REGULARITY OF ORDER BOUNDED OPERATORS ON A SEPARABLE BANACH LATTICE WITH AN EXTENDED

REGULARITY OF ORDER BOUNDED OPERATORS ON A SEPARABLE BANACH LATTICE WITH AN EXTENDED

a-NORM

Cuijie Zhang ¹ and Hongyun Xiong ²

(Received November 2006)

Abstract. Let H be a separable Banach lattice equipped with an extended a- norm and Y be an arbitrary Banach lattice with the Cantor property. In this paper, we shall show that every order bounded linear operator T : H → Y is regular, i.e., there exists a positive operator S : H → Y with S ≥ T , −T .

1. Introduction

Let E be an Archimedean Riesz space, and let a ∈ E be a point. For any x ∈ E, define

kxk a = inf{λ ∈ R ⁺ : |x| ≤ λ · |a|},

with inf ∅ = +∞. Then, k · k a is called an extended a-norm on E and E is called an extended a-normed Riesz space, see [2]. Let E be an extended a-normed Riesz space. The sequence {x n : n ∈ N} in E is called an extended a-normed Cauchy sequence, if for every ε > 0 there exists N (ε) ∈ N such that kx m − x n k a < ε for all m, n ≥ N (ε). If every extended a-normed Cauchy sequence is convergent in E, then E is called an extended a-normed Banach lattice.

For two Riesz spaces E and F , if F is Dedekind complete, then it has been known that each order bounded linear operator T : E → F is regular (i.e., T is the difference of two positive linear operators from E into F , or equivalently there exists a positive linear operator S : E → F with S ≥ T ). For the case of E = C(X) and F = C(Y ), where X and Y are compact Hausdorff spaces, some results on the regularity of order bounded linear operators from E into F need to be mentioned here. If there exists a nontrivial convergent sequence in Y , several necessary and sufficient conditions under which each order bounded linear operator from C(X) into C(Y ) is regular were obtained in Theorem 2.12 and Theorem 2.14 of [6]. Furthermore, Wickstead [4], [5] also obtained some results on the regularity of order bounded operators into C(Y ). The authors in [1] proved that every order bounded linear operator T : C(S) → Y is regular, where S is any compact metric space and Y is an arbitrary Banach lattice with the Cantor property. Recall that an Archimedean Riesz space E is said to have the Cantor

1991 Mathematics Subject Classification 46B42, 47B60.

Key words and phrases: Banach lattice; order bounded operator; regular operator; the Cantor property.

1

The first-named author’s research was supported by the scientific research fund in the Civil Aviation University of China, 04-CAUC-24S..

2

The second-named author’s research research was supported by Liu Hui Research Centre of

Applied Mathematics, Nakai University and Tianjin University.

property (or the countable interpolation property, or the σ-interpolation property) if {a n } and {b n } are arbitrary sequences in E such that a n ≤ b m for all m, n ∈ N, then there exists an element c ∈ E satisfying a n ≤ c ≤ b n for all n ∈ N. Since C(S) is separable if and only if S is metrizable, the authors in [1] proposed the following open problem: Does the above conclusion still hold if we replace the domain space C(S) by an arbitrary separable Banach lattice? In Section 3, we shall present a partial answer to this open problem by claiming that the conclusion still holds if we replace the domain space C(S) by a separable extended a-normed Banach lattice.

For undefined terminology and basic facts used in this paper, refer to [3], [7] and [8].

2. Operators on Extended a-Normed Riesz Spaces

In this section, E will always be an extended a-normed Riesz space. If there is no special explanation, generally we suppose a 6= 0.

Definition 2.1. Let E be an extended a-normed Riesz space and F be a normed Riesz space, T : E → F be a linear operator. If there exists a real number C > 0 such that kT xk ≤ Ckxk a for all x ∈ E, then the linear operator T is said to be bounded in the extended a-norm.

Definition 2.2. Let E be an extended a-normed Riesz space and F be a normed Riesz space, T : E → F be a bounded linear operator in the extended a-norm. The extended a-norm kT k a of T is defined by

kT k a = sup

x∈E, x6=0

kT xk kxk a .

Obviously for any bounded linear operators T and S in the extended a-norm, the following statements hold.

(1) kT k a = 0 ⇐⇒ T | I

(a) = 0, where I d (a) is the principal ideal generated by a in E;

(2) kαT k a = |α| · kT k a for any α ∈ R;

(3) kS + T k a ≤ kSk a + kT k a ;

(4) kT k a = sup _x∈E,kxk

₌₁ kT xk = sup _x∈E,kxk

_≤1 kT xk.

Also, we have the following lemma without difficulty.

Lemma 2.3. Let E be an extended a-normed Riesz space and F be a normed Riesz space, T : E → F be a linear operator. Then T is continuous if and only if it is bounded in the extended a-norm.

3. The Main Result

In this section, before showing the main theorem of this paper, we shall give some lemmas which are similar to those in [8].

Let {x n } be a sequence in an extended a-normed Riesz space E. In what follows,

we shall use x n → x (extended a-norm) to denote that {x n } is convergent to x in

the extended a-norm. If {x _n } is monotonously increasing and x is the least upper

bound of the sequence {x }, then we write x ↑ x.

Lemma 3.1. Let {x n } be a sequence in an extended a-normed Riesz space E. If x n → x (extended a-norm) and x n ≥ y holds for all n ∈ N, then x ≥ y. If x n → x (extended a-norm) and x n ≤ y holds for all n ∈ N, then x ≤ y.

Proof. Assume y = 0. We have |x ⁻ − x ⁻ _n | ≤ |x − x n | by Birkhoff inequalities, then kx ⁻ − x ⁻ _n k a ≤ kx − x n k a . So, x ⁻ _n → x ⁻ (extended a-norm). Since x n ≥ 0 for all n ∈ N (i.e., x ⁻ _n = 0), we get x ⁻ = 0 (i.e., x ≥ 0). Now we assume that y 6= 0. Since x n − y → x − y (extended a-norm) and x n − y ≥ 0 holds for all n ∈ N, it follows that x − y ≥ 0, i.e., x ≥ y. Similarly, one can prove the case of “x n ≤ y”. ¤ Lemma 3.2. If {x n } is monotonously increasing in the extended a-normed Riesz space E and x n → x (extended a-norm), then x n ↑ x.

Proof. For any fixed m ∈ N, since x n ≥ x m for all n with n ≥ m, we have x ≥ x m by Lemma 3.1. So, x is an upper bound of the sequence {x n }. Let y be an arbitrary upper bound of the sequence {x n }. Then, by Lemma 3.1 again, x n ≤ y for all n ∈ N. So, x ≤ y. It follows that x is the least upper bound of the sequence

{x n }, i.e., x n ↑ x. ¤

Recall that a normed Riesz space (E, k · k) is said to have the Riesz-Fischer property, if for any sequence {u n } ⊆ E ⁺ such that Σ ^∞ _n=1 ku n k converges, then the partial sum sequence {s n } (where s n = Σ ⁿ _k=1 u k for all n ∈ N) is also norm convergent in (E, k · k).

Lemma 3.3. Suppose that (E, k · k a ) is an extended a-normed Banach lattice.

Then (E, k · k a ) has the Riesz-Fischer property.

Proof. Let u n ∈ E ⁺ for n ∈ N and Σ ^∞ _n=1 ku n k a converges, i.e., Σ ^∞ _n=1 ku n k a < ∞.

Then for the partial sum sequence {s n }, we have ks m − s n k a = k

X m k=n+1

u k k a

≤ X m k=n+1

ku k k a

≤ X ∞ k=n+1

ku k k a → 0 (m > n → ∞).

So, the sequence {s _n } is a Cauchy sequence in the extended a-normed Banach lattice

E. Therefore, there exists an s ∈ E such that s n → s (extended a-norm). Since

{s n } is increasing as n increases, it follows that s n ↑ s by Lemma 3.2, and hence

Σ ^∞ _n=1 u n is equal to s. This shows that (E, k·k a ) has the Riesz-Fischer property. ¤

Lemma 3.4. Let E be an extended a-normed Banach lattice and F be a normed

Riesz space. If T : E → F is an order bounded linear operator, then T is bounded

in the extended a-norm (and hence it is continuous). It follows that every regu-

lar operator from E into F is bounded in the extended a-norm (and hence it is

continuous).

Proof. First, it is obvious that any order interval [f 1 , f 2 ] in F is norm bounded.

Assume now that T : E → F is order bounded but not bounded in the extended a-norm. Then there exists a sequence {x n } ⊆ E such that kT x n k > 2n ³ kx n k a . Note that x n 6= 0 and kx n k a < +∞, then kT _kx ^x

n

k

k > 2n ³ . Since

° °

° °T x ⁺ _n kx n k a

° °

° ° +

° °

° °T x ⁻ _n kx n k a

° °

° ° ≥

° °

° °T x n

kx n k a

° °

° ° > 2n ³ ,

it follows that at least one of kT _kx ^x

n

k

k and kT _kx ^x

n

k

k exceeds n ³ . Hence, we may assume that x n > 0, kx n k a ≤ 1 and kT x n k > n ³ for all n ∈ N. Put u n = n ⁻² · x n . Then, we get u _n > 0, ku _n k _a ≤ n ⁻² and kT u _n k > n for n ∈ N. Thus, Σ ^∞ _n=1 ku _n k _a converges. Since E is an extended a-normed Banach lattice, by Lemma 3.3, E has the Riesz-Fischer property. So, the element s = Σ ^∞ _n=1 u n exists in E. Since u n ∈ [0, s] and T is an order bounded, then all terms of the sequence {T u n } are contained in some order interval in F . Hence {T u n } is norm bounded, which contradicts with the fact that kT u n k > n for all n ∈ N. Hence, the operator T is

bounded in the extended a-norm. ¤

By using the technique similar to that in Theorem 3.1 of [1], now we are ready to present the main result of this paper.

Theorem 3.5. Let H be an extended a-normed separable Banach lattice and Y be a Banach lattice which has the Cantor property. Then every order bounded linear operator T : H → Y is regular, i.e., there exists a positive linear operator S : H → Y with S ≥ T, −T .

Proof. By Lemma 3.4, the order bounded linear operator T is continuous. Since T is order bounded, there exists y 0 ∈ Y such that y 0 > 0 and |T x| ≤ y 0 for all x ∈ H with |x| ≤ |a|. Since Y ^∗∗ is also Dedekind complete, there exists the modulus

|T | : H → Y ^∗∗ such that |T |x = sup{T x ⁰ : |x ⁰ | ≤ x} for any x ∈ H ⁺ . It follows that

|T |(|a|) = sup{T x ⁰ : |x ⁰ | ≤ |a|} ≤ y 0 .

Let Ω be the set of all pairs hM, Si, where M is a linear subspace of H containing

|a| and S : M → Y is a linear operator with S ≥ |T | | M and S(|a|) = y 0 . We partially order the set Ω by defining

hM 1 , S 1 i ¹ hM 2 , S 2 i if and only if M 1 ⊆ M 2 and S 2 | M

= S 1 . Let M = {α · |a| : α ∈ R} and define S : M → Y by letting S(α · |a|) = α · y 0 for all α ∈ R. So, if α ≥ 0, then

S(α|a|) = αy 0 ≥ α|T |(|a|) = |T |(α|a|).

Hence S ≥ |T | | M . It follows that Ω 6= ∅. Suppose that Ω 0 is an arbitrary chain in Ω. Let M 0 = S

hM,Si∈Ω

M and S 0 : M 0 → Y be defined such that for every

x ∈ M ₀ , S ₀ x = Sx if x ∈ M , where hM, Si ∈ Ω ₀ . Hence S ₀ ≥ |T | | _M

and

S (|a|) = y . This implies that hM , S i ∈ Ω and hM , S i is an upper bound of

Ω 0 . By Zorn’s Lemma, Ω has a maximal element hE, U i. It follows from kU k a = sup

kxk

≤1

kU xk = sup

|x|≤|a|

kU xk

≤ sup

|x|≤|a|

kU |x|k ≤ kU |a|k = ky 0 k

that the operator U is bounded in the extended a-norm. Hence, U is continuous on E by Lemma 2.3.

Next, we shall show E = H, and U actually satisfies the requirement of the theorem. Suppose E 6= H. Then there exists a positive element g ∈ H r E. Let G = {x + αg : x ∈ E, α ∈ R} be the linear span of E and g. Consider the sets B = {b ∈ E : b ≥ g} and C = {c ∈ E : c ≥ −g}. It is obvious that C 6= ∅. In the sequel, we shall consider two distinct cases.

Case (i). B 6= ∅. Since H is separable, both B and C are separable. Let B ⁰ and C ⁰ be countable dense subsets of B and C respectively. For b ∈ B ⁰ , let P b be a countable dense subset of the set {u ∈ H : |u| ≤ b − g}. Similarly, for c ∈ C ⁰ , let Q c be a countable dense subset of the set {v ∈ H : |v| ≤ c + g}. If b ∈ B ⁰ , c ∈ C ⁰ , u ∈ P b and v ∈ Q c , then b − g ≥ 0 and c + g ≥ 0. So, b + c ≥ 0. Hence

U b + U c = U (b + c) ≥ |T |(b + c)

= |T |(b − g + c + g)

= |T |(b − g) + |T |(c + g)

≥ |T |(|u|) + |T |(|v|)

≥ T u + T v.

It follows that U b − T u ≥ T v − U c. Since there are only countably many choices of b ∈ B ⁰ , c ∈ C ⁰ , u ∈ P b and v ∈ Q c , we may use the Cantor property in Y to find h ∈ Y such that

U b − T u ≥ h ≥ T v − U c

for all b ∈ B ⁰ , c ∈ C ⁰ , u ∈ P b and v ∈ Q c . Now we define a linear operator V : G → Y by V (x + αg) = U x + αh for all x ∈ E and all α ∈ R. If x ∈ C ⁰ , then for each v ∈ Q x , we have

V (x + g) = U x + h ≥ U x + T v − U x = T v.

By continuity of T , we have T v ≤ V (x + g) for all v with |v| ≤ x + g. Hence, we have V (x + g) ≥ |T |(x + g) for all x ∈ C ⁰ . Again using the density of C ⁰ in C together with continuity of U and |T |, we actually have V (x + g) ≥ |T |(x + g) for all x with x + g ≥ 0. Similarly, if x − g ≥ 0, then we also have V (x − g) ≥ |T |(x − g).

Using the linearity of V and |T |, it is clear that if x + αg ≥ 0 then V (x + αg) ≥ |T |(x + αg)

whenever α > 0, α < 0 or α = 0 (in which case it is clear). Therefore, V ≥ |T || G .

Since V (|a|) = U (|a|) = y 0 , we have hG, V i ∈ Ω. But hG, V i 6= hE, U i and

hE, U i ¹ hG, V i.

Case (ii). B = ∅. Let C ⁰ be a countable dense subset of C. For c ∈ C ⁰ , let Q c be a countable dense subset of the set {v ∈ H : |v| ≤ c + g}. If c ∈ C ⁰ and v ∈ Q c , then T v + |U c| ≥ T v − U c. Since there are only countably many choices of c ∈ C ⁰ and v ∈ Q c , we may use the Cantor property in Y to find h ⁰ ∈ Y such that

T v + |U c| ≥ h ⁰ ≥ T v − U c.

Now, we can define a linear operator V ⁰ : G → Y by letting V ⁰ (x + αg) = U x + αh ⁰ for all x ∈ E and all α ∈ R. Similar to Case (i), it can be verified that if x + αg ≥ 0 then

V ⁰ (x + αg) ≥ |T |(x + αg)

for any α > 0. Since B = ∅ and E is a linear subspace of H, then {x ∈ E : x ≥ αg, α > 0} = {x ∈ E : x + αg ≥ 0, α < 0} = ∅.

Hence, if x + αg ≥ 0, then V ⁰ (x + αg) ≥ |T |(x + αg), whenever α > 0, α < 0 or α = 0. It follows that V ⁰ ≥ |T | on G. Since V ⁰ (|a|) = U (|a|) = y 0 , we have hG, V ⁰ i ∈ Ω. It is clear that hG, V ⁰ i 6= hE, U i and hE, U i ¹ hG, V ⁰ i.

Therefore, we have shown that in either case there always exists hG, V i ∈ Ω such that hG, V i 6= hE, U i and hE, U i ¹ hG, V i. This leads to a contradiction, which implies that E = H and thus U actually satisfies the requirement of the

theorem. ¤

Recall that a compact Hausdorff space S is said to be an F -space if any two disjoint open F σ -sets have disjoint closures.

Corollary 3.6. Let H be an extended a-normed separable Banach lattice and S be a compact Hausdorff space. If S is an F -space, then every continuous linear operator T : H → C(S) with U ≥ T, −T and kU k a = kT k a .

Proof. Since T : H → C(S) is continuous, then for all x ∈ H with |x| ≤ |a|

(i.e., kxk a ≤ 1), we have

kT xk ≤ kT k a · kxk a ≤ kT k a .

This implies that |T x| ≤ kT k a · 1 S . Now letting y 0 = kT k a · 1 S in Theorem 3.5, we can show that T is regular and there exists a positive continuous linear op- erator U : H → C(S) with U ≥ |T | and kU k a ≤ ky 0 k = kT k a . It follows that kU k a ≥ k|T |k a . Since kT k a ≤ k|T |k a , we have kU k a = kT k a . ¤ If a is a strong unit of H, then the extended a-norm on H is a norm. It implies that the extended a-normed Banach lattice H is a Banach lattice. Since any two different norms in Banach lattice are equivalent (see exercise 101.10 in [7]), we have the following corollary.

Corollary 3.7. If H is a separable Banach lattice with a strong unit and Y is a Banach lattice with the Cantor property, then every order bounded linear operator T : H → Y is regular.

Suppose H = C(S) (where S is a compact Hausdorff space). Since C(S) is

separable if and only if S is metrizable, then Theorem 3.1 in [1] can be taken as a

Corollary 3.8. Let S be any compact metric space and Y be an arbitrary Banach lattice which has the Cantor property. Then every order bounded linear operator T : C(S) → Y is regular, i.e., there is a positive operator U : C(S) → Y with U ≥ T, −T .

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Cuijie Zhang School of Sciences

The Civil Aviation University of China Tianjin 300300

P.R. CHINA

zhang cui [email protected]

Hongyun Xiong

Department of Mathematics School of Sciences

Tianjin University Tianjin 300072 P.R. CHINA

[email protected]