Definition 1. See [27]. Let E 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

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 set E with a partial ordering is called a partially ordered set, denoted as (E, ≤). With out of generality, we denote (E, ≤) as E.

Definition 3. See [28]. Let E 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⁺,

• (1)

  E is a lattice

Assume that E is a Riesz space, and f, g ∈ E, we denote the following notations as

If |f|∧|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

Definition 4. See [28]. Let E be a Riesz space, a, b ∈ E, and a ≤ b. Then, the order interval is defined as

Definition 5. See [28]. Let E 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. Let E 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]. Let E be a Riesz space.

• (1)

  The linear subspace V 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 subspace S of E is called an ideal in E if for any f ∈ S and |g| ≤ |f|, and then g ∈ S

• (3)

  The ideal B in E is called a band if, whenever a subset of B possesses a supremum in E, this supremum is a member of B

Definition 8. See [28]. Let E be a Riesz space, and {f_n} be a sequence in E. Then,

• (1)

  {f_n} is said to be increasing (decreasing), if f₁ ≤ f₂ ≤ ⋯(f₁ ≥ f₂≥⋯).This will denoted by f_n↑(f_n↓). If f_n↑ and supf_n = f, we say f_n↑f. Similarly, if f_n↓ and inff_n = f, we say f_n ↓ f

• (2)

  The {f_n} is said to converge in order to f if there exists a sequence {p_n} and p_n ↓ θ, such that |f − f_n| ≤ p_n for all n, we call f the order limit of f_n. We shall denote this by $$ or lim_n⟶∞f_n = f

• (3)

  A subset V of E is said to be an order closed set if for any sequence {f_n} in V and $$, 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)

  If $$, $$, then f = g

• (2)

  If $$, $$, then for any α, β ∈ R, we have $$. Therefore, $$, $$, and $$

• (3)

  If $$ and f_n ≥ g, then f ≥ g

• (4)

  If $$, then for any subsequence $$, we have $$

Definition 10. Let E be a Dedekind complete Riesz space, ∅≠S ⊂ E, f ∈ E.The quantity

Proposition 11. If S is a nonempty subset of Riesz space E, x ∈ E, and $$, then $$ for all y ∈ [x∧z, x∨z].

Proof. We first prove |y − z| = |x − z| − |x − y|, for any y ∈ [x∧z, x∨z].

In fact, we have

Then for any k ∈ S, by Definition 10, there is

So, |y − z| ≤ inf_k∈S|y − k|. Because $$, then |y − z| = inf_k∈S|y − k|, $$.

Obviously, according to the definition of the order best approximation set $$, there are following three cases: $$, $$, and $$; $$ (where $$ is the cardinality of the set $$). Based on above cases of the order best approximation set $$, we introduce the following definitions.

Definition 12. Let S is a nonempty subset of Riesz space E.

• (1)

  S is said to be an order proximinal set if $$ for every f ∈ E

• (2)

  S is said to be an order uniqueness set if $$ or $$ for every f ∈ E

• (3)

  S is said to be an order Cheybeshev set if $$ for every f ∈ E

Example 13. Let X ≠ ∅ and μ be the nonnegative countable addable measure. $$ is the finite or countable union of the set X with finite measure, and $$ is the real vector space formed by all the measurable functions defined on $$ is a set with a measure zero on $$. For all $$, f and g are not equal only on the set of measure zero. Obviously, “≈” is an equivalence relationship. Denote by $$, the space formed by equivalence classes in $$. Define the partial ordering “≤“on $$ with [f] ≤ [g], if f(x) ≤ g(x) for $$. Then, $$ is Dedekind complete (ref. [28]).

Theorem 14. The order proximinal set is an order closed lattice.

Proof. Let E 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 {f_n} ⊂ S such that $$ and f ∉ S. Thus, there is a sequence {p_n} and p_n ↓ θ such that |f_n − f| ≤ p_n for all n. Then,

Remark 15. Theorem 14 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 space ℝ² 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, x₂)|x₁ = 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

Lemma 17. See [28] (minimal decomposition theorem).

Let E 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. Let E 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 inf_g∈S|x − g| = inf_m∈M|m|. For any m ∈ M, we have

Taking u = (x − b)⁺ − (x − a)⁻, it is truly that u ∈ [x − b, x − a]. From Lemma 17, there is a minimum positive decomposition

And for u ∈ [x − b, x − a], there exists a g₀ ∈ S such that u = x − g₀. Then,

Lemma 19. For every order bounded sequence in linearly ordered spaces, there exists an order convergent subsequence.

Proof. Assuming that {x_k} is an order bounded sequence in the strictly ordered space E, then there exists an order interval [a, b] such that {x_k}⊆[a, b]. Taking c = 1/2(a + b) ∈ [a, b], we can find that all points of {x_k} 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 {x_k} and denoted as [a₁, b₁]. Obviously, b₁ − a₁ = 1/2(b − a) holds. Continuing the above process, we can get an ordered interval sequence {[a_n, b_n]} satisfying

• (1)

  [a_n, b_n] contains infinite number of points of {x_k}

Since E is Dedekind complete, we have a_n ≤ b_n, n = 1, 2, ⋯, which means {a_n} has a supremum denoted as ξ. Choosing a subsequence {k_n} of {k} and $$, then

With the Dedekind completeness of E, we obtain $$. Therefore, $$ is order convergent to ξ.

Theorem 20. Every order closed set in a linearly ordered space is an order proximinal set.

Proof. Let E be a linearly ordered space and S be an order closed set in E. For all x ∈ E, there exists a sequence {p_n} ⊂ E, and p_n ↓ θ, such that e(x, S) ≤ e(x, S) + p_n. Since e(x, S) = inf_u∈S|x − u|, there exists a sequence {u_n} ⊂ S such that |x − u_n| ≤ e(x, S) + p_n. In fact, for all u ∈ S and e(x, S) + p_n ≤ |x − u| holds. Then, e(x, S) + p_n ≤ inf_u∈S|x − u|, which shows the contradiction. Hence, we have

Then, {u_n} is an order bounded sequence. According to Lemma 9 (4), there exists an order convergent subsequence $$ of {u_n} and denotes the order limit by ξ. Since the sequence {|x − u_n|} converges in order to e(x, S), we obtain that the subsequence $$ also order converges to e(x, S) by Lemma 19. Thus, by Lemma 9 (44), we have

Hence, $$ holds. Therefore, we obtain e(x, S) = ∣x − ξ∣, and it follows that $$.

With the arbitrariness of x, 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]. If f⊥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| − |g||

Theorem 22. A convex set in Riesz space is an order uniqueness set.

Proof. Let E 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, $$, f₁ ≠ f₂, i.e., e(f, 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

Then, |f − f₁ + f₂/2| = e(f, S). Therefore, we have

So, 2|f − f₁|∧|f₂ − f₁| = θ. Hence, by Lemma 21, there is

Therefore, we obtain |f₂ − f₁| = θ⇒f₁ = f₂, and the conclusion is proved.

Remark 23. Theorem 22 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. If S is an order uniqueness set and the elements of S are disjoint with each other, then S must be a single point set.

Proof. Let E 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)

  If S = {a, b}, then $$ is not unique

• (2)

  If there exists c ∈ S such that c⊥a, c⊥b, then

Meanwhile, since |a − a + b/2| = |b − a + b/2| = |a| + |b|/2, we have

Therefore, S cannot be an order uniqueness set, which contradicts the suppose of the theorem.

Lemma 27. See [28]. Let E be a Dedekind complete Riesz space. Then, E = B ⊕ B^d 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 ∈ B^d.

Theorem 28. The band in Riesz space is order Cheybeshev set.

Proof. Let E 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 ∈ B^d. Then,

Therefore, there exists g₀ = x₁ ∈ B, such that

Hence, $$.

Because of the uniqueness of the order direct sum decomposition of E, the conclusion is proved.

From Theorem 18 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. Let E 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 − α₀)y ∉ S. Thus, there exists m ∈ S by the definition of the order Chebyshev set, such that

E is a linearly ordered space; so, m and α₀x + (1 − α₀)y are comparable. Without loss of generality, we assume that m ≤ α₀x + (1 − α₀)y. Taking m₁ = α₀x + (1 − α₀)y + q, by the definition of order Chebyshev set, then m ≠ m₁ (if m ≥ α₀x + (1 − α₀)y, take m₁ = α₀x + (1 − α₀)y − 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

There is a contradiction with the definition of the order Chebyshev set; so, the conclusion is proved.

Definition 31. Let E be a Riesz space and S be an order proximinal set of E. Define the mapping $$ as

Definition 32. Let X 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 = f(x) ∈ [−M, M] for all x ∈ [x₀ − p₀, x₀ + p₀], i.e., f([x₀ − p₀, x₀ + p₀]) is order bounded in Y.

Definition 33. Let X 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 {x_n} of X, $$, then $$. If f is order continuous at every point of X, then f is said to be order continuous at X.

Lemma 34. Let E be a Riesz space and S be a nonempty subset of E; then for any x, y ∈ E, we have

Proof. For any x, y ∈ E, z ∈ S, we have

Theorem 35. Let E be a Riesz space and S be an order proximinal set in E, and then the order best approximation projection $$ is locally order bounded.

Proof. Taking x₀ ∈ E, p₀ ∈ E⁺. For all x in [x₀ − p₀, x₀ + p₀], $$, we have |y| + |y − x| + |x − x₀| + |x₀| = e(x, S) + |x − x₀| + |x₀|. From Lemma 34, we obtain |y| ≤ e(x₀, S) + 2|x − x₀| + |x₀| < e(x₀, S) + 2p₀ + |x₀|, which shows that the upper bound does not depend on x or y. Thus, $$ is order bounded.

Remark 36. As previously studied, if S is an order proximinal set in E, then the order best approximation element set $$ 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. Let E be a Riesz space, S be an order proximinal set in E and {x_n}, {y_n} be two sequences of E. If

$$ for ∀n ∈ N, $$, and $$, then $$

Proof. From Theorem 14, the order proximinal set S is order closed in E. So, for the sequence $$, according to Lemma 34, we have

Theorem 38. Let E be a Riesz space and S be a convex order Chebyshev set in E, and then the order best approximation projection $$ is order continuous.

Proof. For ∀x ∈ E, there exists a sequence {x_n} of E converges in order to x, i.e., there exists {p_n} and p_n ↓ θ such that |x_n − x| ≤ θ. By the definition of order Chebyshev set, there is a unique $$ for all n, and $$. Therefore,