[PDF] o?-Continuous, Lebesgue, KB, and Levi operators between vector

28530_4index.pdf arXiv:2105.01810v2 [math.FA] 15 Mar 2022 oτ-Continuous, Lebesgue, KB, and Levi operators between vector lattices and topological vector spaces

March 16, 2022

Safak Alpay, Eduard Emelyanov, Svetlana Gorokhova

Abstract

We investigateoτ-continuous/bounded/compact and Lebesgue opera- tors from vector lattices to topological vector spaces; theKantorovich- Banach operators between locally solid lattices and topological vector spaces; and the Levi operators from locally solid lattices to vector lattices. The main idea of operator versions of notions related to vector lattices lies in redistributing topological and order properties of a topological vector lattice between the domain and range of an operator under investigation. Domination properties for these classes of operators are studied. keywords:topological vector space, locally solid lattice, Banach lat- tice, order convergence, domination property, adjoint operator

MSC2020:46A40, 46B42, 47L05

1 Introduction and preliminaries

In the present paper, all vector spaces are supposed to be real, operators linear, vector topologies Hausdorff, and vector lattices Archimedean. For any vector latticeX, the Dedekind complete vec- tor lattice of all order bounded linear functionals onX is calledorder dualofXand is denoted byX≂. Recall that the order continuous partX≂nis a band ofX≂. Some vector lattices may have trivial order duals, for example X ≂={0}wheneverX=Lp[0,1] with 0< p <1 (cf. [2,

Thm.5.24]).

IfTis an operator from a vector spaceXto a vector

spaceY, thealgebraic adjointT#is an operator from the 1 algebraic dualY#toX#, defined by (T#f)(x) :=f(Tx) for allx?Xand allf?Y#. In the case whenXandY are vector lattices andTis order bounded, the restriction T ≂ofT#to the order dualY≂ofYis called theorder ad- jointofT. The operatorT≂:Y≂→X≂is not only order bounded, but even order continuous (cf. [3, Thm.1.73]). Clearly,T≂:Y≂n→X≂nwhenTis order continuous. In the case when (X,ς) and (Y,τ) are topological vec- tor spaces andTis continuous, the restrictionT?ofT#to thetopological dualY?(= the collection of allς-continuous linear functionals onY) is called thetopological adjointof T. For every locally solid lattice (X,ς), we haveX??X≂ sinceς-continuous functionals are bounded onς-bounded subsets and hence are order bounded. The topological dualX?of a locally convex-solid lattice (X,ς) is an ideal of X ≂(and henceX?is Dedekind complete) (cf. [3, Thm.3.49]). Every Fr´echet lattice (X,ς) satisfiesX?=X≂(cf. [2,

Thm.5.23]).

A net (xα)α?Ain a vector latticeXis said to be: a)order convergent(o-convergent) tox?X, if there exists a net (zβ)β?BinXsuch thatzβ↓0 and, for anyβ?B, there existsαβ?Awith|xα-x| ≤zβfor allα≥αβ. In this case, we writexαo-→x; b)uo-convergenttox?X, if|xα-x| ?uo-→0 for every u?X+.

Alocally solid lattice(X,ς) is called:

c)Lebesgue/σ-Lebesgueifxα↓0 impliesxας-→0 for ev- ery net/sequencexαinX. d)pre-Lebesgueif 0≤xn↑≤xinXimplies thatxnis aς-Cauchy sequence inX. e)Levi/σ-Leviif every increasingς-bounded net/sequen- ce inX+has a supremum inX. f)Fatou/σ-Fatouif the topologyςhas a base at zero consisting of order/σ-order closed solid sets. The assumptionxα↓0 on the netxαinc) can be replaced byxαo-→0. A Lebesgue lattice (X,ς) is Dedekind complete 2 iff order intervals ofXareς-complete [31, Prop.3.16]. Ev- ery Lebesgue lattice is pre-Lebesgue [2, Thm.3.23] and Fa- tou [2, Lem.4.2]. Furthermore, (X,ς) is pre-Lebesgue iff the topological completion (

ˆX,ˆς) of (X,ς) is Lebesgue [2,

Thm.3.26] iff (

ˆX,ˆς) is pre-Lebesgue and Fatou [2, Thm.4.8]. By [2, Thm.3.22],d) is equivalent to each of the following two conditions: d ?) if 0≤xα↑≤xholds inX, thenxαis aς-Cauchy net; d ??) every order bounded disjoint sequence inXisς- convergent to zero. The next well known fact follows directly fromd?). Proposition 1.1.Everyς-complete pre-Lebesgue lattice (X,ς)is Dedekind complete. A normed lattice (X,?·?) is calledKantorovich-Banach orKB-spaceif every norm bounded upward directed set in X +converges in the norm. EachKB-space is a Levi lattice with order continuous complete norm; each order contin- uous Levi normed lattice is Fatou; and each Levi normed lattice is Dedekind complete. Lattice-normed versions of

KB-spaces were recently studied in [9, 10, 12].

g) We call a locally solid lattice (X,ς) by aKB/σ-KB latticeif every increasingς-bounded net/sequence in X +isς-convergent. Clearly, eachKB/σ-KBlattice is Levi/σ-Levi and each Levi/σ-Levi lattice is Dedekind complete/σ-complete.

Recall that a continuous operatorT:

h) between two Banach spaces is said to beDunford-

PettisifTtakes weakly null sequences to norm null

sequences. It is well known that every weakly compact operator onL1(μ) is Dunford-Pettis and that an op- erator is Dunford-Pettis iff it takes weakly Cauchy se- quences to norm convergent sequences [3, Thm.5.79]. i) from a Banach latticeXto a Banach spaceYis calledM-weakly compactif?Txn?Y→0 holds for every norm bounded disjoint sequencexninX. 3 j) from a Banach spaceYto a Banach latticeXis called L-weakly compactwhenever?xn?X→0 holds for ev- ery disjoint sequencexnin the solid hull ofT(UY), whereUYis the closed unit ball ofY. An operatorTfrom a vector latticeXto a topological vector space (Y,τ) is called k)oτ-continuous/σoτ-continuousifTxατ-→0 for every net/sequencexαsuch thatxαo-→0 [29]. Replacement ofo-null nets/sequences byuo-null ones above gives the definitions ofuoτ-continuous/σuoτ-continuousop- erators. L-/M-weakly compact operators are weakly compact (cf. [3, Thm.5.61]) and the norm limit of a sequence ofL-/M- weakly compact operators is againL-/M-weakly compact (cf. [3, Thm.5.65]). For further unexplained terminology and notions, we refer to [2, 3, 4, 5, 27, 30, 35, 36]. Various versions of Banach lattice properties like a prop- erty to be aKB-space were investigated recently (see, e.g., [1, 6, 7, 8, 9, 10, 13, 14, 15, 18, 19, 21, 22, 23, 24, 25, 27, 29,

31, 32, 33]). In the present paper we continue the study of

operator versions of several topological/order properties, focusing on locally solid lattices. The main idea behind operator versions consists in a redistribution of topolog- ical and order properties of a topological vector lattice between the domain and range of the operator under in- vestigation (like in the case of Dunford-Pettis andL-/M- weakly compact operators). As the order convergence is not topological in general [14, 28], the most important op- erator versions emerge when botho- andς-convergences are involved simultaneously. Definition 1.1.LetTbe an operator from a vector lattice Xto a topological vector space (Y,τ). We say that: (a)Tisτ-Lebesgue/στ-LebesgueifTxατ-→0 for every net/ sequencexαsuch thatxα↓0;Tisquasiτ- Lebesgue/στ-LebesgueifTxαisτ-Cauchy for every net/sequencexαinX+satisfyingxα↑≤x?X. If there is no confusion with the choice of the topology τonY, we callτ-Lebesgue operators byLebesgueetc. 4 (b)Tisoτ-bounded/oτ-compactifT[0,x] is aτ-bounded/

τ-totally bounded subset ofYfor eachx?X+.

If additionallyX= (X,ς) is a locally solid lattice, (c)TisKB/σ-KBif, for everyς-bounded increasing net/sequencexαinX+, there exists (not necessarily unique)x?Xsuch thatTxατ-→Tx. (d)TisquasiKB/quasiσ-KBifTtakesς-bounded increasing nets/sequences inX+toτ-Cauchy nets. IfXandYare vector lattices with (X,ς) locally solid, (e)TisLevi/σ-Leviif, for everyς-bounded increasing net/sequencexαinX+, there exists (not necessarily unique)x?Xsuch thatTxαo-→Tx. (f)Tisquasi Levi/quasiσ-LeviifTtakesς-bounded increasing nets/sequences inX+too-Cauchy nets. Replacement of decreasingo-null nets/sequences byuo- null ones in (a) ando-convergent (o-Cauchy) nets/sequen- ces by (uo-Cauchy)uo-convergent ones in (e) and (f) above gives the definitions ofuoτ-continuousand of (quasi) uo-Levi operatorsrespectively.

In our approach, we focus on:

?modification of nets/sets of operators domains in (a), (b), (d), (e), and (f); ??information which operators provide about conver- gences in their domains/ranges in (c), (e), and (f). Remark 1.1.a) The identity operator in a locally solid lattice (Y,τ) is Lebesgue/KB/Levi iff (Y,τ) is Lebes- gue/KB/Levi. This motivates the terminology. Cle- arly, everyoτ-continuous/σoτ-continuous operator is τ-Lebesgue/στ-Lebesgue. By Lemma 2.1, any posi- tive operator to a locally solid lattice is Lebesgue/σ- Lebesgue iffTisoτ-continuous/σoτ-continuous; and every positive Lebesgue operator to a locally solid lat- tice is quasi Lebesgue by Theorem 2.2. b) Each regular operator from a vector latticeXto a lo- cally solid lattice (Y,τ) isoτ-bounded by Proposition

2.1. In the case of normed range spaceY,oτ-bounded

5 operators are also known asinterval-bounded(cf. [30, Def.3.4.1]). Like in [30, Lem.3.4.2], eachoτ-bounded operatorTfrom a vector latticeXto a topological vector space (Y,τ) possessesadjointT◦:Y?→X≂ given byT◦y?=y?◦Tfor ally??Y?. c) In the case whenXis also a topological vector lat- tice, theτ-continuity of operatorTis not assumed in (b) of Definition 1.1. For example, the rank one dis- continuous operatorTx:= (?∞k=1xk)e1in (c00,?·?) is oτ-compact andoτ-continuous yet not compact. Each continuous operatorTfrom a discrete Dedekind com- plete locally convex Lebesgue lattice to a topological vector space isoτ-compact by the Kawai theorem (cf. [2, Cor.6.57)]. Every Dunford-Pettis operator from a Banach lattice to a Banach space iso-weakly-compact (cf. [3, Thm 5.91]). d) EachKB-operator is quasiKB; each quasi (σ-)KB- operator is quasi (σ-) Lebesgue; and each continu- ous operator from aKB-space to a topological vector space isKB. It is well known that the identity op- eratorIin a Banach lattice isKBiffIisσ-KBiff

Iis quasiKB. Proposition 1.2 shows that the no-

tions of quasiKBand quasiσ-KBoperator coincide.

The most important reason for using the termKB-

operatorfor (c) of Definition 1.1 is existence of limits ofς-bounded increasing nets. Some authors (see, e.g., [8, 13, 33]) use the term "KB-operator" for (d) of Definition 1.1 which is slightly confusing because it says nothing about existence of limits of topologically bounded increasing nets and all continuous finite-rank operators in every Banach lattice satisfy (d). Each or- der bounded operator from a Banach lattice to aKB- space is quasiKB. However, if we takeX= (c00,?·?l) andY= (c00,?·?p) withl,p?[1,∞], the identity op- eratorIis quasiKBiffl≤p <∞. e) It is clear that every compact operatorTfrom a Ba- nach latticeXto a Banach spaceYisoτ-compact.

However, compact operator need not to be Lebesgue

(cf. Example 3.1). In particular,oτ-compact opera- 6 tors are not necessarilyoτ-continuous. Example 1.1.Let(c(R),?·?∞)be the Banach lattice of all R-valued functions onRsuch that for everyf?c(R)there existsaf?Rfor which the set{r?R:|f(r)-af| ≥ε} is finite for eachε >0. Then the identity operatorIin c(R)isσoτ-continuous and quasiσ-Lebesgue yet neither

Lebesgue nor quasi Lebesgue.

Example 1.2.Recall that, for a nonempty setH, the vec- tor spacec00(H)of all finitely supportedR-valued functions onHis a Dedekind complete vector lattice. Furthermore, any vector spaceXis linearly isomorphic toc00(H), where His a Hamel basis forX. As each order interval ofc00(H) lies in a finite-dimensional subspace ofc00(H), every op- erator fromc00(H)to any topological vector space(Y,τ)is oτ-continuous,oτ-bounded, andoτ-compact. On the other hand, the identity operator in(c00,? ·?∞)is neither quasi

σ-KBnor quasiσ-Levi.

Remark 1.2.It is well known that, if a netyαof a topo- logical vector space (Y,τ) is notτ-Cauchy, then there ex- istU?τ(0) and an increasing sequenceαnsuch that y αn+1-yαn??Ufor eachn(see, e.g., [2, Lem.2.5]). Proposition 1.2.An operatorTfrom a locally solid lat- tice(X,ς)to a topological vector space(Y,τ)is quasiKB iffTis quasiσ-KB. Proof.The necessity is trivial. For the sufficiency, sup- poseTis not quasiKB. Then there exists aς-bounded increasing netxαinX+such thatTxαis notτ-Cauchy in Y. It follows from Remark 1.2 that for some increasing sequenceαnand someU?τ(0) Tx

αn+1-Txαn??U(?n?N) (1)

Since the sequencexαnis increasing andς-bounded, the condition (1) implies thatTis not quasiσ-KB. Corollary 1.1.Everyς-completeσ-KBlattice(X,ς)is aKBlattice.

Proof.The identity operatorIonXisσ-KBand hence

quasiσ-KB. By Proposition 1.2,Iis quasiKB. Then 7 everyς-bounded increasing net inX+isς-Cauchy, and hence isς-convergent. Remark 1.3.The following fact (cf., e.g., [11, Prop.1.1]) is well known: a)ρ(yα,y)→0 in a metric space (Y,ρ) iff, for every subnetyαβof the netyα, there exists a (not necessary increasing) sequenceβkof indices withρ(yαβk,y)→0. Application ofa) to Cauchy nets/sequences in (Y,ρ) gives that: b) a net/sequenceyαof a metric space (Y,ρ) is Cauchy iff, for every subnet/subsequenceyαβofyα, there ex- ists a sequenceβkof indices such that the sequence y

αβkisρ-Cauchy.

Application ofb) and Proposition 1.2 gives that, for an operatorTfrom a locally solid lattice (X,ς) to a metric vector space (Y,ρ), the following are equivalent: (i)Tis quasiKB; (ii) Everyς-bounded increasing sequencexninX+has a subsequencexnksuch thatTxnkisρ-Cauchy inY. Another application ofb) to an operatorTfrom a locally solid lattice (X,ς) to a metric vector space (Y,ρ) gives the equivalence of the following conditions: (i)?TisKB; (ii)?For anyς-bounded increasing netxαinX+, there exist an elementx?Xand a sequenceβkof indices such thatρ(Txαβk,Tx)→0. In many cases, like for (quasi)KBor Levi operators, the lattice structure in the domain/range of operator can be relaxed to the ordered space structure [8, 12, 20, 23], or combined with the lattice-norm structure [1, 9, 10, 12]. Such generalizations are not included in the present pa- per. In Section 2 we investigate operators whose domains and/or ranges are locally solid lattices. Section 3 is de- voted to operators between Banach lattices. 8

2 Operators between locally solid latticesIn this section, we study mostly operators whose domainsand/or ranges are locally solid lattices. Observe first thatthe setsLLeb(X,Y),Loτ(X,Y),Loτb(X,Y), andLoτc(X,Y)

of Lebesgue,oτ-continuous,oτ-bounded, andoτ-compact operators respectively from a vector latticeXto a topo- logical vector space (Y,τ) are vector spaces and: (?)Loτ(X,Y)?LLeb(X,Y); (??)Loτc(X,Y)?Loτb(X,Y) since every totally bounded subset inYis bounded.

Theorem 2.1.LetXbe a vector lattice and(Y,τ)be

a locally convex Lebesgue lattice which is either Dedekind complete orτ-complete. ThenLoτc(X,Y)?Lr(X,Y)is a band of the latticeLr(X,Y)of all regular operators from XtoY. Proof.Since every Lebesgue lattice is pre-Lebesgue by [2, Thm.3.23], and every topologically complete pre-Lebesgue lattice is Dedekind complete by Proposition 1.1, the lat- tice (Y,τ) is Dedekind complete in either case. By [3,

Thm.5.10], for eachx?X+, the set

C(x) ={T?Lb(X,Y) :T[0,x] isτ-totally bounded}

is a band of the Dedekind complete latticeLb(X,Y) of all order bounded operators fromXtoY. SinceLr(X,Y) = L b(X,Y), the setLoτc(X,Y)?Lr(X,Y) =? x?X+C(x) is also a band ofLr(X,Y) as desired.

The following two propositions might be known. We

include their elementary proofs as we did not find appro- priate references. Proposition 2.1.Each regular operatorTfrom a vector latticeXto a locally solid lattice(Y,τ)isoτ-bounded. Proof.Without lost of generality, assumeT≥0. Letx? X +,U?τ(0), andVbe a solidτ-neighborhood of zero ofYwithV?U. ThenTx?λVfor allλ≥λ0for someλ0>0 and henceT[0,x]?[0,Tx]?λV?λU 9 for allλ≥λ0. SinceU?τ(0) was arbitrary,Tisoτ- bounded. Proposition 2.2.An order continuous positive operator Tfrom a vector latticeXto a locally convex-solid lattice (Y,τ)is Lebesgue iffTis weakly Lebesgue. Proof.The necessity is trivial. For the sufficiency, assume thatTis weakly Lebesgue (i.e.,T:X→Yis Lebesgue with respect to the weak topologyσ(Y,Y?) onY). Let x α↓0 inX. ThenTxα↓0 inY. SinceTisσ(Y,Y?)- Lebesgue,Txασ(Y,Y?)----→0. By using Dini-type arguments (cf. [3, Thm.3.52]), we concludeTxατ-→0, as desired. We do not know any example of an order continuousoτ- bounded weakly Lebesgue operator from a vector lattice to a locally solid lattice that is not Lebesgue. Proposition 2.3.A weakly continuous positive operator Tfrom a Lebesgue(σ-Lebesgue)lattice(X,ς)to a locally solid lattice(Y,τ)is Lebesgue(σ-Lebesgue). Proof.As arguments are similar, we restrict ourselves to the Lebesgue case. Letxα↓0 inX. Since (X,ς) is Lebesgue,xας-→0 and hencexασ(X,X?)----→0. The weak con- tinuity ofTimpliesTxασ(Y,Y?)----→0. SinceTxα↓0 inY, it followsTxατ-→0, as in the proof of Proposition 2.2.

Theorem 2.2.Each positive Lebesgue operatorTfrom

a vector latticeXto a locally solid lattice(Y,τ)is quasi

Lebesgue.

Proof.Letxα↑≤x?X, souα:=x-xα↓≥0. Therefore (uα-z)α;z?L↓0, whereL={z?X: (?α?A)[z≤uα]} is the upward directed set of lower bounds of the netuα (cf. [17, Thm.1.2]). SinceTis Lebesgue andT≥0, (Tuα-Tz)α;z?Lτ-→0 & (Tuα-Tz)α;z?L↓≥0.(2) It follows from (2), that (Tuα-Tz)α;z?L↓0. ThenTuα and henceTxαisτ-Cauchy, as desired. 10 We do not know any example of anoτ-bounded Lebesgue operator which is not quasi Lebesgue. In contrast to The- orem 3.24 of [2], the converse of Theorem 2.2 is false even ifY=Rdue to Example 3.1. Recall that an Archimedean vector latticeXis called laterally(σ-)completewhenever every (countable) subset of pairwise disjoint vectors ofX+has a supremum. Every laterally complete vector latticeXcontains a weak order unit and every band of such anXis a principal band (cf. [2, Thm.7.2]). Furthermore, by the Veksler - Geiler theorem (cf. [2, Thm.7.4]), if a vector latticeXis laterally (σ-) complete, thenXsatisfies the (principal projection) projection property. A vector lattice that is both laterally and Dedekind (σ- ) complete is referred to asuniversally(σ-)complete. It follows from [2, Thm.7.4] that a vector latticeXis univer- sally complete iffXis Dedekindσ-complete and laterally complete iffXis uniformly complete and laterally com- plete (cf. [2, Thm.7.5]). Similarly, a laterallyσ-complete vector latticeXis Dedekindσ-complete iffXis uniformly complete. A universal completion of a vector latticeXis a later- ally and Dedekind complete vector latticeXuwhich con- tainsXas an order dense sublattice. Every vector lattice has unique universal completion (cf. [2, Thm.7.21]). A laterally complete vector latticeXis discrete iffX is lattice isomorphic toRSfor some nonempty setSiffX admits a Hausdorff locally convex-solid Lebesgue topol- ogy iff the spaceX≂nof order continuous functionals on XseparatesX(cf. [2, Thm.7.48]); in each of these cases X=Xu. Definition 2.1.A topological vector lattice (Y,τ) is called τ-laterally(σ-)completewhenever everyτ-bounded (count- able) subset of pairwise disjoint vectors ofX+has a supre- mum. Every laterally (σ-) complete topological vector lattice (Y,τ) isτ-laterally (σ-) complete. Every Dedekind com- pleteAM-spaceXwith an order unit isτ-laterally com- plete with respect to the norm. 11 Example 2.1.LetXbe a vector lattice of real functions onRsuch that eachf?Xmay differ from a constant say a fon a countable subset ofRandf-afIR??1(R) for eachf?X. (i) The vector latticeXis not Dedekindσ-complete, as f n:=IR\{1,2,...,n}↓≥0 yet infn?Nfndoes not exist inX. (ii) The vector latticeXis a Banach space with respect to the norm?f?:=|af|+?f-afIR?1. (iii) The vector latticeXis notτ-laterallyσ-complete with respect to the norm topology onX. Indeed, the norm bounded countable set of pairwise disjoint orths e n=I{n}ofX+has no supremum inX. (iv) The identity operatorIonXis notσ-Lebesgue, as 1 nIR\{1,2,...,n}↓0 inXyet?1nIR\{1,2,...,n}?=n+1n?→0. Proposition 2.4.Any Levi(σ-Levi)lattice isτ-laterally (σ-)complete and Dedekind(σ-)complete. Proof.We consider the Levi case only, because theσ-Levi case is similar. The Dedekind completeness ofXis trivial. LetDbe aτ-bounded subset of pairwise disjoint positive vectors of a Levi lattice (Y,τ). Then the collectionD? of suprema of finite subsets ofDforms an increasingτ- bounded net indexed by the setPfin(D) of all finite subsets ofDdirected by inclusion. Since (Y,τ) is Levi,D?↑d for somed?Y. It follows from supD= supD?=dthat (Y,τ) isτ-laterally complete. Example 2.2.The locally solid lattice(RS,τ), whereτis the product topology on the vector latticeRSof real func- tions on a setSis locally convex, Lebesgue, Levi, and universally complete. Proposition 2.5.Each regular operatorTfrom a later- allyσ-complete vector latticeXto aσ-Lebesgue lattice (Y,τ)isσ-Lebesgue. Proof.Without lost of generality, we can supposeT≥0. Letxn↓0 inX. By Theorem 7.8 of [2],Tisσ-order continuous, and henceTxn↓0 inY. Since (Y,τ) isσ-

Lebesgue,Txnτ-→0.

12 We do not know whether or not every positive operatorT from a laterally complete vector latticeXto a Lebesgue lattice (Y,τ) is Lebesgue. Each laterallyσ-complete locally solid lattice isσ-Le- besgue and pre-Lebesgue [2, Thm.7.49]. Proposition 2.6.LetTbe a continuous operator from a laterallyσ-complete locally solid lattice(X,ς)to a topolog- ical vector space(Y,τ). ThenTisσ-Lebesgue in each of the following cases: (i)Xis discrete. (ii) (X,ς)is metrizable. Proof.Letxn↓0 inX. In case (i), (X,ς) isσ-Lebesgue by [2, Thm.7.49]. The result follows becauseTisςτ- continuous. In case (ii), the result follows from (i) as every infinite dimensional laterallyσ-complete metrizable locally solid lattice is lattice isomorphic toRN. Since (X,uς) is pre-Lebesgue iff (X,ς) is pre-Lebesgue [32, Cor.4.3], the statement of Proposition 2.6 is also true when (X,ς) is replaced by (X,uς). Remark 2.1.LetTbe a positive operator from a locally solid lattice (X,ς) to a laterallyσ-complete locally solid lattice (Y,τ) and let 0≤xα↑≤xinX. Since (Y,τ) is pre-Lebesgue by [2, 7.49], it follows from 0≤Txα↑≤Tx thatTxαisτ-Cauchy inYand henceTis quasi Lebesgue. Remark 2.2.LetTbe a continuous operator from a lat- erallyσ-complete locally solid lattice (X,ς) to a locally solid lattice (Y,τ). a) SupposeT≥0 and 0≤xα↑≤xinX. Since (X,ς) is pre-Lebesgue by [2, 7.49] thenxαisς-Cauchy inX and henceTxαisτ-Cauchy inYmeaning thatTis quasi Lebesgue. b) Suppose (X,ς) is Fatou, andxα↓0 inX. Then by [2,

7.50] (X,ς) is Lebesgue, soxας-→0 and henceTxατ-→0

meaning thatTis Lebesgue. c) Supposeςis a metrizable locally solid topology and x α↓0 inX. Then (X,ς) is Lebesgue by [2, Thm.7.55]. 13 Thusxας-→0 and henceTxατ-→0 meaning thatTis

Lebesgue.

d) If (X,ς) is a Fr´echet lattice thenςis uniquely defined and every regularT: (X,ς)→(Y,τ) is continuous (cf. [2, Thm.5.19 and Thm.5.21]). In particular, every regular operator from a laterallyσ-complete Fr´echet latticeX(ς) to a locally solid lattice (Y,τ) is Lebesgue. Proposition 2.7.Any continuous operatorTfrom a lo- cally solid lattice(X,ς)to a topological vector space(Y,τ) isoτ-bounded. Proof.Letx?X+,U?τ(0), andVbe a solidς-neighbor- hood of zero ofXwithV?T-1U. Then there exists λ

0>0 such thatx?λVfor allλ≥λ0and hence [0,x]?

λV?λT-1(U) for allλ≥λ0. ThusT[0,x]?λUfor all λ≥λ0, which shows thatTisoτ-bounded. It is worth mentioning here that anoτ-bounded operator Tfrom Dedekindσ-complete vector latticeXto a normed latticeYisσ-Lebesgue iffTis order-weakly compact (cf. [29, Cor.1]). The following lemma can be considered as an extension of [29, Lem.1] to locally solid lattices. Lemma 2.1.A positive operatorTfrom a vector lattice Xto a locally solid lattice(Y,τ)is Lebesgue/σ-Lebesgue iffTisoτ-continuous/σoτ-continuous. Proof.As theσ-Lebesgue case is similar, we consider only Lebesgue operators. The sufficiency is routine. For the necessity, assumeTis Lebesgue andxαo-→0 inX. Take a netzβinXwithzβ↓0 such that, for eachβ, there existsαβwith|xα| ≤zβforα≥αβ. It follows fromT≥0 that|Txα| ≤T|xα| ≤Tzβforα≥αβ. AsTis Lebesgue, Tz βτ-→0, which impliesTxατ-→0, because the topologyτ is locally solid. Corollary 2.1.An order bounded operatorTfrom a vec- tor lattice to a Dedekind complete locally solid lattice is Lebesgue/σ-Lebesgue iffTisoτ-continuous/σoτ-continu- ous. 14 Now we investigate some properties of adjoint opera- tors. Recall that, for a locally solid lattice (X,ς), the absolute weak topology|σ|(X≂,X) onX≂is the locally convex-solid topology generated by the collection of Riesz seminorms{ρx(f) =|f|(|x|) :x?X}. The locally solid lattice (X≂,|σ|(X≂,X)) is Lebesgue, Levi, and Fatou [26, Prop.81C]. We begin with the following technical lemma. Lemma 2.2.LetT: (X,ς)→(Y,τ)be an order bounded continuous operator from a locally solid lattice(X,ς)to a Dedekind complete locally solid lattice(Y,τ). Then the topological adjointT?: (Y?,|σ|(Y≂,Y))→(X?,|σ|(X≂,X)) is also continuous. Proof.Indeed, forfα|σ|(Y≂,Y)------→0 inY≂, it follows from ρ x(T?fα) =?|T?fα|,|x|? ≤ ?|T?||fα|,|x|? ≤ ?|T|?|fα|,|x|?= ?|fα|,|T||x|?=|fα|(|T||x|) =ρ|T||x|(fα)→0 (3) thatT?fα|σ|(X≂,X)------→0. The second inequality in (3) fol- lows from the existence of the modulus ofTand from the observation±T≤ |T| ? ±T?≤ |T|?(cf. [3, p.67]) which implies|T?| ≤ |T|?. Theorem 2.3.LetT:X→Ybe an order bounded oper- ator from a vector latticeXto a Dedekind complete vector latticeY, andT?: (Y≂,|σ|(Y≂,Y))→(X≂,|σ|(X≂,X)) the correspondent topological adjoint ofT. ThenT?isoτ- bounded,oτ-continuous, Levi, andKB. Proof.Theoτ-boundedness ofT?follows from Lemma 2.2 by Proposition 2.7. SinceTis regular, without lost of generality, we can supposeT≥0 (and henceT?≥0). Letfαo-→0 inY≂. AsT?is order continuous,T?fαo-→0 inX≂, and since (X≂,|σ|(X≂,X)) is a Lebesgue lattice, we obtainT?fα|σ|(X≂,X)------→0 and henceT?isoτ-continuous. Letfαbe a positive|σ|(Y≂,Y)-bounded increasing net inY≂. Since (Y≂,|σ|(Y≂,Y)) is Levi,fα↑ffor some f?Y≂. AsT?is order continuous,T?fα↑T?finX≂, in particularT?is Levi. Since the locally solid lattice (X≂,|σ|(X≂,X)) is Lebesgue,T?fα|σ|(X≂,X)------→T?fand 15 henceT?: (Y≂,|σ|(Y≂,Y))→(X≂,|σ|(X≂,X)) isKB.

The natural embeddingJof a vector latticeXto (X≂n)≂nis defined by (Jx)(f) =f(x) for allf?X≂n,x?X. The

mappingJis an order continuous lattice homomorphism (cf. e.g. [3, Thm.1.70]). By the Nakano theorem (cf. [3, Thm.1.71]),Jis one-to-one and onto (i.e.Xisperfect) iffX≂nseparatesXand whenever a netxαofX+satisfies x α↑and supf(xα)<∞for each 0≤f?X≂n, then there exists somex?Xsatisfyingxα↑xinX.

Lemma 2.3.LetTbe an order bounded operator from a

vector latticeXto a vector latticeY. The second order adjointT≂≂, while restricted to(X≂n)≂,

((X≂n)≂,|σ|((X≂n)≂,X≂n))T≂≂--→((Y≂)≂,|σ|((Y≂n)≂,Y≂n))

is continuous.

Proof.Let (X≂n)≂?xα|σ|((X≂n)≂,X≂n)---------→0. By use of the fact

|T≂≂| ≤ |T≂|≂it follows ?|T≂≂xα|,|y|? ≤ ?|T≂|≂|xα|,|y|? ≤ ?|xα|,|T≂||y|?(4) for ally?Y≂n. Sincexα|σ|((X≂n)≂,X≂n)---------→0, it follows from (4)

thatT≂≂xα|σ|((Y≂n)≂,Y≂n)--------→0 and hence the operatorT≂≂is

|σ|((X≂n)≂,X≂n) to|σ|((Y≂n)≂,Y≂n) continuous.

Theorem 2.4.LetTbe an order bounded operator from

a vector latticeXto a vector latticeY. Then

((X≂n)≂n,|σ|((X≂n)≂n,X≂n))T≂≂--→((Y≂n)≂n,|σ|((Y≂n)≂n,Y≂n)).

is a continuous, Lebesgue, Levi, andKBoperator. Proof.The continuity follows from Lemma 2.3 by restrict- ing of the second order adjointT≂≂to (X≂n)≂n. Letxα↓0 in (X≂n)≂n. AsT≂≂is order continuous, T

≂≂xα↓0 in (Y≂n)≂n. Since ((Y≂n)≂n,|σ|((Y≂n)≂n,Y≂n)) is

Lebesgue,T≂≂xα|σ|((Y≂n)≂n,Y≂n)--------→0 showing that the opera- tor is Lebesgue. 16 Letxαbe a positive increasing|σ|((X≂n)≂n,X≂n)-bounded net in (X≂n)≂n. Since ((X≂n)≂n,|σ|((X≂n)≂n,X≂n)) is Levi, the netxαhas a supremum in (X≂n)≂n, sayx. Thusxα↑x andT≂≂xα↑T≂≂xasT≂≂is order continuous. Thus the operator is Levi.

Since ((X≂n)≂n,|σ|((X≂n)≂n,X≂n)) is Levi,xα↑xin (X≂n)≂n.

ThenT≂≂xαo-→T≂≂xby order continuity of the operator T ≂≂. As ((Y≂n)≂n,|σ|((Y≂n)≂n,Y≂n)) is a Lebesgue lattice, T

≂≂xα|σ|((Y≂n)≂n,Y≂n)--------→T≂≂xand the operator isKB.

A vector sublatticeZof a vector latticeXis called

regular, if the embedding ofZintoXpreserves arbitrary suprema and infima. Order ideals and order dense vector sublattices of a vector latticeXare regular [2, Thm.1.23]. Corollary 2.2.LetX≂=X≂n, andYbe a vector lattice. Then each order bounded operatorT:X→(Y,|σ|(Y,Y≂n)) is Lebesgue. Proof.Letxα↓0 inX. Since the natural embeddingJ: X→X≂≂is one-to-one andJ(X) is a regular sublattice ofX≂≂(cf. [2, Thm.1.67]), thenT=T≂≂◦JandJxα↓0

inX≂≂. ThusTxα=T≂≂(Jxα)|σ|((X≂n)≂n,X≂n)---------→0 sinceT≂≂

is Lebesgue by Theorem 2.4. We also mention the following question. LetTbe Lebes- gue,oτ-continuous,oτ-bounded,oτ-compact,KB, or Levi. DoesT≂≂satisfy the same property? The answer is nega- tive, whenTmaps a Banach latticeXto a Banach lattice

Y, in the following cases:

(?) for Lebesgue and foroτ-continuous operators, since the identityI:c0→c0isoτ-continuous yet its second order adjointI:?∞→?∞is not even Lebesgue; (??) foroτ-compact operators, since the natural embed- dingJ:c0→?∞isoτ-compact butJ≂≂:?∞→???∞is notoτ-compact. Next we discuss the domination properties of positive operators. LetTandSbe positive operators between vector latticesXandYsatisfying 0≤S≤T. 17 Question 2.1.When does the assumption thatTis Lebes- gue,oτ-continuous,oτ-bounded,oτ-compact,KB, or Levi imply thatShas the same property? Question 2.1 has positive answers in the following cases: < a >Trivially, for Lebesgue;σ-Lebesgue; andoτ-bounded operators from a vector lattice to a locally solid lattice. < b >Foroτ-compact operators from a vector latticeXto a locally convex Lebesgue lattice (Y,τ) which is either

Dedekind complete orτ-complete by Theorem 2.1.

< c >Foro|σ|(Y,Y?)-compact operators from a Banach lat- ticeXto a Banach latticeY[3, Thm.5.11]. < d >ForKB/σ-KBlattice homomorphisms between lo- cally solid lattices by Corollary 2.3. < e >For quasiKBoperators between locally solid lattices by Theorem 2.6. < f >For quasi Levi operators from a locally solid lattice to a vector lattice by Theorem 2.7.

We remind that the modulus|T|of an order bounded

disjointness preserving operatorTbetween vector lattices XandYexists and satisfies|T||x|=|T|x||=|Tx|for all x?X[3, Thm.2.40]; there exist lattice homomorphisms R

1,R2:X→YwithT=R1-R2(cf. [3, Exer.1, p.130].

Theorem 2.5.LetTbe an order bounded disjointness

preservingKB/σ-KBoperator between locally solid lat- tices(X,ς)and(Y,τ). If|S| ≤ |T|thenSisKB/σ-KB. Proof.Take aς-bounded increasing net/sequencexαin X +. ThenT(xα-x)τ-→0 for somex?X. It follows from |S(xα-x)| ≤ |S||xα-x| ≤ |T||xα-x|=|Txα-Tx|τ-→0 thatSxατ-→Sx, as desired. Since 0≤S≤Twith a lattice homomorphismTim- plies thatSis also a lattice homomorphism, the next result is a direct consequence of Theorem 2.5.

Corollary 2.3.LetTbe aKB/σ-KBlattice homomor-

phism between locally solid lattices(X,ς)and(Y,τ). Then 18 eachSsatisfying0≤S≤Tis also aKB/σ-KBlattice homomorphism.

Theorem 2.6.LetTbe a positive quasiKBoperator

between locally solid lattices(X,ς)and(Y,τ). Then each operatorS:X→Ywith0≤S≤Tis also quasiKB. Proof.Let 0≤S≤Tandxαbe an increasingς-bounded net inX+. SinceT≥0 thenTxα↑. SinceTis quasi KBthenTxαisτ-Cauchy. Take aU?τ(0) and a solid neighborhoodV?τ(0) withV-V?U. There existsα0 satisfyingT(xα-xβ)?Vfor allα,β≥α0. In particular, T(xα-xα0)?Vfor allα≥α0. Since 0≤S≤TandV is solid,S(xα-xα0)?Vfor allα≥α0. Thus, we obtain Sx α-Sxβ=S(xα-xα0)-S(xβ-xα0)?V-V?U(5) for allα,β≥α0. SinceU?τ(0) was taken arbitrary, (5) implies thatSxαis alsoτ-Cauchy.

Theorem 2.7.LetTbe a positive quasi Levi operator

from a locally solid lattice(X,ς)to a vector latticeY. Then each operatorS:X→Ywith0≤S≤Tis also quasi Levi. Proof.LetT:X→Ybe quasi Levi andxαbe a positive ς-bounded increasing net in (X,ς). ThenTxαiso-Cauchy inY. So, there exists a netzβ↓inYsuch that, for each β, there existsαβsuch that|Txα1-Txα2| ≤zβfor all α

1,α2≥αβ. Choosingα1,α2≥αβfor a fixedαβwe have

Sx α1-Sxα2≤S(xα1-xαβ)≤T(xα1-xαβ)≤zβ, and similarly Sx α2-Sxα1≤S(xα2-xαβ)≤T(xα2-xαβ)≤zβ. Thus|Sxα1-Sxα2| ≤zβfor allα1,α2≥αβshowing that the operatorSis also quasi Levi. Any order dense sublattice is regular by [2, Thm.1.27]. Also a locally solid lattice (X,ς) is a regular sublattice of itsς-completionˆXiff everyς-Cauchy netxαinX+such thatxαo-→0 inXsatisfiesxας-→0 [2, Thm.2.41]. 19

Lemma 2.4.LetZbe a regular sublattice of a vector

latticeXandTbe anoτ-continuous operator fromX to a topological vector space(Y,τ). Then the restriction

T|Z:Z→Yis alsooτ-continuous.

Proof.Under the assumptions of the lemma, for each net z αinZ,zαo-→0 inXifzαo-→0 inZ, by [27, Lem.2.5]. Therefore the result follows directly from the definition of anoτ-continuous operator. Similarly foruo-null nets. LetZbe a regular sublattice of Xthenxαuo-→0 inZiffxαuo-→0 inXby [27, Thm.3.2], and hence, for auoτ-continuous operatorTfrom a vector lat- ticeXto a topological vector space (Y,τ), the restriction

T|Z:Z→Yis alsouoτ-continuous.

Lemma 2.5.LetTbe a positive Lebesgue operator from a vector latticeXto a locally solid lattice(Y,τ). IfZis a regular sublattice ofX, then the restrictionT|Z:Z→Y isoτ-continuous. Proof.Tisoτ-continuous by Lemma 2.1. It follows from

Lemma 2.4 thatT|Z:Z→Yisoτ-continuous.

Since a vector latticeXis order dense inXδ, and hence is regular inXδ, the next result follows immediately from

Lemma 2.5.

Proposition 2.8.If an operatorTfrom the Dedekind

completionXδof a vector latticeXto a locally solid lattice (Y,τ)is Lebesgue then the operatorT|Xisoτ-continuous.

Each order continuous lattice homomorphismTfrom

a Fatou lattice (X,ς) to a Lebesgue lattice (Y,τ) isςτ- continuous on order intervals ofXby [2, Thm.4.23]. Since Tis also Lebesgue under the given conditions, the follow- ing can be considered as a generalization of [2, Thm.4.23]. Theorem 2.8.Let(X,ς)be a Fatou lattice,(Y,τ)a Lebes- gue lattice, andT:X→Ya Lebesgue lattice homomor- phism. ThenTisςτ-continuous when restricted to order bounded subsets inX. 20 Proof.LetNbe a base of solid neighborhoods forτ(0). SinceTis a Lebesgue lattice homomorphism, the family {T-1(V)}V?Nis a base of solid neighborhoods of zero for a Lebesgue (not necessary Hausdorff) topologyζonX. Sinceςis Fatou andζis Lebesgue, then on any order bounded subset ofXthe topology induced byςis finer than the topology induced byζdue to Theorem 4.21 of [2]. Clearly,T:X→Yisζτ-continuous. Thus,Tis also

ςτ-continuous on order bounded subsets ofX.

Recall that, for a locally solid lattice (X,ς), a vector ˆv?ˆX+is called anupper elementofXwhenever there exists an increasing netuαinX+such thatuαˆς→ˆv(cf. [2, Def.5.1]). In a metrizable locally solid lattice (X,ς) upper elements can be described by sequences: a vector ˆv?ˆX+ is an upper element ofXiff there exists an increasing sequenceuninX+such thatunˆς→ˆv(cf. [2, Thm.5.2]). For a metrizableσ-Lebesgue lattice (X,ς) its topological completion ( ˆX,ˆς) is alsoσ-Lebesgue (cf. [2, Thm.5.36]). Theorem 2.9.Let(X,ς)be a metrizable locally solid lat- tice,(Y,τ)aσ-Lebesgue lattice, andT: (X,ς)→(Y,τ) a positiveςτ-continuousσ-Lebesgue operator. Then the unique extensionˆT: (ˆX,ˆς)→(ˆY ,ˆτ)is positive,ˆςˆτ-conti- nuous, andσ-Lebesgue. Proof.Clearly,ˆTis positive and ˆςˆτ-continuous. To show thatˆTisσ-Lebesgue, assume ˆxn↓0 in (ˆX,ˆς). LetW?

τ(0). We have to showˆTˆxn?

Wfor large enoughn. Take

a solid neighborhoodW1?τ(0) withW1+W1?Wand choose a neighborhoodV?ς(0) withT(V)?W1. Let {Vn}∞n=1be a base forς(0) satisfyingVn+1+Vn+1?Vn for all naturalnand alsoV1+V1?V. At this point we borrow the second paragraph of the proof of [2, Thm.5.36]. For eachn, let ˆvn?ˆXbe an upper element ofXsuch that ˆxn≤ˆvnand ˆvn-ˆxn?

Vn; see [2, Thm.5.4]. Put

ˆwn:=?ni=1ˆvnfor allnand note that each ˆwnis an upper element ofX, ˆwn↓and ˆwn-ˆxn?

Vn. Since ˆxn↓0, it

follows from [2, Thm.2.21(e)] that ˆwn↓0 also holds inˆX. 21
Thus, we can assume without loss in generality that ˆxnis a sequence of upper elements ofX.

Takeun?X+withun≤ˆxnand ˆxn-un?

Vn. Application of [2, Thm.2.21(e)] to the nets ˆxn↓0 and z n:=?ni=1un↓with ˆxn-znˆς-→0 giveszn↓0 inX. AsTis σ-Lebesgue,Tznτ-→0. So, there exists somen1such that ˆ

Tzn=Tzn?W1?

W1(?n≥n1).(6)

Observe that

0≤ˆxn-zn= ˆxn-?ni=1ui=?ni=1(ˆxn-ui)≤ ?ni=1(ˆxi-ui)≤

n ? i=1(ˆxi-ui)?

V1+V2+...+Vn?V(?n?N).(7)

It follows from (6), (7), and

ˆT(

V)?W1that

ˆ

T(ˆxn)≤ˆT(zn)+ˆT(ˆxn-zn)?

W1+W1?W(?n≥n1).

SinceW?τ(0) was arbitrary,ˆTisσ-Lebesgue. Since Theorem 5.36 of [2] follows from Theorem 2.9 when (X,ς) coincides with (Y,τ) andTis the identity operator onX, Theorem 2.9 can be also considered as a generaliza- tion of [2, Thm.5.36]. One more question deserves being mentioned. LetTbe Lebesgue,oτ-continuous,oτ-bounded,oτ-compact,KB, or Levi. Does|T|satisfy the same property? The answer is positive, whenTmaps a vector latticeXto a Dedekind complete locally solid lattice (Y,τ), in the following cases: (?) for a Lebesgue/oτ-continuous order continuous op- eratorT, when (Y,τ) is Lebesgue, since|T|is order continuous by [3, Thm.1.56] and hencexα↓0/xαo-→0 inXimply|T|xαo-→0 and (since (Y,τ) is Lebesgue) |T|xατ-→0; (??) for anoτ-bounded operatorT, by Proposition 2.1. Remark 2.3.Let (X,ς) and (X,τ) be locally solid lattices. 22
a) Recall that (Y,τ) is said to beboundedly order-boun- ded(BOB) whenever increasingτ-bounded nets inY+ are order bounded inY(cf. [31, Def.3.19]).

Assume (Y,τ) to be Dedekind complete and BOB,

Then each positive quasiKBoperator from (X,ς) to (Y,τ) is quasi Levi. Letxαbe an increasingς-bounded net inX+. ThenTxαis aτ-Cauchy increasing net in Y +. ThenTxαisτ-bounded and henceo-bounded since (Y,τ) is BOB. The Dedekind completeness ofY implies thatTxα↑yfor somey?Y. ThereforeTxα iso-Cauchy as desired. b) Recall that (X,ς) satisfies theB-property(cf. [31,

Def.3.1]) whenever the identity operatorIonXis

quasiσ-KB. LetT: (X,ς)→(Y,τ) be a continu- ous operator and assume the topologyςin (X,ς) to be minimal. ThenTis Lebesgue and quasiKB. The first part follows as (X,ς) is Lebesgue by [2, Thm.7.67] and henceT: (X,ς)→(Y,τ) is Lebesgue. The sec- ond part follows as (X,ς) satisfies theB-property [31,

Prop.3.2] and henceTisσ-KB, and thusTis quasi

KBby Proposition 1.2.

Similarly, each regular operator from a vector latticeX to a locally solid lattice (Y,τ) with minimal topologyτis quasi Lebesgue by [31, Cor.3.3].

3 Operators between Banach lattices

In this section we investigate operators whose domains and ranges are Banach lattices. It is clear that every con- tinuous operator from an order continuous Banach lattice to a topological vector space is Lebesgue. The following proposition generalizes [9, Prop.2]. Proposition 3.1.An operatorTfrom aσ-order contin- uous Banach latticeXto a metrizable topological vector space(Y,τ)is continuous iffTisσoτ-continuous.

Proof.The necessity is trivial.

For the sufficiency: letTbeσoτ-continuous,xα?·?

X--→x,

andxαβbe a subnet of the netxα. Bya) of Remark 1.3, 23
there exists a sequenceβnof indices with?xαβn-x?X→0. SinceXis a Banach lattice, there is a subsequencexαβnk ofxαβnsatisfyingxαβnko-→xinX(see [34, Thm.VII.2.1]). SinceTisσoτ-continuous thenTxαβnkτ-→Tx. Remark 1.3 impliesTxατ-→Tx, as required. In the next proposition we transfer theσ-condition from the domainXofTto the operatorT. Proposition 3.2.Eachσoτ-continuous operatorTfrom a Banach latticeXto a Banach latticeYis continuous. Proof.It is sufficient to show that?Txn? →0 for every norm-null sequencexninX. Let?xn? →0 inX. For proving?Txn? →0, in view ofb) of Remark 1.3, it is enough to show that, for each subsequenceTxnkofTxn, there exists a sequencenkiof indices with?Txnki? →0. LetTxnkbe a subsequence ofTxn. By [34, Thm.VII.2.1], x nkhas a further subsequencexnkiwithxnkio-→0, and since Tisσoτ-continuous,?Txnki? →0 as desired.

Theorem 3.1.LetT:X→Ybe an order bounded op-

erator between Banach latticesXandY, the norms inX? andYbe order continuous, andYbe weakly sequentially complete. ThenTis a quasiKB-operator. Proof.SinceYhas order continuous norm,Yis Dedekind complete, and hence the operatorTis regular. Therefore, we may assumeT≥0. By Proposition 1.2, it suffices to prove thatTis quasiσ-KB. Letxnbe a? · ?X-bounded increasing sequence inX+. ThenTxnis? · ?Y-bounded and 0≤Txn↑. It follows from [3, Thm.5.28] thatTxn has a weak Cauchy subsequenceTxnk. ThenTxnis weak Cauchy and henceTxnw-→yfor somey?Y. SinceTxn↑, it follows from [3, Thm.3.52] thatTxn?·?

Y--→y, and hence

Tx nis? · ?Y-Cauchy as required. Asc0is not aKB-space, the identity operatorI:c0→ c

0is not quasiKBby Remarka);d). Thus, the weak

sequential completeness ofEis essential in Theorem 3.1. 24
Example 3.1.LetX= (cω(R),?·?∞) be the Banach lat- tice of all boundedR-valued functions onRsuch that each f?Xdiffers from a constant sayafon a countable subset ofR. It is easy to see thatXis Dedekindσ-complete yet not Dedekind complete. We define a positive operatorT:X→Xas follows. Tfis the constant functionaf·IR?X, for which the set {d?R:f(d)?=af}is countable. Clearly,Tis quasiKB and hence quasi Lebesgue. (i) For the sequencefn:=I{1,2,...n}?Xwe have that f no-→IN?X, yet?fn-IN?∞= 1 for alln. Thus, the norm inXis notσ-order continuous. In particular,

Proposition 3.1 is not applicable toT.

(ii)Tis rank-one continuous (therefore quasiKB, com- pact, andoτ-compact) operator inX. Letfno-→0. Since for everyε >0 there existsnεsuch that the set ? n≥nε{d?R:|fn(d)| ≥ε}is countable,?Tfn?∞< ε for alln≥nε. Thus,Tisσoτ-continuous and hence σ-Lebesgue with respect to the norm topology onX. (iii) OperatorTis not Lebesgue. Indeed, for the net f α:=IR\α?Xindexed by the family Δ of all finite subsets ofRordered by inclusion, we havefα↓0 yet?Tfα?∞=?IR?∞= 1 for allα?Δ. The same argument shows thatTis not weakly Lebesgue. Since Tf k?= 0 for at most onekfor each disjoint sequence f ninX, operatorTisM-weakly compact. It should be clear thatTis notL-weakly compact. (iv)TisKB. Indeed, letX?fα↑and?fα? ≤M?R.

Thenafα↑≤M. Takef:=a·IR?Xfora=

sup αafα?R. Clearly,?T(f-fα)? ↓0, as desired. (v) Clearly,Tis a Dunford-Pettis lattice homomorphism. By Corollary 2.3, the span of [0,T] is a vector sublat- tice ofKB-operators in the ordered spaceLr(X) of all regular operators inX. (vi) The vector latticeXisτ-laterallyσ-complete yet not τ-laterally complete with respect to the norm topol- ogy onX. 25
Observe that aKB-operatorT:?1→?∞defined by [T(a)]k:=?∞n=1anfor allk?Nis neitherL- norM- weakly compact (cf., e.g., [3, p.322]). In the present paper we do not investigate conditions under thatKB-operators are (weakly) Lebesgue. The following lemma generalizes the observation that the identity operator in a Banach latticeXis quasiKBiffXis aKB-space.

Lemma 3.1.Ifc0does not embed in a Banach spaceY,

then every continuous operatorTfrom any Banach lattice

XtoYis a quasiKBoperator.

Proof.It follows from [3, Thm.4.63] thatT=S◦Q, where

Qis a lattice homomorphism fromXto aKB-spaceZ

(and henceQis a quasiKB-operator byd) of Remark 3) andSis a continuous operator fromZtoY. Now apply the fact that the class of quasiKB-operators is closed under the left composition with continuous operators.

Theorem 3.2.LetXbe a Banach lattice andYa Banach

space. TFAE. (i)Every continuous operatorT:X→Yis quasiKB. (ii)c0does not embed inY.

Proof.(ii)?(i) is Lemma 3.1.

(i)?(ii) It suffices to prove that any embeddingJ: c

0→Yis not a quasiKB-operator. Assume by way

of contradiction that such aJis quasiKBand letxn=?nk=1ek?c0. The increasing sequencexnis norm bounded,

yet the sequenceJxnis not? · ?Y-Cauchy because ?Jxn-Jxm?Y=????? Jn? k=me k????? ≥1 ?J-1|J(c0)?>0 (n?=m).

The obtained contradiction completes the proof.

Among other things, Theorem 3.2 asserts that, in the Banach space setting, the concept of continuous quasi

KB-operator does not depend on the domainXand is

completely determined by the property that the range spaceYnot containing an isomorphic copy ofc0. In light of this observation, the last sentence in itemd) of Remark 26

3 is reduced to the condition 1≤p <∞under which

c00?·?p=?pis aKB-space and to the condition 1≤l≤p under which the identity operatorIfrom (c00,? · ?l) to (c00,? · ?p) is continuous.

Now we pass to a discussion ofKBoperators.

Theorem 3.3.LetT:X→Ybe regular operator between Banach lattices. IfX?has order continuous norm andc0 does not embed inYthenTis KB. Proof.Without lost of generality, we supposeT≥0. By [3, Thm.4.63],Tadmits a factorizationT=S◦Q, whereZ is aKB-space andQ:X→Zis a lattice homomorphism. Letxαbe an increasing norm bounded net inX+. Since Q≥0 thenQxαis an increasing norm bounded net in Z +, and sinceZis aKB-space, there existz?Zwith ?Qxα-z? →0 and hence?Txα-Sz? →0 as desired. The following theorem gives conditions under which each positive weakly compact operator from a Banach lat- ticeXto an arbitraryKB-spaceYisσoτ-continuous.

Theorem 3.4.For a Banach latticeXTFAE.

(i)The imagei(X)under the natural embeddingi:X→ X ??is a regular sublattice ofX??. (ii)Each positive weakly compact operatorTfromXto aKB-spaceYisσoτ-continuous. (iii)Each positive compact operatorTfromXto aKB- spaceYisσoτ-continuous. (iv)T2isσoτ-continuous for every positive weakly com- pact operatorTinX. (v)T2isσoτ-continuous for every positive compact op- eratorTinX. (vi)T2is Lebesgue for every positive compact operatorT inX.

Proof.(i)?(ii) LetTbe a positive weakly compact

operator fromXto aKB-spaceY. By [3, Thm.5.42],

T=R◦Swith bothS:X→ZandR:Z→Yare

positive andZis a reflexive Banach lattice. Letxαo-→0 inX. Sincei(X) is regular inX??theni(xα)o-→0 inX?? 27
(e.g., by [2, Thm.1.20]). SinceS??:X??→Z??is order continuous (e.g., by [3, Thm.1.73]),S??(i(xα))o-→0 inZ??, and henceSxα=S??(i(xα))o-→0 inZ. The order continuity of norm inZimplies?Sxα?Z→0 and hence?Txα?Y= ?R◦S(xα)?Y→0, showing thatTisσoτ-continuous. (ii)?(iii) and (iv)?(v)?(vi) are trivial. (iii)?(i) Ifi(X) is not regular inX??, then (e.g., by [2, Thm.3.12]) there exists a positive functionalf? X ?which is not order continuous. Thusf:X→Ris positive compact yet notσoτ-continuous asf(xα) does not converge to 0 in theKB-spaceY=Rfor some net withxαo-→0 inX. The obtained contradiction proves that i(X) must be regular inX??. (i)?(iv) The proof is similar to the proof of (i)?(ii) above with the only difference that the factorization ofT2 by [3, Cor.5.46] must be used instead of the factorization ofTby [3, Thm.5.42]. (vi)?(i) The proof is again similar to the proof of (iii)?(i) above. Ifi(X) is not regular inX??, then there exists a positive functionalf?X?which is not order continuous. Take anyz?X+withf(z) = 1 and define a positive compact operatorTinXbyTx:=f?z. Asf is not order continuous, there is a netxαwithxα↓0 and f(xα)≥1 for allα. Therefore T

2(xα) =T(f(xα)z) =f(xα)Tz=f(xα)z≥z >0 (?α)

violating thatT2is Lebesgue. The obtained contradiction completes the proof. Proposition 3.3.LetT:X→Ybe aoτ-bounded oper- ator from a Banach latticeXto a Banach spaceY. Ifc0 does not embed inY, thenTisσ-KB. Proof.By [30, Thm.3.4.6],Tfactors over aKB-spaceZas T=S◦Q, whereQis a lattice homomorphismQ:X→Z, andS:Z→Yis continuous. Letxnbe a norm bounded increasing sequence inX+. SinceQis positive,Qxnis positive norm bounded increasing sequence in theKB- spaceZ. Thus,Qxnis norm convergent inZ. AsSis continuous,Txn=SQxnis norm convergent and henceT isσ-KB. 28
Proposition 3.4.LetT:X→Ybe a continuous opera- tor from a Banach latticeXto a Banach spaceY. If either T ??is order-weakly compact or the norm inXis order con- tinuous andTdoes not preserve a sublattice isomorphic to c

0, thenTisσ-KB.

Proof.The operatorT:X→Yfactors over aKB-space ZasT=S◦Q, whereQ:X→Zis a lattice homomor- phism andS:Z→Ycontinuous by [30, Thm.3.5.8]. Let x nbe a norm bounded increasing sequence inX+, then Qx nis also norm bounded increasing inZ+. AsZis a KB-space,Qxnis norm convergent inZ. Hence (S◦Q)xn is convergent inYandTis aσ-KB operator.

Remark that ifT:X→Yis a continuous opera-

tor from a Banach latticeXto a Banach spaceYand Tpreserves no subspace (or no sublattice) isomorphic to c

0, thenTxnis norm convergent for each increasing norm

bounded sequence inX+by [30, Thm.3.4.11] and therefore

Tisσ-KB.

It follows from [35, Thm.1] that everyuoτ-continuous operatorTfrom an atomic Banach latticeXto a Banach spaceYis a Dunford-Pettis operator. By Theorem 5.57 of [3], a continuous operatorT:X→Yfrom a Banach latticeXto a Banach spaceYiso-weakly compact iff, for each order bounded disjoint sequencexninX, we have ?Txn? →0. Proposition 3.5.Each positive weakly compact operator

Tbetween Banach latticesXandYis Levi.

Proof.Letxαbe a norm bounded increasing net inX+. Weak compactness ofTensures thatTxαβw-→yfor some y?Yand some subnetxαβofxα. SinceTxαβ↑then Tx αβw-→yimpliesTxαβ↑yand henceTxα↑ywhich means thatTis a Levi operator. Remark 3.1.a) It follows directly from [30, Cor.3.4.12] that every continuous operator from a Banach lattice

Xto a Banach spaceYthat does not containc0is

quasiKB. 29
b) LetXbe a Banach lattice andYbe a Banach space.

Then every continuous operatorT:X→Ythat does

not preservec0is quasi-KB by [30, Thm.3.4.11]. Un- der the same settings, it was observed in [30, 3.4.E4, p. 203] that ifT??(BX)?Y, whereBXis the band generated byXinX??, thenTdoes not preserve a subspace isomorphic toc0. Note thatb-weakly com- pact operators satisfyT??(BX)?Y, wheneverXhas order continuous norm. It follows that under these conditions, eachb-weakly compact operator is quasi

KB (see [4, Prop 2.11]).

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