Factorizing operators on Banach function spaces through spaces of multiplication operators

Factorizing operators on Banach function spaces through spaces of multiplication operators J.M. Calabuig 1,O. Delgado ∗,2, E.A. Sánchez Pérez 3 Instituto Universitario de Matemática Pura y Aplicada, Universitat Politècnica de València, València 46022, Spain abstract Keywords: Banach function spaces Factorization of operators Multiplication operators Vector measures In order to extend the theory of optimal domains for continuous operators on a Banach function space X(μ)over a finite measure μ,weconsideroperatorsTsatisfying other type of inequalities than the one given by the continuity which occur in several well-known factorization theorems (for instance, Pisier Factorization Theorem through Lorentz spaces, pth-power factorable operators . . . ). We prove that such a Tfactorizes through a space of multiplication operators which can be understood in a certain sense as the optimal domain for T. Our extended optimal domain technique does not need necessarily the equivalence between μand the measure defined by the operator Tand, by using δ-rings, μis allowed to be infinite. Classical and new examples and applications of our results are also given, including some new results on the Hardy operator and a factorization theorem through Hilbert spaces. 1. Introduction Let (Ω, Σ) be a measurable space, X(μ)a Banach function space related to a positive measure μon Σand T:X(μ)→E a linear operator with values in a Banach space E. Supposing that Tsatisfies a certain property, natural questions arise: Can Tbe extended to a larger domain in a way that the extension (still with values in E)preservesthesameproperty?And,for a positive answer, which is the largest domain in this sense? These questions have been solved for the continuity of Tin the case when μis finite and satisfies a compatibility property with Tand X(μ)is order continuous and contains the simple functions. Namely, in this case, we can consider the vector measure mT:Σ→Edefined by Tas mT(A)=T(χA)and then the space L1(mT)of integrable functions with respect to mTis the largest within all the order continuous Banach function spaces related to μto which Tcan be extended as a continuous operator still with values in E, see [4, Corollary 3.3]. Note that the continuity property of T,i.e.thereexists K>0 such that T(f)E⩽KfX(μ)for all f∈X(μ), can be rewritten as T(fϕ)E⩽KfX(μ)ϕL∞(μ)(1) for all f∈X(μ)and ϕsimple function. Then, L1(mT)is the largest domain for Tsatisfying the inequality (1). *Corresponding author. E-mail addresses: [email protected].es (J.M. Calabuig), [email protected].es (O. Delgado), [email protected].es (E.A. Sánchez Pérez). URLs: http://euler.us.es/~olvido (O. Delgado), http://www.personales.upv.es/~easancpe (E.A. Sánchez Pérez). 1Research supported by Generalitat Valenciana (project GV/2007/191), MEC (project #MTM2008-04594/MTM) (Spain) and FEDER. 2Research supported by Generalitat Valenciana (TSGD-07), MEC (program “Juan de la Cierva” and project #MTM2006-13000-C03-01) (Spain) and FEDER. 3Research supported by MEC (project #MTM2006-11690-C02-01) (Spain) and FEDER. In this paper we consider weaker conditions on X(μ)(e.g. it does not need to be order continuous or contain the simple functions and μdoes not need to be finite) and study the questions above when Tis what we call Y(η)-extensible, i.e. satisfies an inequality of the type T(fϕ)E⩽KfX(μ)ϕY(η)(2) for all f∈X(μ)and ϕsimple function belonging to a Banach function space Y(η)related to another positive measure η on Σ. Of course, this type of domination for Tincludes the continuity given by (1). Other relevant examples of this kind of property are given by the Pisier’s factorization theorem through weighted Lorentz spaces [10,19], the Lp-product extensible operators [3] and the pth power factorable operators [18, §5], all of these are explained in Section 6. Under minimal conditions on X(μ)and Twhich guarantee that mTis a vector measure defined on certain δ-ring (see Section 3), we prove in Section 5 that the largest domain for T(within all the Banach function spaces related to some measure for which every μ-null set is null) satisfying the inequality (2) can be represented as a subspace of L1(mT)which is given by a space of multiplication operators from Yb(η)to L1(mT), where Yb(η)is the closure of the intersection of the simple functions with Y(η). This subspace is studied in Section 4. Actually, from a technical point of view, what we obtain is not a real optimal extension of T(this is only the case when mTand μhave the same null sets) but an optimal factorization of T. This extra complication allows to consider special cases as the Pisier’s factorization theorem. Therefore, the main contributions of the present paper are the following. The vector measure technique for determining the optimal domain of an operator has been used recently in a lot of papers (see for instance [4,5,8,9,18]); these applications usually consider the case of B.f.s. (with weak unit) defined over finite measure spaces. Also, the considered extension that is considered is the one that preserves the continuity of the operator. In this paper we generalize this method in two different directions. (1) We analyze the optimal domain for operators defined on Banach function spaces over measure spaces defined on δ-rings instead of σ-algebras. This allows us to consider also the cases of σ-finite measures and even non-σ-finite ones. We illustrate by means of examples and applications how our extension is meaningful (Section 3). (2) We find the optimal domain for an operator when a different property stronger than continuity is considered. We show that dominations of operators given by norm inequalities involving norms of other spaces can also be extended to a different domain preserving the same domination, and this extension is optimal. For doing this, we consider spaces of multiplication operators (Sections 4–6). All along in the paper, we will consider two examples to illustrate the results obtained. The first one is the Hardy operator S:L1∩L∞→Λψgiven by S(f)(x)=1 xx 0f(y)dy for all x>0, where Λψis a classical Lorentz space. We refer the reader to [9] for information about this operator not explained here. The second one, related to the case μnon-σ-finite, is a certain kind of kernel operator from 1(I)into p(I)with Ibeing a non-countable set. 2. Preliminaries 2.1. Banach function spaces Let (Ω, Σ) be a fixed measurable space. For a measure μ:Σ→[0,∞],wedenotebyL0(μ)the space of all Σ- measurable real valued functions on Ω, where functions which are equal μ-a.e. are identified. By a Banach function space (briefly, B.f.s.) X(μ)we mean a Banach space contained in L0(μ)with norm · X(μ)satisfying that if f∈X(μ)and g∈L0(μ)with |g|⩽|f|μ-a.e. then g∈X(μ)and gX(μ)⩽fX(μ). Note that X(μ)is a Banach lattice for the μ-a.e. pointwise order, in which the convergence in norm of a sequence implies the convergence μ-a.e. for some subsequence. Afunctiong∈X(μ)is a weak unit if g>0μ-a.e. If X(μ)has a weak unit then X(μ)is saturated, i.e. there exists no A∈Σ with μ(A)>0 such that fχA=0μ-a.e. for all f∈X(μ),orequivalently,foreveryA∈Σwith μ(A)>0thereexists B∈Σ∩2Awith μ(B)>0 such that χB∈X(μ). In the case when μis σ-finite, X(μ)is saturated if and only if it has a weak unit. A B.f.s. is order continuous if order bounded increasing sequences are convergent in norm. Denote by S(Σ) the space of the simple functions,i.e.ϕ=n 1αjχAjwith αj∈Rand Aj∈Σ. In the following result, parts (a) and (c) are known for the case μσ-finite (see [1, Theorems I.3.11 and I.3.13]). For a general μ, the arguments are a bit different, so we write the proof for the aim of completeness. Lemma 1. Let X(μ)be a non-trivial B.f.s. and Xb(μ)denote the closure of S(Σ) ∩X(μ)in X(μ). The following claims hold: (a) Xb(μ)is a non-trivial B.f.s. (b) If X(μ)has a weak unit then Xb(μ)also has a weak unit. (c) If X(μ)is order continuous then Xb(μ)=X(μ). Proof. (a) Since X(μ)is a non-trivial B.f.s., there exists A∈Σwith μ(A)>0 such that χA∈X(μ).So,Xb(μ)is non-trivial. Obviously, Xb(μ)is a Banach space with norm · X(μ)contained in L0(μ). To establish the lattice property, let us prove first that Xb(μ)is the closure in the space X(μ)of the ideal {f∈X(μ)∩L∞(μ):χSupp(f)∈X(μ)}.SinceS(Σ) ∩X(μ) is contained in this ideal, we only have to see that given f∈X(μ)∩L∞(μ)with χSupp(f)∈X(μ),thereexists(ϕn)⊂ S(Σ) ∩X(μ)such that ϕn→fin X(μ).Take(ψn)⊂S(Σ) such that ψn→fin L∞(μ).Then f−ψnχSupp fX(μ)⩽f−ψn∞χSupp fX(μ)→0 as n→∞ and ψnχSupp f∈S(Σ) ∩X(μ)as |ψn|χSupp f⩽ψn∞χSupp f∈X(μ). Let f∈Xb(μ)and g∈L0(μ)be such that |g|⩽|f|μ-a.e. In particular, fbelongs to the B.f.s. X(μ)and so g∈X(μ) and gX(μ)⩽fX(μ).Take(fn)∈S(Σ) ∩X(μ)such that fn→fin X(μ)and consider the functions defined by gn(x)= sgn(g(x)) ·min{| fn(x)|,|g(x)|}. It follows that gn∈L∞(μ)∩X(μ)and gn→gin X(μ)(see the proof of [1, Theorem I.3.11]). Moreover, χSupp gn∈X(μ),sinceSuppgn⊂Supp fnand χSupp fn∈X(μ)as fn∈S(Σ) ∩X(μ).So,g∈Xb(μ). (b) Only note that a B.f.s. X(μ)has a weak unit if and only if there exists a sequence (An)⊂Σsuch that χAn∈X(μ) and Ω=nAn. (c) Suppose that X(μ)is order continuous. If 0 ⩽f∈X(μ),taking(ϕn)⊂S(Σ) such that 0 ⩽ϕn↑fpointwise, we have that (ϕn)⊂X(μ)and ϕn→fin X(μ), that is, f∈Xb(μ). For a general f∈X(μ), considering the positive and negative parts of f, we obtain that f∈Xb(μ).2 Given two set functions μ,λ:Σ→[0,∞],wewillwriteλμif λ(A)=0 whenever A∈Σwith μ(A)=0 (i.e. every μ-null set is λ-null). If λμand μλwe will say that μand λare equivalent (i.e. they have the same null sets). Consider two measures μ,λ:Σ→[0,∞] such λμ.Themap[i]:L0(μ)→L0(λ) which takes a μ-a.e. class in L0(μ) represented by finto the λ-a.e. class represented by the same f, is a well-defined linear map. In order to simplify notation [i](f)will be denoted again as f. Of course, in the case when λand μare equivalent we have that L0(μ)=L0(λ) and [i]is the identity map. Note that if X(μ)and Y(λ) are B.f.s.’ such that [i]:X(μ)→Y(λ) is well defined then it is automatically continuous, since it is a positive map between Banach lattices (see [14, p. 2]). Given h∈L0(λ), we can consider the linear map Ph:L0(μ)→L0(λ) defined as Ph(f)=fh (formally [i](f)·h). If X(μ)and Y(λ) are B.f.s.’ such that Ph:X(μ)→Y(λ) is well defined then it is automatically continuous, since we can write Ph=Ph+−Ph−, where h+and h−are the positive and negative parts of hrespectively, and Ph+,Ph−are continuous as they are positive operators between Banach lattices. Denote by M(X(μ), Y(λ)) the space of all h∈L0(λ) such that Ph:X(μ)→Y(λ) is well defined, i.e. MX(μ), Y(λ)=h∈L0(λ):fh∈Y(λ) for all f∈X(μ), which can be endowed with the natural seminorm hM(X(μ),Y(λ)) =sup f∈BX(μ) fhY(λ) where BX(μ)is the open unit ball of X(μ)(i.e. the usual operator norm of Ph). It can be checked that this seminorm is a norm if and only if X(μ)satisfies a variation of the saturation condition, namely, there exists no A∈Σwith λ(A)>0such that fχA=0λ-a.e. for all f∈X(μ),orequivalently,foreveryA∈Σwith λ(A)>0thereexistsB∈Σ∩2Awith λ(B)>0 such that χB∈X(μ).IfX(μ)has a weak unit, this condition holds. Moreover, in this case, M(X(μ), Y(λ)) is complete and so it is a B.f.s. This fact is proved in [15, Proposition 2] in the case when μand λare equivalent and σ-finite. For the general case, the proof is similar. 2.2. Integration with respect to vector measures defined on a δ-ring The classical theory of integration with respect to vector measures defined on a σ-algebra (see for instance [18, Chapter 3]) was extended to the case of vector measures defined on a δ-ring by Lewis [13] and Masani and Niemi [16,17]. Let Rbe a δ-ring of subsets of Ω(i.e. a ring closed under countable intersections) and Rloc the σ-algebra of all subsets Aof Ωsuch that A∩B∈Rfor all B∈R. Note that if Ris a σ-algebra then Rloc =R.DenotebyS(R)the space of all R-simple functions (i.e. simple functions supported in R). Given a real measure λ:R→R, that is λ(An)converges to λ(An)for every pairwise disjoint sequence (An)⊂R with An∈R,thevariation of λis the measure |λ|:Rloc →[0,∞] given by |λ|(A)=supλ(Ai):(Ai)finite disjoint sequence in R∩2A. The space L1(λ) of integrable functions with respect to λis defined as the space L1(|λ|)with the usual norm |f|λ=Ω|f|d|λ|. Note that S(R)is dense in L1(λ). The integral of an R-simple function ϕ=n i=1aiχAiover A∈Rloc is defined in the natural way by Aϕdλ=n i=1aiλ(Ai∩A).For f∈L1(λ) the integral over A∈Rloc is defined by Afdλ=lim Aϕndλ, where (ϕn)is a sequence in S(R)converging to fin L1(λ). Let Ebe a real Banach space and m:R→Eavector measure, that is m(An)converges to m(An)in Efor every pairwise disjoint sequence (An)⊂Rwith An∈R.DenotebyE∗the topological dual of Eand by BE∗its open unit ball. The semivariation of mis the set function m:Rloc →[0,∞] defined by m(A)=sup x∗∈BE∗x∗m(A), where |x∗m|is the variation of the real measure x∗m:R→Rgiven by the composition of mwith x∗. Note that mis finite on R.AsetA∈Rloc is m-null if m(A)=0, or equivalently, m(B)=0 whenever B∈R∩2A.Apropertyholdsmalmost everywhere (briefly, m-a.e.) if it holds except on an m-nullset.From[2,Theorem3.2],therealwaysexistsameasure λ:R→[0,∞] with the same null sets as m(i.e. mand |λ|are equivalent). An Rloc-measurable real function fis integrable with respect to m if (i) f∈L1(|x∗m|)for every x∗∈E∗, and (ii) for each A∈Rloc there exists xA∈Esuch that x∗(xA)=A fdx ∗m,for every x∗∈E∗. The vector xAis unique and will be written as Afdm.WedenotebyL1(m)the space of all integrable functions with respect to mand by L1 w(m)the space of functions satisfying only condition (i). In both spaces, functions which are equal m-a.e. are identified. Taking a measure λequivalent to m,wehavethatL1(m)and L1 w(m)are B.f.s.’ related to (Ω, Rloc,|λ|) with the norm fm=sup x∗∈BE∗ Ω |f|dx∗m,for all f∈L1 w(m). Of course, L1(m)is a closed subspace of L1 w(m). In the case when Edoes not contain an isomorphic copy of c0,wehave that L1(m)=L1 w(m).Moreover,L1(m)is order continuous and contains S(R)as a dense set. Note that · mis defined for any Rloc-measurable function and has the Fatou property, that is, if 0 ⩽fj↑fm-a.e. then fjm↑fm.Actually,L1 w(m) can be described as the space of all Rloc-measurable functions fwith fm<∞. The norm of f∈L1(m)can also be computed by means of the formula fm=sup Ω fϕdmE :ϕ∈SRloc∩BL∞(λ).(3) The integration operator Im:L1(m)→Edefined by Im(f)=Ωfdmfor all f∈L1(m), is a continuous linear operator. Given f∈L1(m),theindefinite integral of f with respect to m is the vector measure mf:Rloc →Edefined by mf(A)=A fdm,for all A∈Rloc. Note that gmf=gfmfor every Rloc-measurable function g. Then, it follows that g∈L1 w(mf)if and only if gf ∈L1 w(m) and g∈L1(mf)if and only if gf ∈L1(m).Moreover,mf(A)=χAmf=fχAmfor every A∈Rloc. For further issues related to integration with respect to vector measures defined on a δ-ring see [7]. Let us end this section by showing two results which will be used in this paper. Lemma 2. Let Rand  Rbe two δ-rings of subsets of Ωsuch that  R⊂R⊂ Rloc =Rloc.Ifm:R→E is a vector measure and  m denotes the restriction of m to  R, then the following assertions hold: (a) For every Rloc-measurable function f we have that f m⩽fmand so [i]:L1 w(m)→L1 w( m)is well defined. (b) If m satisfies that A∈R,e∗∈E∗and sup B∈ R∩2Ae∗m(B)=0⇒e∗m(A)=0(4) then [i]:L1(m)→L1( m)is well defined and Ωfd  m=Ωfdmforeveryf∈L1(m). Proof. (a) Note that for every e∗∈E∗and A∈ Rloc =Rloc we have that |e∗ m|(A)⩽|e∗m|(A).So,foreveryRloc-measurable function fand e∗∈E∗, it follows that Ω|f|d|e∗ m|⩽Ω|f|d|e∗m|and thus f m⩽fm. In particular,  m(A)⩽m(A) for all A∈Rloc (take f=χA). Then,  mmand [i]:L1 w(m)→L1 w( m)is well defined. (b) Take A∈R.SinceχA∈L1(m)⊂L1 w(m)and (a) holds, we have that χA∈L1 w( m)and so condition (i) of the definition of integrable function with respect to  mholds for f=χA. Let us prove that (b) also holds. Let B∈ Rloc =Rloc and e∗∈E∗. Since |e∗ m|(A∩B)=sup{|e∗ m|(H):H∈ R∩2A∩B}<∞, there exists an increasing sequence (Hn)of sets in  R∩2A∩B satisfying that |e∗ m|(A∩B\∪Hn)=0. Condition (4) implies that |e∗m|(A∩B\∪Hn)=0. Thus, χHn↑χA∩Be∗ m-a.e. and e∗m-a.e., and so B χAde∗ m=lim n→∞ χHnde∗ m=lim n→∞ e∗ m(Hn)=lim n→∞ e∗m(Hn) =lim n→∞ χHnde∗m=B χAde∗m=e∗B χAdm. Hence, for every A∈Rwe have that χA∈L1( m)and BχAd m=BχAdm for all B∈Rloc. Now, consider f∈L1(m)⊂ L1 w(m).By(a), f∈L1 w( m).Takingasequence(ϕn)of R-simple functions converging to fin L1(m),wehavethat f−ϕn m⩽f−ϕnm→0. Then, since ϕn∈L1( m)and L1( m)is a closed subspace of L1 w( m), we have that f∈L1( m). Moreover, as the integration operator is continuous,  Ω fd  m=lim n Ω ϕnd m=lim n Ω ϕndm = Ω fdm.2 If the vector measure mconsidered in Lemma 2 satisfies certain σ-finiteness condition, the maps [i]given in (a) and (b) are isometries. Lemma 3. Let Rand  Rbe two δ-rings of subsets of Ωsuch that  R⊂R⊂ Rloc and Ω=nAnwith An∈ R.Ifm:R→Eisavector measure and  m denotes the restriction of m to  R,thenL 1 w(m)=L1 w( m)with equal norms, L1(m)=L1( m)and Ωfdm=Ωfd  m for every f ∈L1(m). Proof. The containments  R⊂R⊂ Rloc imply that Rloc ⊂ Rloc, see [16, Lemma A.3]. For every A∈ Rloc we have that A=nA∩An∈Rloc since A∩An∈ R⊂R.So,Rloc = Rloc. Note that if A∈ Rthen R∩2A= R∩2A.Indeed,ifB∈R with B⊂Athen B=B∩A∈ R. The converse containment is clear. Then, it follows that |x∗m|(A)=|x∗ m|(A)for all A∈ R. Given A∈Rloc,sinceA=nA∩Anwhere A∩An∈ R, noting that (An)can be taken to be pairwise disjoint, we have that x∗m(A)=x∗m(A∩An)=x∗ m(A∩An)=x∗ m(A) for all x∗∈E∗.Hence,f m=fmfor all Rloc-measurable function fand thus L1 w(m)=L1 w( m). Condition (4) in Lemma 2(b) holds. Indeed, if A∈R,e∗∈E∗and sup{|e∗m(B)|:B∈ R∩2A}=0, it follows that e∗m(A)=e∗m(A∩An)=0. Then, L1(m)⊂L1( m)and Ωfd  m=Ωfdm.Moreover,L1( m)⊂L1(m)as S( R)⊂S(R)and ·  m=· m.2 3. Optimal domain for order-w continuous operators Fix (Ω, Σ) a measurable space. Let X(μ)be a B.f.s. and consider the δ-ring of subsets of Ωgiven by ΣX(μ)=A∈Σ:χA∈X(μ), which satisfies Σ⊂Σloc X(μ). Throughout the paper we will assume Σ=Σloc X(μ). This condition holds for instance if X(μ)has a weak unit, which is equivalent to the existence of a sequence (An)⊂ΣX(μ)such that Ω=nAn. Given a linear operator T:X(μ)→Ewith values in a Banach space E, we can consider the finitely additive set function mT:ΣX(μ)→Egiven by mT(A)=T(χA).LetusrequireTto satisfy that there exists A∈ΣX(μ)such that T(χA)= 0 (in particular μ(A)>0), as in other case mTis null. Note that mTμ, that is, mTμor equivalently |λT|μfor any measure λT:ΣX(μ)→[0,∞] with the same null sets as mT.So,themap[i]:L0(μ)→L0(|λT|)is well defined. We will say that Tis order-w continuous if Tf n→Tf weakly in Ewhenever fn,f∈X(μ)with 0 ⩽fn↑fμ-a.e. This property holds for instance if X(μ)is order continuous and Tis continuous. Note that if Tis order-w continuous, then the condition T(χA)= 0 for some A∈ΣX(μ)is equivalent to Tbeing non-null. If Tis order-w continuous, all the conditions required in [8, Proposition 2.3] are satisfied, so mTis a vector measure and for every f∈X(μ)we have that f∈L1(mT)with Ωfdm T=Tf. Therefore, Tfactorizes as follows X(μ)T [i] E L1(mT) ImT (5) where ImTis the integration operator with respect to mT. As a consequence we have that every order-w continuous linear operator T:X(μ)→Eis continuous, since T=ImT◦[i]with [i]and ImTbeing continuous. Note that ImT:L1(mT)→Eis order-w continuous, since L1(mT)is order continuous and ImTis continuous. Also note that if A∈Σwith χA∈L1(mT)and e∗∈E∗satisfy that sup{|e∗ImT(χB)|:B∈ΣX(μ)∩2A}=0, then e∗ImT(χA)=0. The following proposition shows that the factorization (5) is optimal. This outcome is an extended version of the optimal domain result given in [4, Corollary 3.3]. See also [8, Corollary 2.6]. Proposition 4. Let T :X(μ)→E be a non-null order-w continuous linear operator. Suppose that Z(ζ ) is a B.f.s. such that ζμand T factorizes as X(μ)T [i] E Z(ζ) S(6) with S being an order-w continuous linear operator satisfying that, for A ∈Σwith χA∈Z(ζ) and e∗∈E∗, sup B∈ΣX(μ)∩2Ae∗S(χB)=0⇒e∗S(χA)=0.(7) Then, [i]:Z(ζ) →L1(mT)is well defined and S(f)=ImT(f)for all f ∈Z(ζ). Proof. Consider the δ-ring ΣZ(ζ ) ={A∈Σ:χA∈Z(ζ )}.NotethatΣ=Σloc Z(ζ ).Inothercase,thereexistsA∈Σloc Z(ζ ) such that A/∈Σ=Σloc X(μ). Then, there exists B∈ΣX(μ)such that A∩B/∈ΣX(μ).SinceχA∩B∈X(μ)(as χA∩B⩽χB∈X(μ)), it follows that A∩B/∈Σ. In other hand, B∈ΣZ(ζ ) (as ΣX(μ)⊂ΣZ(ζ )), so A∩B∈ΣZ(ζ ) and, in particular, A∩B∈Σwhich is a contradiction. Since S:Z(ζ ) →Eis a non-null order-w continuous linear operator, it factorizes as (5), i.e. Z(ζ) S [i] E L1(mS) ImS (8) where mS:ΣZ(ζ ) →Eis the vector measure given by mS(A)=S(χA)and ImSis the integration operator with respect to mS. By (6) we have that mS(A)=S(χA)=T(χA)=mT(A)for all A∈ΣX(μ), that is, mTis the restriction of mSto ΣX(μ). Condition (7) is just condition (4) in Lemma 2 for mS,thenmTmS,[i]:L1(mS)→L1(mT)is well defined and ImS(f)= ImT(f)for all f∈L1(mS). So, by (8) we have that [i]:Z(ζ ) →L1(mT)is well defined and S(f)=ImT(f)for all f∈Z(ζ ).2 Note that in the case when X(μ)has a weak unit, or equivalently, Ω=nAnfor some (An)⊂ΣX(μ), condition (7) in Proposition 4 always holds for every order-w continuous linear operator Ssatisfying (6). Example 5. Denote by R+the interval [0,∞),byBthe σ-algebra of all Borel subsets of R+and by λthe Lebesgue measure on B. Consider the Hardy operator S defined on L1∩L∞(λ) as S(f)(x)=1 xx 0f(y)dy. Note that L1∩L∞(λ) is a B.f.s. endowed with the usual norm fL1∩L∞(λ) =max{ fL1(λ),fL∞(λ)}and has a weak unit. Let ψ:R+→R+be an increasing concave map with ψ(0)=0, ψ(0+)=0, ψ(∞)=∞ and Λψ(λ) ={f∈L0(λ):fΛψ(λ) =∞ 0f∗(s)dψ(s)<∞} the related Lorentz space ( f∗being the decreasing rearrangement of f), which is an order continuous B.f.s. endowed with the norm fΛψ(λ). For issues related to Lorentz spaces see for instance [11, §II.5]. Assume that θψ(t)=∞ t ψ(s) sds <∞for all t>0, where ψdenotes the derivative of ψ. Then, it can be checked that S:L1∩L∞(λ) →Λψ(λ) is well defined and continuous with S(f)Λψ(λ) ⩽(ψ(a)+θψ(a))fL1∩L∞(λ) for any a>0. Moreover, Sis order-w continuous, since if 0 ⩽fn↑ f∈L1∩L∞(λ) λ-a.e., we have that 0 ⩽S(fn)↑S(f)∈Λψ(λ) pointwise (by the monotone convergence theorem) and so S(fn)→S(f)in Λψ(λ) (by order continuity of the Lorentz space). Therefore, for the δ-ring BL1∩L∞(λ) ={A∈B:λ(A)<∞}, we have that mS:BL1∩L∞(λ) →Λψ(λ),givenbymS(A)=S(χA), is a vector measure with mSλand Soptimally factorizes as L1∩L∞(λ) S [i] Λψ(λ) L1(mS) ImS (9) Note that in this case, [i]is actually the inclusion map i,sincemSand λare equivalent, and ImSis just S,infactL1(mS)= {f∈L0(λ):S(|f|)∈Λψ(λ)}, see [9, Proposition 3.4] and the previous comments. Moreover, if there exists C>0 such that ψ(t) t⩽Cθψ(t)for all t>0, from [9, Theorem 4.4], we can give a precise description of the optimal domain of Sin the sense of Proposition 4. Namely, L1(mS)=L1(θψ(t)dt), that is the space of integrable functions with respect to the Lebesgue measure with density θψ. An example of function satisfying all the above conditions is ψ(t)=t1/pwith 1 <p<∞,for which Λψ(λ) =Lp,1(λ), see for instance [1, §4.4]. Example 6. Let Ibe a non-countable set and K:I×I→[0,∞)a non-null map such that β=(Ki∞)i∈I∈p(I)(1 ⩽p< ∞), where Ki:I→[0,∞)is defined as Ki(j)=K(i,j). Note that p(I)is an order continuous saturated B.f.s. related to (I,2I,μ), where 2Iis the σ-algebra of all parts on Iand μis the counting measure. Also note that (2I)1(I)={A⊂I: Ais finite}and (2I)loc 1(I)=2I. Consider the map T:1(I)→p(I)defined as Tx= j∈I xjK(i,j)i∈I for every x=(xj)j∈I∈1(I).Since  i∈I j∈I xjK(i,j) p1/p ⩽ i∈I Kip ∞ j∈I |xj|p1/p =x1(I)βp(I), the map Tis a well-defined continuous linear operator. Then, since 1(I)is order continuous, Tis order-w continuous. Thus, mT:(2I)1(I)→p(I),givenbymT(A)=T(χA), is a vector measure with mTμand Toptimally factorizes as 1(I)T [i] p(I) L1(mT) ImT (10) Since mTis positive (i.e. mT(A)⩾0forallA∈(2I)1(I)), we have that ImT(f)⩾0forall0⩽f∈L1(mT). Then, by using formula (3), it follows that fmT=ImT(|f|)p(I)for all f∈L1(mT). Note that A∈2Iis mT-null if and only if K(i,j)=0 for all (i,j)∈I×A.Hence,mTand μare equivalent (and so [i]is an inclusion map) if and only if for every j∈Ithere exists i∈Isuch that K(i,j)>0. Since p(I)does not contain any isomorphic copy of c0,wehavethatL1(mT)=L1 w(mT). For every (2I-measurable) function fand e∗∈p(I)=p(I)∗(where 1 p+1 p=1), it is routine to prove that  Ω |f|de∗mT=sup B⊂I finite  j∈Bf(j)e∗,Kj where Kj=(K(i,j))i∈I∈p(I)(as Kj⩽β). Then, L1(mT)=f=f(j)j∈I⊂R:f(j)e∗,Kjj∈I∈1(I)for all e∗∈p(I) and for every f∈L1(mT)we have that fmT=sup e∗∈Bp(I) j∈If(j)e∗,Kj. Moreover, if f∈L1(mT),foreachi∈Iwe have that (f(j)K(i,j))j∈I∈1(I)(take e∗ i=(δi,s)s∈I∈p(I)with δi,s=1ifs=i and δi,s=0 in other case). So, we can consider the element xf=(j∈If(j)K(i,j))i∈I⊂R.Sinceβ∈p(I),thereexistsa countable set J⊂Isuch that K(i,j)=0forall(i,j)∈(I\J)×I.Set J={in}n⩾1. Then, for each nthere exists a countable set Γn⊂Isuch that f(j)K(in,j)=0forall j∈I\Γn. Consider Γ=n⩾1Γn={jm}m⩾1and the (2I)1(I)-simple functions ϕm=fχ{j1,..., jm}.Since0⩽|ϕm|↑|f|χΓ∈L1(mT), by the order continuity of L1(mT), we have that ϕm→fχΓin L1(mT) and so ImT(ϕm)→ImT(fχΓ)in p(I). Note that I\Γ∩Supp fis mT-null, as K(i,j)=0forall(i,j)∈I×(I\Γ∩Supp f). So, ImT(f)=ImT(fχΓ). Then, ImT(f)(i)=lim m→∞ ImT(ϕm)(i)=lim m→∞ T(ϕm)(i) =lim m→∞  j∈I ϕm(j)K(i,j)= j∈I f(j)χΓ(j)K(i,j)=xf(i) for all i∈Iand thus xf=ImT(f)∈p(I). Therefore, for every f∈L1(mT)we have that (f(j)K(i,j))j∈I∈1(I)for all i∈I, (j∈If(j)K(i,j))i∈I∈p(I)and ImT(f)=(j∈If(j)K(i,j))i∈I.Moreover, fmT=ImT|f|p(I)= i∈I j∈If(j)K(i,j)p1/p . Let us show a kernel satisfying all the above conditions. Take I=[0,∞)and φ∈p(I)(e.g. φ(i)=1 irχN(i)with r>1 p). Then K:I×I→[0,∞)defined by K(i,j)=φ(i)χ[0,i](j), satisfies that β=(Ki∞)i∈I∈p(I)(as β=|φ|). For this kernel, the operator T:1(I)→p(I)is given by T(x)=φ(i) j⩽i xji∈I for every x=(xj)j∈I∈1(I). For this “version of Hardy operator”, the optimal factorization (10) holds. 4. Subspaces of L1(m)generated by generalized semivariations Fix (Ω, Σ) a measurable space, Ea Banach space and Y(η)a B.f.s. For a vector measure n:Σ→Ewith nη,we define the Y(η)-semivariation of n by nY(η)=sup Ω ϕdnE :ϕ∈S(Σ) ∩BY(η). This concept generalizes the total semivariation of nwhich is obtained as n(Ω) =nL∞(λ) for any λequivalent to n, see (3) for f=χΩ∈L1(n). Let Rbe a δ-ring of subsets of Ωsuch that Rloc =Σand m:R→Ea vector measure with mη.Forevery f∈L1(m), the set function mf:Σ→E,givenbymf(A)=ΩfχAdm, is a vector measure with mfηand mfY(η)=sup Ω ϕfdm E :ϕ∈S(Σ) ∩BY(η). Let us consider the space L1 Y(η)(m)=f∈L1(m):mfY(η)<∞ equipped with the norm fL1 Y(η)(m)=max{ fm,mfY(η)}. Of course, L1 Y(η)(m)is a subspace of L1(m). The norm in L1(m) and the Y(η)-semivariation are related as the following result shows. Lemma 7. For every f ∈L1(m)and ϕ∈S(Σ) ∩Y(η)we have that fϕm⩽ϕY(η)mfY(η). Proof. For f∈L1(m)and ϕ∈S(Σ) ∩Y(η)we have that fϕ∈L1(m)and so, by (3), fϕm=sup Ω fϕψdmE :ψ∈S(Σ) ∩BL∞(λ), where λ:R→[0,∞] is a measure equivalent to m. Note that for every ψ∈S(Σ) such that |ψ|⩽1λ-a.e., there exists A∈Σsuch that Ω\Ais λ-null (and so m-null) and |ψϕ|χA⩽|ϕ|pointwise. Then, Ωfϕψdm =ΩfϕψχAdm where ϕψχA∈S(Σ) ∩Y(η)and ϕψχAY(η)⩽ϕY(η). So, it follows that fϕm⩽ϕY(η)mfY(η).2 Proposition 8. The space L1 Y(η)(m)is a B.f.s. Proof. Let us see that L1 Y(η)(m)satisfies the Riesz–Fischer property an so we will have that L1 Y(η)(m)is a Banach space. Let (fn)⊂L1 Y(η)(m)be such that fnL1 Y(η)(m)<∞. Then, fnm<∞and since L1(m)is complete, we have that f=fn∈L1(m)and  Ω ϕfdm E ⩽ Ω ϕfndmE ⩽mfnY(η) for all ϕ∈S(Σ) ∩BY(η).So,mfY(η)⩽mfnY(η)⩽fnL1 Y(η)(m)<∞and thus f∈L1 Y(η)(m). In other hand, if fis a measurable function such that |f|⩽|g|m-a.e. for some g∈L1 Y(η)(m),then f∈L1(m)and fm⩽gm. Moreover, for every ϕ∈S(Σ) ∩BY(η), using Lemma 7, it follows  Ω ϕfdm E ⩽ϕfm⩽ϕgm⩽ϕY(η)mgY(η)⩽mgY(η) and so mfY(η)⩽mgY(η)<∞.Hence, f∈L1 Y(η)(m)and fL1 Y(η)(m)⩽gL1 Y(η)(m). Therefore, L1 Y(η)(m)is a B.f.s. 2 The subspace of L1(m)generated by Y(η)=L∞(η)is L1(m)itself. Indeed, for each f∈L1(m), by (3) and since mη, we have that mfL∞(η)⩽fmand by Lemma 7 with ϕ=χΩ,wehavethatfm⩽mfL∞(η).So,L1 L∞(η)(m)=L1(m) with equal norms. Remark 9. In the case when χΩ∈Y(η), by Lemma 7 we have that fm⩽χΩY(η)mfY(η)for every f∈L1(m). Then, it follows that mfY(η)⩽fL1 Y(η)(m)⩽maxχΩY(η),1mfY(η) for every f∈L1 Y(η)(m), that is, mfY(η)is an equivalent norm to fL1 Y(η)(m). Recall that Yb(η)denotes the closure of S(Σ) ∩Y(η)in Y(η), which is a B.f.s. by Lemma 1(a). Since mηwe can consider the space M(Yb(η), L1(m)) endowed with the seminorm fM(Yb(η),L1(m)) =sup{hfm:h∈BYb(η)}.Note that M(Yb(η), L1(m)) may be not a B.f.s. (see Section 2.1). The following result shows that the intersection of this space with L1(m)with the natural norm (the maximum of the seminorm and the norm) coincides isometrically with our space L1 Y(η)(m). Proposition 10. The equality L1 Y(η)(m)=L1(m)∩MYb(η), L1(m) holds and mfY(η)=fM(Yb(η),L1(m)) for all f ∈L1 Y(η)(m). Proof. Let f∈L1 Y(η)(m).Inparticular f∈L1(m)and so, by Lemma 7, we have that fϕm⩽ϕY(η)mfY(η)for every ϕ∈S(Σ) ∩Y(η).Givenh∈Yb(η),wecantake(ϕn)⊂S(Σ) ∩Y(η)such that ϕn→hin Y(η)and η-a.e. pointwise (and so m-a.e.). Since fϕn−fϕkm⩽ϕn−ϕkY(η)mfY(η)→0 as n,k→∞,thereexistsg∈L1(m)such that fϕn→gin L1(m). Then, for some subsequence (ϕnk)we have that fϕnk→g m-a.e. and so fh=gm-a.e. Thus, fh∈L1(m)and fhm=lim nfϕnm⩽mfY(η)lim nϕnY(η)=mfY(η)hY(η). Therefore, f∈M(Yb(η), L1(m)) and fM(Yb(η),L1(m)) ⩽mfY(η). Conversely, let f∈L1(m)∩M(Yb(η), L1(m)).Foreveryϕ∈S(Σ) ∩BY(η)we have that ΩϕfdmE⩽ϕfm⩽ fM(Yb(η),L1(m)). Then, mfY(η)⩽fM(Yb(η),L1(m)) <∞and so f∈L1 Y(η)(m).2 In the case when χΩ∈Y(η),wehavethatM(Yb(η), L1(m)) ⊂L1(m). Then, from Proposition 10 and Remark 9, it follows that L1 Y(η)(m)=M(Yb(η), L1(m)) isomorphically (isometrically if χΩY(η)⩽1). Let us show some examples of subspaces of L1(m)generated by particular B.f.s.’ Y(η). Example 11. Consider Y(η)=Lp(η)with 1 ⩽p<∞.SinceLp(η)is an order continuous B.f.s., from Proposition 10 and Lemma 1(c), we have that L1 Lp(η)(m)=L1(m)∩M(Lp(η), L1(m)) isometrically. If ηis finite (and so χΩ∈Lp(η)), we have that L1 Lp(η)(m)=M(Lp(η), L1(m)). Under certain requirements, this space coincides isomorphically with the space L1 p,η(m)(1 p+1 p=1) studied in [3], given by all functions f∈L1(m)such that the so called p-semivariation of mfwith respect to ηis finite, that is, mfp,η=sup π∈P(Ω) sup x∗∈BE∗ A∈π |Afdx ∗m|p η(A)p−11/p <∞, where P(Ω) is the set of the finite disjoint partitions πof Ω.Namely,ifp>1, R=Σ,mis equivalent to ηand the p-semivariation of mwith respect to ηis finite, then L1 Lp(η)(m)=L1 p,η(m)isomorphically, see [3, Theorem 3]. In particular, the Lp(η)-semivariation and the p-semivariation of mare related as follows mp,η⩽mLp(η)⩽21/pmax1,η(Ω)1/pmp,η. 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