# The Order Continuity of the Regular Norm on Regular Operator Spaces

Authors: Zi Li Chen; Ying Feng; Jin Xi Chen
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We present here some sufficient conditions for the regular norm onℒr(E,F) to be order continuous, and for (ℒr(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 ℒr(E,F) is an Lp-space and is lattice isomorphic to an Lp-space for 1<P<∞. Some related results are also obtained.


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## Body

## 1. Introduction

For Banach latticesE and F, we use ℒ(E,F) to denote the space of all continuous linear operators from E into F, and ℒr(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 ℒr(E,F) is not complete in general (see, e.g., [1]), but there exists a natural norm on ℒr(E,F), the regular norm∥·∥r, which turns ℒr(E,F) into a Banach space (see [2] for details). Namely,
(1)∥T∥r=inf{∥S∥:S∈ℒ+(E,F),±T≤S}.
In particular, ∥T∥≤∥T∥r. If ℒr(E,F) is a vector lattice; then (ℒr(E,F),∥·∥r) is a Banach lattice and ∥T∥r=∥|T|∥ for all T∈ℒr(E,F). For instance, if F is Dedekind complete, then ℒr(E,F) is a Dedekind complete Banach lattice under the regular norm.The natural and important questions are: ifℒr(E,F) is a vector lattice (i.e., a Banach lattice), when is the regular norm ∥·∥r on ℒr(E,F) order continuous? When is ℒr(E,F) a KB-space with respect to the regular norm?Wickstead showed in [3] some characterizations of the space (ℒr(E,F),∥·∥r) being (lattice isomorphic to) an AL- or AM-space. It is natural to ask that when (ℒr(E,F),∥·∥r) is an Lp-space or lattice isomorphic to an Lp-space for 1<p<∞.The purpose of this work is to present some results involving the order continuity of the regular norm onℒr(E,F) and (ℒr(E,F),∥·∥r) being a KB-space. Furthermore we will also present a complete description for (ℒr(E,F),∥·∥r) being (lattice isomorphic to) an Lp-space with 1<p<∞. Some related results are included as well.Recall that an operatorT:E→F is called L-weakly compact if Tball(E) is an L-weakly compact set in F; that is, ∥yn∥→0 for each disjoint sequence (yn)1∞ contained in the solid hull of Tball(E). Also T is called M-weakly compact if ∥Txn∥→0 for each disjoint sequence (xn)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.
LetE and F be Banach lattices. If the regular norm ∥·∥r on ℒr(E,F) is order continuous, then the norms both on E′ and F are order continuous.Proof.
If the regular norm∥·∥r on ℒr(E,F) is order continuous, for each increasing sequence (yn)1∞⊂[0,y]⊂F, taking x′∈E+′ with ∥x′∥=1 and defining S,Sn:E→F by
(2)Snx=x′(x)yn,Sx=x′(x)yforx∈E
then Sn,S∈ℒ+(E,F) and 0≤Sn↑≤S.
The order continuity of the regular norm implies that there isU∈ℒr(E,F) such that ∥Sn-U∥r→0; thus ∥Sn-U∥→0. Choosing x0∈E with x′(x0)=1 we have
(3)∥yn-Ux0∥=∥Snx0-Ux0∥≤∥Sn-U∥∥x0∥⟶0.
It follows from Theorem 2.4.2 of [2] that the norm on F is order continuous.
Similarly, for each increasing sequence(xn′)1∞⊂[0,x′]⊂E′, taking y∈F+ with ∥y∥=1 and defining T,Tn:E→F by
(4)Tnx=xn′(x)y,Tx=x′(x)yforx∈E
then Tn,T∈ℒ+(E,F) and 0≤Tn↑≤T.
Again there isV∈ℒr(E,F) such that ∥Tn-V∥r→0; thus ∥Tn-V∥→0. Choosing y′∈F′ with y′(y)=1, it is easy to verify that
(5)∥xn′-V′y′∥=∥Tn′y′-V′y′∥≤∥Tn′-V′∥∥y′∥=∥Tn-V∥∥y′∥⟶0.
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 latticesE and F, the following statements are equivalent. (1)
ℒr(E,F) is a vector lattice and the regular norm ∥·∥r on ℒr(E,F) is order continuous.(2)
Every positive operatorT:E→F is L- and M-weakly compact.Proof.
(
1
)
⇒
(
2
). If the regular norm ∥·∥r on ℒr(E,F) is order continuous, then by the proposition above, the norms both on E′ and F are order continuous.
For0≤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 (xn)1∞⊂ball(E) such that ∥Txn∥≥δ>0 for all n∈ℕ. Note that 0≤T|xn|→0 weakly as |xn|→0 weakly (see Theorem 2.4.14 of [2]).
By Corollary 2.3.5 of [2] there exists a sequence of naturals (kn) and a disjoint sequence (yn)⊂F+ such that 0≤yn≤T|xkn| and ∥yn∥≥c, where c is any fixed number from (0,δ). Let PnF:→{yn}dd be the band projection; hereby {yn}dd denotes the band generated by yn in F. It is easy to verify that Pi⊥Pj and Pi≤IF-Pj (∀i≠j); it follows that P1+⋯+Pn↑≤IF, and (P1+⋯+Pn)T↑≤T, where IF is the identity operator on F. Now the order continuity of the regular norm implies that ((P1+⋯+Pn)T)1∞ is a ∥·∥r-Cauchy sequence; in particular, ∥PnT∥→0 as n→∞. Therefore
(6)0<c≤∥yn∥=∥Pnyn∥≤∥PnT|xn|∥≤∥PnT∥⟶0
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; ℒr(E,F) is certainly a (Dedekind complete) vector lattice.
For any decreasing sequenceTn∈ℒ+(E,F) with inf{Tn: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 ∥Tn∥r=∥Tn∥→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 latticeE is M-weakly compact if and only if E is finite dimensional. The next result should be no surprise.Corollary 3.
LetE be a Banach lattice. Then ℒr(E) is a vector lattice and the regular norm ∥·∥r on ℒr(E) is order continuous if and only if dimE<∞.Theorem 4.
For Banach latticesE and F, the following statements are equivalent. (1)
(ℒr(E,F),∥·∥r) is a KB-space.(2)
F is a KB-space and ∥·∥r on ℒr(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 (ℒr(E,F),∥·∥r) is a KB-space, then ∥·∥r on ℒr(E,F) certainly is order continuous. For a norm bounded increasing sequence (yn)1∞⊂F+, taking x′∈E+′ with ∥x′∥=1 and defining Sn:E→F by Snx=x′(x)yn for x∈E, then (Sn)1∞⊂ℒ+(E,F) also is increasing and ∥·∥r-bounded, so there is U∈ℒr(E,F) such that ∥Sn-U∥r→0; thus ∥Sn-U∥→0. Choosing x0∈E with x′(x0)=1 we have
(7)∥yn-Ux0∥=∥Snx0-Ux0∥≤∥Sn-U∥∥x0∥⟶0.
It follows that F is a KB-space.
(
2
)
⇒
(
3
) is a consequence of Theorem 2. Now we show that (3)⇒(1). Clearly ℒr(E,F) is a Banach lattice under the regular norm as F is a KB-space. If (Tn)1∞⊂ℒ+(E,F) is a ∥·∥r-bounded increasing sequence, then for each x∈E+, Tnx 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 Tn→T with respect to the strong operator topology; it follows that Tn↑T and by hypothesis T is M-weakly compact. Proposition 3.6.19 of [2] yields that ∥T-Tn∥r=∥T-Tn∥→0 which implies that (ℒr(E,F),∥·∥r) is a KB-space.It is obvious that ifT: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=(∫01‍f(t)rn(t)dt), where rn(t)=sgn(sin2nπt) is the nth Rademacher function on [0,1]. Then T′:ℓ1→L2[0,1], T′(λn)=∑n=1∞‍λnrn, 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 ℒr(E,F), ℒr(E,F′′) and ℒr(F′,E′).Theorem 5.
For Banach latticesE and F, the following assertions are equivalent. (1)
The regular norm∥·∥r on ℒr(E,F′′) is order continuous.(2)
(ℒr(E,F′′),∥·∥r) is a KB-space.(3)
The regular norm∥·∥r on ℒr(F′,E′) is order continuous.(4)
(ℒr(F′,E′),∥·∥r) is a KB-space.Proof.
LetΦ: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.
MoreoverΦ 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.
LetE and F be Banach lattices such that F is reflexive. Then the following statements are equivalent. (1)
The regular norm∥·∥r on ℒr(E,F) is order continuous.(2)
(ℒr(E,F),∥·∥r) is a KB-space.(3)
The regular norm∥·∥r on ℒr(F′,E′) is order continuous.(4)
(ℒr(F′,E′),∥·∥r) is a KB-space.Theorem 7.
LetE 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)
Ifℒr(E,F) is a vector lattice and the regular norm ∥·∥r on ℒr(E,F) is order continuous, then ℒr(H,G) also is a vector lattice and ∥·∥r on ℒr(H,G) is order continuous.(2)
If(ℒr(E,F),∥·∥r) is a KB-space then (ℒr(H,G),∥·∥r) also is a KB-space.Proof.
Suppose thatℒr(E,F) is a vector lattice and the regular norm ∥·∥r on ℒr(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 (yn)1∞ contained in the solid hull of Tball(H) in G, then (yn)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 ∥yn∥→0; that is, T is L-weakly compact. Also for each disjoint sequence (xn)1∞⊂ball(H), (xn)1∞⊂ball(E) is disjoint as H is a sublattice of E; it follows that ∥Txn∥=∥TPxn∥→0, which implies that T is M-weakly compact. Again by Theorem 2  ℒr(H,G) is a vector lattice and the regular norm ∥·∥r on ℒr(H,G) is order continuous; that is, (1) holds.
If(ℒr(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 (ℒr(H,G),∥·∥r) is a Banach lattice with an order continuous norm. Again Theorem 4 yields that (ℒr(H,G),∥·∥r) is a KB-space, so (2) holds.Note that each Banach latticeF 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.
LetE and F be Banach lattices. If (ℒr(F′,E′),∥·∥r) is a KB-space (equivalently the regular norm ∥·∥r on ℒr(F′,E′) is order continuous), then so is (ℒr(E,F),∥·∥r).Remark 9.
The regular norm∥·∥r on ℒr(F′,E′) may fail to be order continuous even if (ℒr(E,F),∥·∥r) is a KB-space.
For example, letE=c0 and F=ℓ1(ℓ∞n)=(c0(ℓ1n))′. Clearly F is a KB-space and F′=ℓ∞(ℓ1n). Define T:ℓ1→F′ by
(8)T(λn)=((λ1),(λ1,λ2),(λ1,λ2,λ3),…),∀(λn)∈ℓ1,
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 Theorem12 (see next) that (ℒr(E,F),∥·∥r) is a KB-space, but the regular norm ∥·∥r on ℒr(E,F′′), and hence on ℒr(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 latticesE and F such that the regular norm ∥·∥r on ℒr(E,F) is order continuous, or (ℒr(E,F),∥·∥r) is a KB-space.Proposition 10.
LetE 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 ℒr(E,F) is order continuous.Proof.
We may assume thatE is equipped with the strong order unit norm and also the norm on F is order continuous; clearly (ℒr(E,F),∥·∥r) is a Banach lattice. For 0≤Tn↑≤T in ℒr(E,F) then 0≤Tnx↑≤Tx for each x∈E+. It follows from the order continuity of the norm on F that (Tnx)1∞ is norm convergent. So there is S∈ℒ+(E,F) such that Tn→S with respect to the strong operator topology and obviously Tn↑S. In particular
(9)∥S-Tn∥r=∥S-Tn∥=∥Se-Tne∥⟶0,
where e is a strong order unit of E. Therefore Theorem 2.4.2 of [2] yields that the regular norm ∥·∥r on ℒr(E,F) is order continuous.Remark 11.
IfE 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 ℒr(E) is not order continuous, compare also Corollary 3.
Recall that Banach latticeE possesses the positive Schur property if every weakly null sequence in E+ is norm convergent to 0.Theorem 12.
LetE be a Banach lattice such that E′ possesses the positive Schur property, F a Banach lattice. Then (ℒr(E,F),∥·∥r) is a KB-space if and only if F is a KB-space.Proof.
The part of “only if” is obvious. IfF 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 (xn)1∞⊂ball(E) with xn≥0 and ∥Txn∥≥2δ>0 for n∈ℕ. Note that Txn→0 weakly as xn→0 weakly (see Theorem 2.4.14 of [2]); by Proposition 2.3.4 of [2] there exists a disjoint sequence (yn′)1∞⊂ball(F′), yn≥0, satisfying
(10)(T′yn′)(xn)=yn′(Txn)>δ,∀n.
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′yn′ is weakly convergent (replacing by a subsequence if necessary); say T′yn′→x′ weakly, then for each x∈E(11)x′(x)=limn→∞T′yn′(x)=limn→∞yn′(Tx)=0
as yn′→0 in σ(F′,F) (see Corollary 2.4.3 of [2]); that is, T′yn′→0 weakly, so the positive Schur property of E′ implies that ∥T′yn′∥→0 and it follows that
(12)0<δ<yn′(Txn)=(T′yn′)(xn)≤∥T′yn′∥⟶0.
This is impossible, thus T is M-weakly compact.The following result is a dual version of Theorem12.Theorem 13.
LetF be a Banach lattice with the positive Schur property, E a Banach lattice. Then (ℒr(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 Theorem2. If the norm on E′ is order continuous, for T∈ℒ+(E,F) and each disjoint sequence (xn)1∞⊂ball(E), then |xn|→0 weakly, and T|xn|→0 weakly. It follows from the positive Schur property of F that ∥Txn∥→0 as |Txn|≤T|xn|; that is, T is M-weakly compact; Theorem 4 yields that (ℒr(E,F),∥·∥r) is a KB-space.For a Banach latticeE and 1≤p≤∞, recall that E has the strongℓp-decomposition property if there exists a constant M such that for all disjoint elements x1,…,xn in E we have (∑i=1n‍∥xi∥p)1/p≤M∥∑i=1n‍xi∥ for p<∞ and max{∥xi∥:i=1,…,n}≤M∥∑i=1n‍xi∥ in case p=∞. The number σ(E)=inf{p≥1:E has the strong ℓp-decomposition property} is call the upper index of E.SimilarlyE has the strongℓp-composition property if there exists a constant M such that for all disjoint elements x1,…,xn in E we have ∥∑i=1n‍xi∥≤M(∑i=1n‍∥xi∥p)1/p for p<∞ and ∥∑i=1n‍xi∥≤Mmax{∥xi∥:i=1,…,n} in case p=∞. The number s(E)=sup{p≥1:E has the strong ℓp-composition property} is called lower index of E.It is known that1≤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 detailsAlso recall that if the norm on a Banach latticeE 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.
LetE and F be Banach lattices. If s(E)>σ(F) then (ℒr(E,F),∥·∥r) is a KB-space.Proof.
The norm onE′ 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.
LetE 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 (ℒr(E,F),∥·∥r)is a KB-space.Remark 16.
It is worth to point out thats(E)>σ(F) fails to be true in general even if ℒr(E,F) is a KB-space, see [7, Example 3.6].ForE and F being Lp- and Lq-spaces, respectively, we have the following characterization.Theorem 17.
LetE and F be infinite dimensional Lp-space, and Lq-space respectively, then (ℒr(E,F),∥·∥r) is a KB-space if and only if q<p.Proof.
The part of “if” is a simple consequence of Corollary15. To see the part of “only if”, we may first assume that
(13)Hn={ℓpifp<∞c0ifp=∞Gq={ℓqifq<∞c0ifq=∞
are sublattices of E and F, respectively. Suppose that p≤q then Hp⊂Gq.
Ifp<∞ there is a positive projection P from E onto Hp (the existence of P is following from Theorem 2.7.11 of [2]), then P:E→Hp⊂Gq⊂F is not M-weakly compact, which, by Theorem 4, implies that (ℒr(E,F),∥·∥r) is not a KB-space.
Also ifp=∞, then q=∞ and F is not a KB-space, Theorem 4 again yields that (ℒr(E,F),∥·∥r) is not a KB-space.The next result shows that under the regular norm the spacesℒr(E,F) are rather rare to be Lp-spaces.Theorem 18.
LetE 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 Lp-space).(b)
dimF=1 and the norm on E is q-additive (i.e., E is an Lq-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
(14)∥y1+y2∥p=∥x′⨂(y1+y2)∥p=∥x′⨂y1∥p+∥x′⨂y2∥p=∥y1∥p+∥y2∥p
which means that F is an Lp-space. A similar argument involving a fixed element of F+ and two elements of E+′ shows that E′ is an Lp-space; hence E is an Lq-space (compare with Theorem 2.7.1 of [2]), where p-1+q-1=1.
Now if bothdim(E)≥2 and dim(F)≥2 hold we will obtain a contradiction. In fact, we may assume that ℓ2q and ℓ2p are 2-dimensional sublattices of E and F, respectively; define T1,T2:ℓ2q→ℓ2p by
(15)T1(λ1,λ2)=(λ1,0),T2(λ1,λ2)=(0,λ2)for(λ1,λ2)∈ℓ2q;
then ∥T1∥=∥T2∥=1. Let P be a positive contractive projection from E onto ℓq2 (see Theorem 2.7.11 of [2]); it follows that
(16)∥T1+T2∥p=∥PT1+PT2∥p=∥PT1∥p+∥PT2∥p=2.
Also it is easy to calculate that ∥T1+T2∥=21/p-1/q; this is impossible. (1)⇒(2) holds.Theorem 19.
LetE 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 ℒr(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 ℒr(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(yn)1∞⊂F+ with ∥yn∥=1, fix x′∈E+′ with ∥x′∥=1. It is easy to verify that (x′⨂yn)1∞⊂ℒ+(E,F) is a disjoint sequence with ∥x′⨂yn∥=1 for all n∈ℕ. Corollary 2.8.12  of [2] yields that (x′⨂yn)1∞ is equivalent to the natural basis of ℓp. Note that
(17)∥∑i=1n‍λi(x′⨂yi)∥=∥x′⨂(∑i=1n‍λiyi)∥=∥∑i=1n‍λiyi∥
for all n∈ℕ and λi∈ℝ. It follows that (yn)1∞ is equivalent to the natural basis of ℓp, 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 ofF+ 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 eitherdimE<∞ 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 ℒr(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 ℓq⊂E and ℓp⊂F are sublattices, respectively. By Theorem 2.7.11 of [2] there is a positive contractive projection P from E onto ℓq. Consider the operators Tn:ℓq→ℓp by Tn(λk)=λnen, where en is the element in ℓq and ℓp with nth entry equals to 1 and all others are 0. Then it is easy to verify that (TnP)1∞ is a disjoint sequence in ℒ+(E,F) with ∥TnP∥=1. Corollary 2.8.12 of [2] yields that (TnP)1∞ is equivalent to the natural basis of ℓp. In particular, we have
(18)Bn1/p≤∥T1+T2+⋯+Tn∥=∥T1P+T2P+⋯+TnP∥≤An1/p
for all n∈ℕ, where A>0 and B>0 are constants. But
(19)∥T1+T2+⋯+Tn∥=sup{∥(λi)1n∥p:∥(λi)1n∥q≤1}
which easily shows that ∥T1+T2+⋯+Tn∥≤1 if q≤p and ∥T1+T2+⋯+Tn∥≤n1/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,…,em} with {e1,e2,…,em}⊂E+ pairwise disjoint and ∥ei∥=1. Then each T∈ℒr(E,F) corresponds to unique (x1(T),x2(T),…,xm(T)); moreover, xi(T)=Tei∈F, satisfying the following conditions.(i)
T≥0⇔xi(T)≥0 for 1≤i≤m.(ii)
xi(|T|)=|xi(T)| for all T∈ℒr(E,F) and 1≤i≤m.(iii)
xi(λT+μS)=λxi(T)+μxi(S) for all T,S∈ℒr(E,F), λ,μ∈ℝ and 1≤i≤m.(iv)
max{∥xi(T)∥:1≤i≤m}≤∥T∥r=∥|T|∥≤∑i=1m∥xi(T)∥.
Since the norm onF is equivalent to a p-additive norm, for each disjoint sequence (yk)1∞⊂F+, by Corollary 2.8.12 of [2] there exist constants A>0, B>0 such that
(*)B∥(λk∥yk∥)1n∥p≤∥∑k=1n‍λkyk∥≤A∥(λk∥yk∥)1n∥p
for all λk∈ℝ and n∈ℕ. Now for any disjoint sequence (Tn)1∞⊂ℒ+(E,F) with ∥Tn∥=1, note that the disjointness of (xi(Tn))n=1∞⊂F+ for each 1≤i≤m; it follows from (iii), (iv), and (*) that
(20)∥∑k=1n‍λkTk∥≤∑i=1m‍∥∑k=1n‍λkxi(Tk)∥≤A∑i=1m‍(∑k=1n‍|λk|p∥xi(Tk)∥p)1/p≤mA∥(λk)1n∥p
as ∥xi(Tk)∥≤∥Tk∥=1. Also
(21)∥∑k=1n‍λkTk∥≥max{∥∑k=1n‍λkxi(Tk)∥:1≤i≤m}≥Bmax{(∑k=1n‍|λk|p∥xi(Tk)∥p)1/p:1≤i≤m}≥Bm-1/p(∑i=1m‍∑k=1n‍|λk|p∥xi(Tk)∥p)1/p=Bm-1/p(∑k=1n‍|λk|p(∑i=1m‍∥xi(Tk)∥p))1/p≥Bm-1∥(λk)1n∥p
as 1≤∑=1m∥xi(Tk)∥≤(∑i=1m∥xi(Tk)∥p)1/pm1/q.
Therefore(22)Bm-1∥(λk)1n∥p≤∥∑k=1n‍λkTk∥≤mA∥(λk)1n∥p
for all λk∈ℝ and n∈ℕ; that is, (Tn)1∞ is equivalent to the natural basis of ℓp. Corollary 2.8.12 of [2] again shows that the regular norm ∥·∥r on ℒr(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.



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*Source: 183786-2013-06-09.xml*