The Order Continuity of the Regular Norm on Regular Operator Spaces

Abstract

We present here some sufficient conditions for the regular norm on ℒ^𝑟(𝐸,𝐹) to be order continuous, and for (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) to be a KB-space. In particular we deduce a characterization of the order continuity of the regular norm using L- and M-weak compactness of regular operators. Also we characterize when the space ℒ^𝑟(𝐸,𝐹) is an 𝐿^𝑝-space and is lattice isomorphic to an 𝐿^𝑝-space for 1<𝑃<∞. Some related results are also obtained.

1. Introduction

For Banach lattices 𝐸 and 𝐹, we use ℒ(𝐸,𝐹) to denote the space of all continuous linear operators from 𝐸 into 𝐹, and ℒ^𝑟(𝐸,𝐹) to denote the space of all regular operators from 𝐸 into 𝐹, which is the linear span of the set ℒ₊(𝐸,𝐹) of all positive operators from 𝐸 into 𝐹. With respect to the operator norm ‖⋅‖ the space ℒ^𝑟(𝐸,𝐹) is not complete in general (see, e.g., [1]), but there exists a natural norm on ℒ^𝑟(𝐸,𝐹), the regular norm ‖⋅‖_𝑟, which turns ℒ^𝑟(𝐸,𝐹) into a Banach space (see [2] for details). Namely, ‖𝑇‖_𝑟=inf‖𝑆‖∶𝑆∈ℒ₊(𝐸,𝐹),±𝑇≤𝑆. (1) In particular, ‖𝑇‖≤‖𝑇‖_𝑟. If ℒ^𝑟(𝐸,𝐹) is a vector lattice; then (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a Banach lattice and ‖𝑇‖_𝑟=‖|𝑇|‖ for all 𝑇∈ℒ^𝑟(𝐸,𝐹). For instance, if 𝐹 is Dedekind complete, then ℒ^𝑟(𝐸,𝐹) is a Dedekind complete Banach lattice under the regular norm.

The natural and important questions are: if ℒ^𝑟(𝐸,𝐹) is a vector lattice (i.e., a Banach lattice), when is the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) order continuous? When is ℒ^𝑟(𝐸,𝐹) a KB-space with respect to the regular norm?

Wickstead showed in [3] some characterizations of the space (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) being (lattice isomorphic to) an AL- or AM-space. It is natural to ask that when (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is an 𝐿^𝑝-space or lattice isomorphic to an 𝐿^𝑝-space for 1<𝑝<∞.

The purpose of this work is to present some results involving the order continuity of the regular norm on ℒ^𝑟(𝐸,𝐹) and (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) being a KB-space. Furthermore we will also present a complete description for (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) being (lattice isomorphic to) an 𝐿^𝑝-space with 1<𝑝<∞. Some related results are included as well.

Recall that an operator 𝑇∶𝐸→𝐹 is called L-weakly compact if 𝑇ball(𝐸) is an L-weakly compact set in 𝐹; that is, ‖𝑦_𝑛‖→0 for each disjoint sequence (𝑦_𝑛)^∞₁ contained in the solid hull of 𝑇ball(𝐸). Also 𝑇 is called M-weakly compact if ‖𝑇𝑥_𝑛‖→0 for each disjoint sequence (𝑥_𝑛)^∞₁⊂ball(𝐸), where ball(𝐸) denotes the unit ball of 𝐸. See, for example, [2].

We refer to [2, 4] for any unexplained terms from the theory of Banach lattices and operators.

2. Some General Results

We start with a necessary condition for the order continuity of the regular norm on spaces of regular operators.

Proposition 1. Let 𝐸 and 𝐹 be Banach lattices. If the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous, then the norms both on 𝐸^′ and 𝐹 are order continuous.

Proof. If the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous, for each increasing sequence (𝑦_𝑛)^∞₁⊂[0,𝑦]⊂𝐹, taking 𝑥^′∈𝐸^′₊ with ‖𝑥^′‖=1 and defining 𝑆,𝑆_𝑛∶𝐸→𝐹 by 𝑆_𝑛𝑥=𝑥^′(𝑥)𝑦_𝑛,𝑆𝑥=𝑥^′(𝑥)𝑦for𝑥∈𝐸 (2) then 𝑆_𝑛,𝑆∈ℒ₊(𝐸,𝐹) and 0≤𝑆_𝑛↑≤𝑆.
The order continuity of the regular norm implies that there is 𝑈∈ℒ^𝑟(𝐸,𝐹) such that ‖𝑆_𝑛−𝑈‖_𝑟→0; thus ‖𝑆_𝑛−𝑈‖→0. Choosing 𝑥₀∈𝐸 with 𝑥^′(𝑥₀)=1 we have 𝑦_𝑛−𝑈𝑥₀=𝑆_𝑛𝑥₀−𝑈𝑥₀≤𝑆_𝑛−𝑈𝑥₀⟶0. (3) It follows from Theorem 2.4.2 of [2] that the norm on 𝐹 is order continuous.
Similarly, for each increasing sequence (𝑥^′_𝑛)^∞₁⊂[0,𝑥^′]⊂𝐸^′, taking 𝑦∈𝐹₊ with ‖𝑦‖=1 and defining 𝑇,𝑇_𝑛∶𝐸→𝐹 by 𝑇_𝑛𝑥=𝑥^′_𝑛(𝑥)𝑦,𝑇𝑥=𝑥^′(𝑥)𝑦for𝑥∈𝐸 (4) then 𝑇_𝑛,𝑇∈ℒ₊(𝐸,𝐹) and 0≤𝑇_𝑛↑≤𝑇.
Again there is 𝑉∈ℒ^𝑟(𝐸,𝐹) such that ‖𝑇_𝑛−𝑉‖_𝑟→0; thus ‖𝑇_𝑛−𝑉‖→0. Choosing 𝑦^′∈𝐹^′ with 𝑦^′(𝑦)=1, it is easy to verify that 𝑥^′_𝑛−𝑉^′𝑦^′=𝑇^′_𝑛𝑦^′−𝑉^′𝑦^′≤𝑇^′_𝑛−𝑉^′𝑦^′=𝑇_𝑛−𝑉𝑦^′⟶0. (5) Theorem  2.4.2 of [2] yields that the norm on 𝐸^′ is order continuous.

Next result is a characterization of the order continuity of the regular norm on spaces of regular operators.

Theorem 2. For Banach lattices 𝐸 and 𝐹, the following statements are equivalent. (1)ℒ^𝑟(𝐸,𝐹) is a vector lattice and the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous. (2)Every positive operator 𝑇∶𝐸→𝐹 is L- and M-weakly compact.

Proof. (1)⇒(2). If the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous, then by the proposition above, the norms both on 𝐸^′ and 𝐹 are order continuous.
For 0≤𝑇∶𝐸→𝐹, it suffices to show that 𝑇 is M-weakly compact (see Theorem 3.6.17 of [2]). Otherwise, there is a disjoint sequence (𝑥_𝑛)^∞₁⊂ball(𝐸) such that ‖𝑇𝑥_𝑛‖≥𝛿>0 for all 𝑛∈ℕ. Note that 0≤𝑇|𝑥_𝑛|→0 weakly as |𝑥_𝑛|→0 weakly (see Theorem 2.4.14 of [2]).
By Corollary 2.3.5 of [2] there exists a sequence of naturals (𝑘_𝑛) and a disjoint sequence (𝑦_𝑛)⊂𝐹₊ such that 0≤𝑦_𝑛≤𝑇|𝑥_𝑘𝑛| and ‖𝑦_𝑛‖≥𝑐, where 𝑐 is any fixed number from (0,𝛿). Let 𝑃_𝑛𝐹∶→{𝑦_𝑛}^𝑑𝑑 be the band projection; hereby {𝑦_𝑛}^𝑑𝑑 denotes the band generated by 𝑦_𝑛 in 𝐹. It is easy to verify that 𝑃_𝑖⟂𝑃_𝑗 and 𝑃_𝑖≤𝐼_𝐹−𝑃_𝑗 (∀𝑖≠𝑗); it follows that 𝑃₁+⋯+𝑃_𝑛↑≤𝐼_𝐹, and (𝑃₁+⋯+𝑃_𝑛)𝑇↑≤𝑇, where 𝐼_𝐹 is the identity operator on 𝐹. Now the order continuity of the regular norm implies that ((𝑃₁+⋯+𝑃_𝑛)𝑇)^∞₁ is a ‖⋅‖_𝑟-Cauchy sequence; in particular, ‖𝑃_𝑛𝑇‖→0 as 𝑛→∞. Therefore 0<𝑐≤𝑦_𝑛=𝑃_𝑛𝑦_𝑛≤𝑃_𝑛𝑇𝑥_𝑛≤𝑃_𝑛𝑇⟶0 (6) This is impossible, so (1)⇒(2) holds.
(2)⇒(1). For any 0<𝑦∈𝐹₊ and 0<𝑥^′∈𝐸^′₊, let 𝑇∶𝐸→𝐹 by 𝑇𝑥=𝑥^′(𝑥)𝑦. Clearly 𝑇≥0 and the L- and M-weak compactness of 𝑇 yield the relatively weak compactness of both [−𝑦,𝑦] and [−𝑥^′,𝑥^′]. It follows from Theorem 2.4.2 of [2] that the norms both on 𝐸^′ and 𝐹 are order continuous; ℒ^𝑟(𝐸,𝐹) is certainly a (Dedekind complete) vector lattice.
For any decreasing sequence 𝑇_𝑛∈ℒ₊(𝐸,𝐹) with inf{𝑇_𝑛∶𝑛∈ℕ}=0, Proposition 3.6.19 of [2] yields that the operator norm, and hence the regular norm, on order interval [0,𝑇₁] is order continuous, which implies that ‖𝑇_𝑛‖_𝑟=‖𝑇_𝑛‖→0. Now the order continuity of the regular norm is following from Theorem 2.4.2 of [2].

It is clear that the identity operator on a Banach lattice 𝐸 is M-weakly compact if and only if 𝐸 is finite dimensional. The next result should be no surprise.

Corollary 3. Let 𝐸 be a Banach lattice. Then ℒ^𝑟(𝐸) is a vector lattice and the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸) is order continuous if and only if dim𝐸<∞.

Theorem 4. For Banach lattices 𝐸 and 𝐹, the following statements are equivalent. (1)(ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space. (2)𝐹 is a KB-space and ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous. (3)𝐹 is a KB-space and every positive operator 𝑇∶𝐸→𝐹 is M-weakly compact.

Proof. (1)⇒(2). If (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space, then ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) certainly is order continuous. For a norm bounded increasing sequence (𝑦_𝑛)^∞₁⊂𝐹₊, taking 𝑥^′∈𝐸^′₊ with ‖𝑥^′‖=1 and defining 𝑆_𝑛∶𝐸→𝐹 by 𝑆_𝑛𝑥=𝑥^′(𝑥)𝑦_𝑛 for 𝑥∈𝐸, then (𝑆_𝑛)^∞₁⊂ℒ₊(𝐸,𝐹) also is increasing and ‖⋅‖_𝑟-bounded, so there is 𝑈∈ℒ^𝑟(𝐸,𝐹) such that ‖𝑆_𝑛−𝑈‖_𝑟→0; thus ‖𝑆_𝑛−𝑈‖→0. Choosing 𝑥₀∈𝐸 with 𝑥^′(𝑥₀)=1 we have 𝑦_𝑛−𝑈𝑥₀=𝑆_𝑛𝑥₀−𝑈𝑥₀≤𝑆_𝑛−𝑈𝑥₀⟶0. (7) It follows that 𝐹 is a KB-space.
(2)⇒(3) is a consequence of Theorem 2. Now we show that (3)⇒(1). Clearly ℒ^𝑟(𝐸,𝐹) is a Banach lattice under the regular norm as 𝐹 is a KB-space. If (𝑇_𝑛)^∞₁⊂ℒ₊(𝐸,𝐹) is a ‖⋅‖_𝑟-bounded increasing sequence, then for each 𝑥∈𝐸₊, 𝑇_𝑛𝑥 is norm convergent as it is a norm bounded increasing sequence in 𝐹. It is easy to see that there is a 𝑇∈ℒ₊(𝐸,𝐹) such that 𝑇_𝑛→𝑇 with respect to the strong operator topology; it follows that 𝑇_𝑛↑𝑇 and by hypothesis 𝑇 is M-weakly compact. Proposition 3.6.19 of [2] yields that ‖𝑇−𝑇_𝑛‖_𝑟=‖𝑇−𝑇_𝑛‖→0 which implies that (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space.

It is obvious that if 𝑇∶𝐸→𝐹 is regular then 𝑇^′ is also regular, and the converse is false in general. For example, let 𝑇∶𝐿²[0,1]→𝑐₀ defined by 𝑇𝑓=(∫¹₀𝑓(𝑡)𝑟_𝑛(𝑡)𝑑𝑡), where 𝑟_𝑛(𝑡)=sgn(sin2^𝑛𝜋𝑡) is the 𝑛th Rademacher function on [0,1]. Then 𝑇^′∶ℓ₁→𝐿²[0,1], 𝑇^′(𝜆_𝑛)=∑^∞_𝑛=1𝜆_𝑛𝑟_𝑛, is regular (as it is order bounded) but 𝑇 is not regular. The following results will show some relationships between the order continuity of the regular norms in ℒ^𝑟(𝐸,𝐹), ℒ^𝑟(𝐸,𝐹^′′) and ℒ^𝑟(𝐹^′,𝐸^′).

Theorem 5. For Banach lattices 𝐸 and 𝐹, the following assertions are equivalent. (1)The regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹^′′) is order continuous. (2)(ℒ^𝑟(𝐸,𝐹^′′),‖⋅‖_𝑟) is a KB-space. (3)The regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐹^′,𝐸^′) is order continuous. (4)(ℒ^𝑟(𝐹^′,𝐸^′),‖⋅‖_𝑟) is a KB-space.

Proof. Let Φ∶ℒ(𝐸,𝐹^′′)→ℒ(𝐹^′,𝐸^′) by Φ(𝑇)=𝑇^′𝑗 for 𝑇∈ℒ(𝐸,𝐹^′′), where 𝑗∶𝐹^′→𝐹^′′′ is the natural embedding. According to Theorem 5.6 of [5] the operator 𝑇∈ℒ(𝐸,𝐹^′′) is regular if and only if Φ(𝑇) is regular and ‖Φ(𝑇)‖_𝑟=‖𝑇‖_𝑟.
Moreover Φ is an order continuous isometric lattice isomorphism from (ℒ(𝐸,𝐹^′′),‖⋅‖_𝑟) onto (ℒ(𝐹^′,𝐸^′),‖⋅‖_𝑟). Thus (1)⇔(3) is a simple consequence of these facts. Also the equivalences of (1) and (2), (3) and (4) easily follow from Theorem 4 and the proof of Theorem 2 (remembering that the norm on 𝐸^′ is order continuous if and only if 𝐸^′ is a KB-space; compare Theorem 2.4.14 of [2]).

Corollary 6. Let 𝐸 and 𝐹 be Banach lattices such that 𝐹 is reflexive. Then the following statements are equivalent. (1)The regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous. (2)(ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space. (3)The regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐹^′,𝐸^′) is order continuous. (4)(ℒ^𝑟(𝐹^′,𝐸^′),‖⋅‖_𝑟) is a KB-space.

Theorem 7. Let 𝐸 and 𝐹 be Banach lattices, 𝐻⊂𝐸 and 𝐺⊂𝐹 closed sublattices. Supposing that there is a positive projection 𝑃 from 𝐸 onto 𝐻 then the following statements hold.(1)If ℒ^𝑟(𝐸,𝐹) is a vector lattice and the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous, then ℒ^𝑟(𝐻,𝐺) also is a vector lattice and ‖⋅‖_𝑟 on ℒ^𝑟(𝐻,𝐺) is order continuous. (2)If (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space then (ℒ^𝑟(𝐻,𝐺),‖⋅‖_𝑟) also is a KB-space.

Proof. Suppose that ℒ^𝑟(𝐸,𝐹) is a vector lattice and the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous. For 0≤𝑇∶𝐻→𝐺, then 0≤𝑇𝑃∶𝐸→𝐺⊂𝐹, Theorem 2 yields that 𝑇𝑃 is L- and M-weakly compact. For any disjoint sequence (𝑦_𝑛)^∞₁ contained in the solid hull of 𝑇ball(𝐻) in 𝐺, then (𝑦_𝑛)^∞₁ is a disjoint sequence in 𝐹 as 𝐺 is a sublattice of 𝐹, which is contained in the solid hull of (𝑇𝑃)ball(𝐸) as 𝑇ball(𝐻)⊂(𝑇𝑃)ball(𝐸), so that ‖𝑦_𝑛‖→0; that is, 𝑇 is L-weakly compact. Also for each disjoint sequence (𝑥_𝑛)^∞₁⊂ball(𝐻), (𝑥_𝑛)^∞₁⊂ball(𝐸) is disjoint as 𝐻 is a sublattice of 𝐸; it follows that ‖𝑇𝑥_𝑛‖=‖𝑇𝑃𝑥_𝑛‖→0, which implies that 𝑇 is M-weakly compact. Again by Theorem 2  ℒ^𝑟(𝐻,𝐺) is a vector lattice and the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐻,𝐺) is order continuous; that is, (1) holds.
If (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space then it follows from Theorem 4 and (1) that 𝐹 is a KB-space, and hence 𝐺, as a closed sublattice of a KB-space, also is a KB-space, and that (ℒ^𝑟(𝐻,𝐺),‖⋅‖_𝑟) is a Banach lattice with an order continuous norm. Again Theorem 4 yields that (ℒ^𝑟(𝐻,𝐺),‖⋅‖_𝑟) is a KB-space, so (2) holds.

Note that each Banach lattice 𝐹 can be identified with a closed sublattice of 𝐹^′′, and so, as a consequence of Theorems 4 and 7 we have the following result.

Corollary 8. Let 𝐸 and 𝐹 be Banach lattices. If (ℒ^𝑟(𝐹^′,𝐸^′),‖⋅‖_𝑟) is a KB-space (equivalently the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐹^′,𝐸^′) is order continuous), then so is (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟).

Remark 9. The regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐹^′,𝐸^′) may fail to be order continuous even if (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space.
For example, let 𝐸=𝑐₀ and 𝐹=ℓ₁(ℓ^𝑛_∞)=(𝑐₀(ℓ^𝑛₁))^′. Clearly 𝐹 is a KB-space and 𝐹^′=ℓ_∞(ℓ^𝑛₁). Define 𝑇∶ℓ₁→𝐹^′ by 𝑇𝜆_𝑛=𝜆₁,𝜆₁,𝜆₂,𝜆₁,𝜆₂,𝜆₃,…,∀𝜆_𝑛∈ℓ₁, (8) it is easy to see that 𝑇 is an isometric lattice homomorphism; that is, 𝐹^′ contains a closed sublattice isometrically lattice isomorphic to ℓ₁. Thus Theorem 2.4.14 of [2] implies that 𝐹^′′ fails to be a KB-space (i.e., the norm on 𝐹^′′ is not order continuous).

Now it follows from Theorem 12 (see next) that (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space, but the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹^′′), and hence on ℒ^𝑟(𝐹^′,𝐸^′), is not order continuous as the norm on 𝐹^′′ is not order continuous (see the proof of Theorem 2).

3. Some Concrete Sufficient Conditions

In this section we will present some sufficient conditions on Banach lattices 𝐸 and 𝐹 such that the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous, or (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space.

Proposition 10. Let 𝐸 be an AM-space with a strong order unit and 𝐹 a Banach lattice with an order continuous norm. Then the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous.

Proof. We may assume that 𝐸 is equipped with the strong order unit norm and also the norm on 𝐹 is order continuous; clearly (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a Banach lattice. For 0≤𝑇_𝑛↑≤𝑇 in ℒ^𝑟(𝐸,𝐹) then 0≤𝑇_𝑛𝑥↑≤𝑇𝑥 for each 𝑥∈𝐸₊. It follows from the order continuity of the norm on 𝐹 that (𝑇_𝑛𝑥)^∞₁ is norm convergent. So there is 𝑆∈ℒ₊(𝐸,𝐹) such that 𝑇_𝑛→𝑆 with respect to the strong operator topology and obviously 𝑇_𝑛↑𝑆. In particular 𝑆−𝑇_𝑛𝑟=𝑆−𝑇_𝑛=𝑆𝑒−𝑇_𝑛𝑒⟶0, (9) where 𝑒 is a strong order unit of 𝐸. Therefore Theorem 2.4.2 of [2] yields that the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is order continuous.

Remark 11. If 𝐸 fails to possess a strong order unit the above result is false even if 𝐸 is an AM-space; 𝐸, 𝐸^′ and 𝐹 are atomic with an order continuous norm. For example, let 𝐸=𝐹=𝑐₀ and then the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸) is not order continuous, compare also Corollary 3.
Recall that Banach lattice 𝐸 possesses the positive Schur property if every weakly null sequence in 𝐸₊ is norm convergent to 0.

Theorem 12. Let 𝐸 be a Banach lattice such that 𝐸^′ possesses the positive Schur property, 𝐹 a Banach lattice. Then (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space if and only if 𝐹 is a KB-space.

Proof. The part of “only if” is obvious. If 𝐹 is a KB-space, by Theorem 4 it suffices to show that each positive operator 𝑇∶𝐸→𝐹 is M-weakly compact. Indeed, if 𝑇 is not M-weakly compact then there is a disjoint sequence (𝑥_𝑛)^∞₁⊂ball(𝐸) with 𝑥_𝑛≥0 and ‖𝑇𝑥_𝑛‖≥2𝛿>0 for 𝑛∈ℕ. Note that 𝑇𝑥_𝑛→0 weakly as 𝑥_𝑛→0 weakly (see Theorem 2.4.14 of [2]); by Proposition 2.3.4 of [2] there exists a disjoint sequence (𝑦^′_𝑛)^∞₁⊂ball(𝐹^′), 𝑦_𝑛≥0, satisfying 𝑇^′𝑦^′_𝑛𝑥_𝑛=𝑦^′_𝑛𝑇𝑥_𝑛>𝛿,∀𝑛. (10) Also by Theorem 2.5.6 and 3.4.18 of [2], 𝑇 is weakly compact and so is 𝑇^′ by Gantmacher’s theorem (see Theorem 17.2 of [4]), so we may assume that 𝑇^′𝑦^′_𝑛 is weakly convergent (replacing by a subsequence if necessary); say 𝑇^′𝑦^′_𝑛→𝑥^′ weakly, then for each 𝑥∈𝐸𝑥^′(𝑥)=lim_𝑛→∞𝑇^′𝑦^′_𝑛(𝑥)=lim_𝑛→∞𝑦^′_𝑛(𝑇𝑥)=0 (11) as 𝑦^′_𝑛→0 in 𝜎(𝐹^′,𝐹) (see Corollary 2.4.3 of [2]); that is, 𝑇^′𝑦^′_𝑛→0 weakly, so the positive Schur property of 𝐸^′ implies that ‖𝑇^′𝑦^′_𝑛‖→0 and it follows that 0<𝛿<𝑦^′_𝑛𝑇𝑥_𝑛=𝑇^′𝑦^′_𝑛𝑥_𝑛≤𝑇^′𝑦^′_𝑛⟶0. (12) This is impossible, thus 𝑇 is M-weakly compact.

The following result is a dual version of Theorem 12.

Theorem 13. Let 𝐹 be a Banach lattice with the positive Schur property, 𝐸 a Banach lattice. Then (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space if and only if the norm on 𝐸^′ is order continuous.

Proof. The part of “only if” easily follows from the proof of Theorem 2. If the norm on 𝐸^′ is order continuous, for 𝑇∈ℒ₊(𝐸,𝐹) and each disjoint sequence (𝑥_𝑛)^∞₁⊂ball(𝐸), then |𝑥_𝑛|→0 weakly, and 𝑇|𝑥_𝑛|→0 weakly. It follows from the positive Schur property of 𝐹 that ‖𝑇𝑥_𝑛‖→0 as |𝑇𝑥_𝑛|≤𝑇|𝑥_𝑛|; that is, 𝑇 is M-weakly compact; Theorem 4 yields that (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space.

For a Banach lattice 𝐸 and 1≤𝑝≤∞, recall that 𝐸 has the strong ℓ_𝑝-decomposition property if there exists a constant 𝑀 such that for all disjoint elements 𝑥₁,…,𝑥_𝑛 in 𝐸 we have (∑^𝑛_𝑖=1‖𝑥_𝑖‖^𝑝)^1/𝑝≤𝑀‖∑^𝑛_𝑖=1𝑥_𝑖‖ for 𝑝<∞ and max{‖𝑥_𝑖‖∶𝑖=1,…,𝑛}≤𝑀‖∑^𝑛_𝑖=1𝑥_𝑖‖ in case 𝑝=∞. The number 𝜎(𝐸)=inf{𝑝≥1∶𝐸 has the strong ℓ_𝑝-decomposition property} is call the upper index of 𝐸.

Similarly 𝐸 has the strong ℓ_𝑝-composition property if there exists a constant 𝑀 such that for all disjoint elements 𝑥₁,…,𝑥_𝑛 in 𝐸 we have ‖∑^𝑛_𝑖=1𝑥_𝑖‖≤𝑀(∑^𝑛_𝑖=1‖𝑥_𝑖‖^𝑝)^1/𝑝 for 𝑝<∞ and ‖∑^𝑛_𝑖=1𝑥_𝑖‖≤𝑀max{‖𝑥_𝑖‖∶𝑖=1,…,𝑛} in case 𝑝=∞. The number 𝑠(𝐸)=sup{𝑝≥1∶𝐸 has the strong ℓ_𝑝-composition property} is called lower index of 𝐸.

It is known that 1≤𝑠(𝐸)≤𝜎(𝐸)≤∞ for any Banach lattice 𝐸. If 𝜎(𝐸)<∞ then 𝐸 has an order continuous norm. If 𝑠(𝐸)>1 then the norm on 𝐸^′ is order continuous. See [6] for details

Also recall that if the norm on a Banach lattice 𝐸 is 𝑝-superadditive then 𝜎(𝐸)≤𝑝; and if 𝐸 has a 𝑝-subadditive norm then 𝑠(𝐸)≥𝑝; see Proposition 2.8.2 of [2].

Theorem 14. Let 𝐸 and 𝐹 be Banach lattices. If 𝑠(𝐸)>𝜎(𝐹) then (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space.

Proof. The norm on 𝐸^′ clearly is order continuous. Note that if 𝜎(𝐹)<∞ then 𝐹 is a KB-space. Indeed, if 𝐹 is not a KB-space, then 𝐹 contains a sublattice 𝐻 lattice isomorphic to 𝑐₀, which implies that 𝜎(𝐹)=∞ as 𝜎(𝑐₀)=∞. Now the rest is a simple consequence of Theorem 4, Theorem 6.7 of [6], and Theorem 3.6.17 of [2].

Corollary 15. Let 𝐸 and 𝐹 be Banach lattices. If the norm of 𝐸 is 𝑝-subadditive, the norm of 𝐹 is 𝑞-superadditive and 1≤𝑞<𝑝≤∞, then (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟)is a KB-space.

Remark 16. It is worth to point out that 𝑠(𝐸)>𝜎(𝐹) fails to be true in general even if ℒ^𝑟(𝐸,𝐹) is a KB-space, see [7, Example 3.6].

For 𝐸 and 𝐹 being 𝐿^𝑝- and 𝐿^𝑞-spaces, respectively, we have the following characterization.

Theorem 17. Let 𝐸 and 𝐹 be infinite dimensional 𝐿^𝑝-space, and 𝐿^𝑞-space respectively, then (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is a KB-space if and only if 𝑞<𝑝.

Proof. The part of “if” is a simple consequence of Corollary 15. To see the part of “only if”, we may first assume that 𝐻_𝑛=ℓ_𝑝𝑐if𝑝<∞₀𝐺if𝑝=∞^𝑞=ℓ_𝑞𝑐if𝑞<∞₀if𝑞=∞ (13) are sublattices of 𝐸 and 𝐹, respectively. Suppose that 𝑝≤𝑞 then 𝐻_𝑝⊂𝐺_𝑞.
If 𝑝<∞ there is a positive projection 𝑃 from 𝐸 onto 𝐻_𝑝 (the existence of 𝑃 is following from Theorem 2.7.11 of [2]), then 𝑃∶𝐸→𝐻_𝑝⊂𝐺_𝑞⊂𝐹 is not M-weakly compact, which, by Theorem 4, implies that (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is not a KB-space.
Also if 𝑝=∞, then 𝑞=∞ and 𝐹 is not a KB-space, Theorem 4 again yields that (ℒ^𝑟(𝐸,𝐹),‖⋅‖_𝑟) is not a KB-space.

The next result shows that under the regular norm the spaces ℒ^𝑟(𝐸,𝐹) are rather rare to be 𝐿^𝑝-spaces.

Theorem 18. Let 𝐸 and 𝐹 be non-zero Banach lattices and 1<𝑝, 𝑞<∞ with (1/𝑝)+(1/𝑞)=1. Then the following assertions are equivalent. (1)The regular norm ‖⋅‖_𝑟 is 𝑝-additive on ℒ₊(𝐸,𝐹). (2)One of following two conditions holds.(a)dim𝐸=1 and the norm on 𝐹 is 𝑝-additive (i.e., 𝐹 is an 𝐿^𝑝-space). (b)dim𝐹=1 and the norm on 𝐸 is 𝑞-additive (i.e., 𝐸 is an 𝐿^𝑞-space).

Proof. (2)⇒(1) is obvious. To see that (1)⇒(2), we assume that ‖⋅‖_𝑟 is 𝑝-additive on ℒ₊(𝐸,𝐹). For any 𝑦₁,𝑦₂∈𝐹₊, pick 𝑥^′∈𝐸^′₊ with ‖𝑥^′‖=1; thus the 𝑝-additivity of the regular norm yields that 𝑦₁+𝑦₂^𝑝=𝑥^′𝑦₁+𝑦₂^𝑝=𝑥^′𝑦₁^𝑝+𝑥^′𝑦₂^𝑝=𝑦₁^𝑝+𝑦₂^𝑝 (14) which means that 𝐹 is an 𝐿^𝑝-space. A similar argument involving a fixed element of 𝐹₊ and two elements of 𝐸^′₊ shows that 𝐸^′ is an 𝐿^𝑝-space; hence 𝐸 is an 𝐿^𝑞-space (compare with Theorem 2.7.1 of [2]), where 𝑝⁻¹+𝑞⁻¹=1.
Now if both dim(𝐸)≥2 and dim(𝐹)≥2 hold we will obtain a contradiction. In fact, we may assume that ℓ^𝑞₂ and ℓ^𝑝₂ are 2-dimensional sublattices of 𝐸 and 𝐹, respectively; define 𝑇₁,𝑇₂∶ℓ^𝑞₂→ℓ^𝑝₂ by 𝑇₁𝜆₁,𝜆₂=𝜆₁,𝑇,0₂𝜆₁,𝜆₂=0,𝜆₂for𝜆₁,𝜆₂∈ℓ^𝑞₂; (15) then ‖𝑇₁‖=‖𝑇₂‖=1. Let 𝑃 be a positive contractive projection from 𝐸 onto ℓ²_𝑞 (see Theorem 2.7.11 of [2]); it follows that 𝑇₁+𝑇₂^𝑝=𝑃𝑇₁+𝑃𝑇₂^𝑝=𝑃𝑇₁^𝑝+𝑃𝑇₂^𝑝=2. (16) Also it is easy to calculate that ‖𝑇₁+𝑇₂‖=2^{1/𝑝−1/𝑞}; this is impossible. (1)⇒(2) holds.

Theorem 19. Let 𝐸 and 𝐹 be non-zero Banach lattices and 1<𝑝, 𝑞<∞ with (1/𝑝)+(1/𝑞)=1. Then the following assertions are equivalent.(1)The regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is equivalent to a 𝑝-additive norm. (2)One of following two conditions holds.(a)dim𝐸<∞ and the norm on 𝐹 is equivalent to a 𝑝-additive norm. (b)dim𝐹<∞ and the norm on 𝐸 is equivalent to a 𝑞-additive norm.

Proof. (1)⇒(2). Suppose that the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is equivalent to a 𝑝-additive norm. We first show that the norms both on 𝐸^′ and 𝐹 are equivalent to 𝑞-additive and 𝑝-additive norms, respectively, where 𝑝⁻¹+𝑞⁻¹=1.
For each disjoint sequence (𝑦_𝑛)^∞₁⊂𝐹₊ with ‖𝑦_𝑛‖=1, fix 𝑥^′∈𝐸^′₊ with ‖𝑥^′‖=1. It is easy to verify that (𝑥^′⨂𝑦_𝑛)^∞₁⊂ℒ₊(𝐸,𝐹) is a disjoint sequence with ‖𝑥^′⨂𝑦_𝑛‖=1 for all 𝑛∈ℕ. Corollary 2.8.12  of [2] yields that (𝑥^′⨂𝑦_𝑛)^∞₁ is equivalent to the natural basis of ℓ^𝑝. Note that ^𝑛_𝑖=1𝜆_𝑖𝑥^′𝑦_𝑖=𝑥^′𝑛_𝑖=1𝜆_𝑖𝑦_𝑖=^𝑛_𝑖=1𝜆_𝑖𝑦_𝑖 (17) for all 𝑛∈ℕ and 𝜆_𝑖∈ℝ. It follows that (𝑦_𝑛)^∞₁ is equivalent to the natural basis of ℓ^𝑝, which by Corollary 2.8.12 of [2] implies that the norm on 𝐹 is equivalent to a 𝑝-additive norm.
A similar argument involving a fixed element of 𝐹₊ and a disjoint sequence of elements of 𝐸^′₊ shows that the norm on 𝐸^′ is equivalent to a 𝑝-additive norm; hence the norm on 𝐸 is equivalent to a 𝑞-additive norm.
Now we show that either dim𝐸<∞ or dim𝐹<∞. Otherwise, both 𝐸 and 𝐹 are infinite dimensional. Renorming 𝐸 and 𝐹 with equivalent 𝑞-additive and 𝑝-additive norms, respectively, the regular norm on ℒ^𝑟(𝐸,𝐹) is still equivalent to a 𝑝-additive norm. Thus we may assume that the norms on 𝐸 and 𝐹 are 𝑞- and 𝑝-additive, and that ℓ^𝑞⊂𝐸 and ℓ^𝑝⊂𝐹 are sublattices, respectively. By Theorem 2.7.11 of [2] there is a positive contractive projection 𝑃 from 𝐸 onto ℓ^𝑞. Consider the operators 𝑇_𝑛∶ℓ^𝑞→ℓ^𝑝 by 𝑇_𝑛(𝜆_𝑘)=𝜆_𝑛𝑒_𝑛, where 𝑒_𝑛 is the element in ℓ^𝑞 and ℓ^𝑝 with 𝑛th entry equals to 1 and all others are 0. Then it is easy to verify that (𝑇_𝑛𝑃)^∞₁ is a disjoint sequence in ℒ₊(𝐸,𝐹) with ‖𝑇_𝑛𝑃‖=1. Corollary 2.8.12 of [2] yields that (𝑇_𝑛𝑃)^∞₁ is equivalent to the natural basis of ℓ^𝑝. In particular, we have 𝐵𝑛^1/𝑝≤𝑇₁+𝑇₂+⋯+𝑇_𝑛=𝑇₁𝑃+𝑇₂𝑃+⋯+𝑇_𝑛𝑃≤𝐴𝑛^1/𝑝 (18) for all 𝑛∈ℕ, where 𝐴>0 and 𝐵>0 are constants. But 𝑇₁+𝑇₂+⋯+𝑇_𝑛=sup𝜆_𝑖^𝑛_1𝑝∶𝜆_𝑖^𝑛_1𝑞≤1 (19) which easily shows that ‖𝑇₁+𝑇₂+⋯+𝑇_𝑛‖≤1 if 𝑞≤𝑝 and ‖𝑇₁+𝑇₂+⋯+𝑇_𝑛‖≤𝑛^{1/𝑝−1/𝑞} for 𝑞>𝑝. This is impossible for either 𝑞≤𝑝 or 𝑞>𝑝. So (1)⇒(2) holds.
(2)(a)⇒(1). Let 𝐸=span{𝑒₁,𝑒₂,…,𝑒_𝑚} with {𝑒₁,𝑒₂,…,𝑒_𝑚}⊂𝐸₊ pairwise disjoint and ‖𝑒_𝑖‖=1. Then each 𝑇∈ℒ^𝑟(𝐸,𝐹) corresponds to unique (𝑥₁(𝑇),𝑥₂(𝑇),…,𝑥_𝑚(𝑇)); moreover, 𝑥_𝑖(𝑇)=𝑇𝑒_𝑖∈𝐹, satisfying the following conditions.(i)𝑇≥0⇔𝑥_𝑖(𝑇)≥0 for 1≤𝑖≤𝑚.(ii)𝑥_𝑖(|𝑇|)=|𝑥_𝑖(𝑇)| for all 𝑇∈ℒ^𝑟(𝐸,𝐹) and 1≤𝑖≤𝑚.(iii)𝑥_𝑖(𝜆𝑇+𝜇𝑆)=𝜆𝑥_𝑖(𝑇)+𝜇𝑥_𝑖(𝑆) for all 𝑇,𝑆∈ℒ^𝑟(𝐸,𝐹), 𝜆,𝜇∈ℝ and 1≤𝑖≤𝑚.(iv)max{‖𝑥_𝑖(𝑇)‖∶1≤𝑖≤𝑚}≤‖𝑇‖_𝑟=‖|𝑇|‖≤∑^𝑚_𝑖=1‖𝑥_𝑖(𝑇)‖.
Since the norm on 𝐹 is equivalent to a 𝑝-additive norm, for each disjoint sequence (𝑦_𝑘)^∞₁⊂𝐹₊, by Corollary 2.8.12 of [2] there exist constants 𝐴>0, 𝐵>0 such that 𝐵(𝜆_𝑘𝑦_𝑘)^𝑛_1𝑝≤^𝑛_𝑘=1𝜆_𝑘𝑦_𝑘≤𝐴(𝜆_𝑘𝑦_𝑘)^𝑛_1𝑝 (∗) for all 𝜆_𝑘∈ℝ and 𝑛∈ℕ. Now for any disjoint sequence (𝑇_𝑛)^∞₁⊂ℒ₊(𝐸,𝐹) with ‖𝑇_𝑛‖=1, note that the disjointness of (𝑥_𝑖(𝑇_𝑛))^∞_𝑛=1⊂𝐹₊ for each 1≤𝑖≤𝑚; it follows from (iii), (iv), and (∗) that ^𝑛_𝑘=1𝜆_𝑘𝑇_𝑘≤^𝑚_𝑖=1^𝑛_𝑘=1𝜆_𝑘𝑥_𝑖𝑇_𝑘≤𝐴^𝑚_𝑖=1^𝑛_𝑘=1𝜆_𝑘^𝑝𝑥_𝑖𝑇_𝑘^𝑝1/𝑝≤𝑚𝐴𝜆_𝑘^𝑛_1𝑝 (20) as ‖𝑥_𝑖(𝑇_𝑘)‖≤‖𝑇_𝑘‖=1. Also ^𝑛_𝑘=1𝜆_𝑘𝑇_𝑘≥max^𝑛_𝑘=1𝜆_𝑘𝑥_𝑖𝑇_𝑘∶1≤𝑖≤𝑚≥𝐵max^𝑛_𝑘=1𝜆_𝑘^𝑝𝑥_𝑖𝑇_𝑘^𝑝1/𝑝∶1≤𝑖≤𝑚≥𝐵𝑚^{−1/𝑝𝑚𝑛}_𝑖=1𝑘=1𝜆_𝑘^𝑝𝑥_𝑖𝑇_𝑘^𝑝1/𝑝=𝐵𝑚^{−1/𝑝𝑛}_𝑘=1𝜆_𝑘^𝑝𝑚_𝑖=1𝑥_𝑖𝑇_𝑘^𝑝1/𝑝≥𝐵𝑚⁻¹𝜆_𝑘^𝑛_1𝑝 (21) as 1≤∑^𝑚₌₁‖𝑥_𝑖(𝑇_𝑘)‖≤(∑^𝑚_𝑖=1‖𝑥_𝑖(𝑇_𝑘)‖^𝑝)^1/𝑝𝑚^1/𝑞.
Therefore 𝐵𝑚⁻¹𝜆_𝑘^𝑛_1𝑝≤^𝑛_𝑘=1𝜆_𝑘𝑇_𝑘≤𝑚𝐴𝜆_𝑘^𝑛_1𝑝 (22) for all 𝜆_𝑘∈ℝ and 𝑛∈ℕ; that is, (𝑇_𝑛)^∞₁ is equivalent to the natural basis of ℓ^𝑝. Corollary 2.8.12 of [2] again shows that the regular norm ‖⋅‖_𝑟 on ℒ^𝑟(𝐸,𝐹) is equivalent to a 𝑝-additive norm, so (2)(a)⇒(1) holds.
The proof of (2)(b)⇒(1) is similar with (2)(a)⇒(1). This completes the proof.

Acknowledgments

One of the authors would like to thank Professor A. W. Wickstead for useful discussions about the subjects of this paper. The authors were supported in part by the Fundamental Research Funds for the Central Universities (SWJTU11CX154, SWJTU12ZT13).