LSa+b/2=a,b is not unique(2)
If there existsc∈S such that c⊥a,c⊥b, then(27)c−a+b2=c+a2+b2≥a2+b2.
Meanwhile, sincea−a+b/2=b−a+b/2=a+b/2, we have
(28)ea+b2,S=a+b2.
Therefore,S cannot be an order uniqueness set, which contradicts the suppose of the theorem.Lemma 27. See [28].
LetE be a Dedekind complete Riesz space. Then, E=B⊕Bd holds for every band B in E, i.e., for any u∈E, there exists a unique decomposition.
u=u1+u2, where u1∈B,u2∈Bd.Theorem 28.
The band in Riesz space is order Cheybeshev set.Proof.
LetE be a Riesz space and B be a band in E. for any x∈E, considering the Lemma 27, we have x=x1+x2, where x1∈B and x2∈Bd. Then,
(29)infg∈Bx−g=infg∈Bx1+x2−g=infg∈Bx1−g+x2holds. Since x1−g∈B, we have x1−g∧x2=θ. So,
(30)infg∈Bx1−g+x2=infg∈Bx1−g+x2≥x2.
Therefore, there existsg0=x1∈B, such that
(31)ex,B=infg∈Bx−g=x−x1=x2.
Hence,x1∈LBx.
Because of the uniqueness of the order direct sum decomposition ofE, the conclusion is proved.
From Theorem18 and Corollary 24, we can further obtain the following corollary.Corollary 29.
Every order closed interval in Riesz space is an order Chebyhsev set.Theorem 30.
The order Cheybeshev set in linearly ordered space is a convex set.Proof.
LetE be the linearly ordered space, and S be an order Chebyshev set in E. According to Theorem 14, the order proximinal set must be an order closed lattice. Therefore, we can infer that S is a convex set. In fact, if S is a nonconvex set, then there exists x,y∈S, and α0∈0,1, such that α0x+1−α0y∉S. Thus, there exists m∈S by the definition of the order Chebyshev set, such that
(32)eα0x+1−α0y,S=m−α0x+1−α0y≔q.
E is a linearly ordered space; so, m and α0x+1−α0y are comparable. Without loss of generality, we assume that m≤α0x+1−α0y. Taking m1=α0x+1−α0y+q, by the definition of order Chebyshev set, then m≠m1 (if m≥α0x+1−α0y, take m1=α0x+1−α0y−q.). Therefore, the order best approximation element of m0 in S exists (notation m1), and then we have m,m1\m,m1∩S=∅. Taking m2=m+m1/2, we get
(33)em2,S=m−m2=m1−m2.
There is a contradiction with the definition of the order Chebyshev set; so, the conclusion is proved.


## 4. Order Best Approximation Projection

In this section, we would further study the properties of the order best approximation from the perspective of operator by constructing a projection operator from the elements to the order proximinal set and study the related properties such as the local order boundedness and order continuity.Definition 31.
LetE be a Riesz space and S be an order proximinal set of E. Define the mapping LS:x⟶LSx as
(34)LSx≔y∈S:x−y=ex,S,and then LS is called the order best approximation projection.Definition 32.
LetX and Y be two Riesz spaces and f be a mapping from X to Y (the mapping here could be multivalued.). Then, f is said to be locally order bounded, and if for all x0∈X, there exists p0∈X+and M∈Y+, such that y=fx∈−M,M for all x∈x0−p0,x0+p0, i.e., fx0−p0,x0+p0 is order bounded in Y.Definition 33.
LetX and Y be two Riesz spaces and f be the mapping from X to Y. Given x∈X, then f is said to be order continuous at x, if f is single-valued, and for all xn of X, xn⟶ox, then fxn⟶ofx. If f is order continuous at every point of X, then f is said to be order continuous at X.The main purpose of this section is to discuss the properties of the order best approximation projection from two aspects: local order boundedness and order continuity. In order to discuss local order boundedness of the order best approximation projection, we introduce the following lemma.Lemma 34.
LetE be a Riesz space and S be a nonempty subset of E; then for any x,y∈E, we have
(35)ex,S−ey,S≤x−y.Proof.
For anyx,y∈E,z∈S, we have
(36)ex,S≤x−z≤x−y+y−z⇒ex,S−x−y≤y−z.Hence, from the definition ofey,S,
(37)ex,S−ey,S≤x−y.Similarly,ey,S−ex,S≤x−y. Therefore, ex,S−ey,S≤x−y.Theorem 35.
LetE be a Riesz space and S be an order proximinal set in E, and then the order best approximation projection LS:x⟶LSx is locally order bounded.Proof.
Takingx0∈E,p0∈E+. For all x in x0−p0,x0+p0, y∈LSx, we have y+y−x+x−x0+x0=ex,S+x−x0+x0. From Lemma 34, we obtain y≤ex0,S+2x−x0+x0<ex0,S+2p0+x0, which shows that the upper bound does not depend on x or y. Thus, LSx0−p0,x0+p0 is order bounded.Remark 36.
As previously studied, ifS is an order proximinal set in E, then the order best approximation element set LSx can be multi-point set. Therefore, we would discuss the continuity of order best approximation projection defined on the set that is the order Chebyshev set.Proposition 37.
LetE be a Riesz space, S be an order proximinal set in E and xn,yn be two sequences of E. If
yn∈LSxn for ∀n∈N, xn⟶ox, and yn⟶oy, then y∈LSx.Proof.
From Theorem14, the order proximinal set S is order closed in E. So, for the sequence yn⊂S,yn⟶oy⇒y∈S, according to Lemma 34, we have
(38)xn−yn−ex,S=exn,S−ex,S≤xn−x.Taking the limits of both sides of above inequality, we obtain(39)x−y−ex,S≤θ.Thus,x−y=ex,S⇒y∈LSx.Theorem 38.
LetE be a Riesz space and S be a convex order Chebyshev set in E, and then the order best approximation projection LS:x⟶LSx is order continuous.Proof.
For∀x∈E, there exists a sequence xn of E converges in order to x, i.e., there exists pn and pn↓θ such that xn−x≤θ. By the definition of order Chebyshev set, there is a unique yn∈LSxn for all n, and y∈LSx. Therefore,
(40)xn−yn−x−y≤xn−y−x−y≤xn−x≤pn,x−y−xn−yn≤x−yn−xn−yn≤x−xn≤pn.So,(41)limn⟶∞yn−xn=limn⟶∞y−xn=y−x.By the definition of the order best approximationyn−xn≤y−xn,
(42)yn−xn∨y−xn−yn−xn≤yn−xn−yn−xn∨y−xn−yn−xn≤y−xn−yn−xnholds. And we know that S is convex; so, yn−xn≤yn+y/2−xn. Hence,
(43)limn⟶∞yn−xn≤limn⟶∞yn+y2−xn≤limn⟶∞yn−xn2+y−xn2=y−x.Therefore,(44)limn⟶∞yn+y2−xn−yn−xn=θ.This means that there isqn and qn↓θ such that
(45)yn+y2−xn−yn−xn≤qn.So,(46)yn−y=yn−xn+xn−y=yn−xn−y−xn=2yn−xn∨y−xn−yn+y−2xn=2yn−xn∨y−xn−yn−xn+2yn−xn−yn+y−2xn≤2y−xn−yn−xn+yn−xn−yn+y2−xn.By the equations (41), (44), and definition of order convergence, there is yn−y≤2pn+qn. From the Lemma 9, we can conclude that yn order converges to y. So, LS:x⟶LSx is continuous.



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