Positive Representations of L1 of a Vector Measure
Abstract
We characterize the vector measures n on a Banach lattice such that the map |·|dn provides a quasi-norm which is equivalent to the canonical norm · n of the space L1(n) of integrable functions as an specific type of transformations of positive vector measures that we call cone-open transformations. We also prove that a vector measure m on a Banach space X constructed as a cone-open transformation of a positive vector measure can be considered in some sense as a positive vector measure by defining a new order on X.
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Positive Representations of L1 of a Vector Measure Ricardo del Campo and Enrique A. S´anchez-P´erez Abstract. We characterize the vector measures non a Banach lattice such that the map |·|dnprovides a quasi-norm which is equivalent to the canonical norm · nof the space L1(n) of integrable functions as an specific type of transformations of positive vector measures that we call cone-open transformations. We also prove that a vector measure mon a Banach space Xconstructed as a cone-open transformation of a positive vector measure can be considered in some sense as a positive vector measure by defining a new order on X. Mathematics Subject Classification (2000). 46G10; 46E30. Keywords. Vector measures, integration, positive operators. 1. Introduction and notation Let m:Σ→Xbe a (countably additive) vector measure, where Xis a Banach lattice. It is said that mis positive if m(A)∈C+ Xfor every A∈Σ, where C+ Xis the positive cone of X. It is well-known that in this case the norm of the space L1(m) of integrable functions with respect to mcan be computed by the formula fm= |f|dm ,f∈L1(m),(1) (see Lemma 1.2 for a short proof of this result). The aim of this paper is to characterize the class of vector measures for which this formula gives an equivalent norm – or at least a quasi-norm – for L1(m). A problem related to this one has been recently studied in [5, Section 4], where a technique to construct optimal representations of L1(m) of a positive vector measure mhas been introduced. The motivation of the problem is given by the fact that any order continuous Banach function space with a weak unit can be written as L1(m) of a vector measure m ([3, Theorem 8]). Therefore it seems natural to analyze when there is a representation of L1(m) using a vector measure n(i.e. L1(m)=L1(n)) such that the norm
can be computed in an easy way using formula (1). In order to develop our analysis, we define and study a particular class of transformations of vector measures defined by means of linear operators preserving certain order relations that we call cone-open transformations. Actually, we prove in Section 2 that the class of vector measures msatisfying that formula (1) gives a quasi-norm for L1(m) coincides with the class of cone-open transformations of positive vector measures. It is easy to find examples of vector measures for which the expression given in (1) is not a norm for L1(m). For instance, if ν:Σ→Ris a signed measure (with non-trivial positive and negative parts) then there is always a non-zero function f∈L1(ν) such that |f|dν = 0, and then the map f→ |f|dν does not give a norm for L1(ν). Although there are a lot of examples of non-positive vector measures msuch that formula (1) gives a norm for L1(m) (see for instance Example 2.1), we study if in the case that this property holds it is possible to define a suitable order on X such that mcan be considered in some sense as a positive vector measure. We also answer in the positive this question, although further requirements for the coneopen transformations involved are needed; in particular, we need to introduce the notion of lattice generating operator. This is done in Section 3. In the rest of this section we give some definitions and basic results. Let Xbe a real Banach space with dual X∗,(Ω,Σ) a measurable space and m:Σ−→ Xa vector measure. For any x∗∈X∗we define m, x∗to be the scalar measure given by m, x∗(A):=m(A),x ∗, for all A∈Σ. Following Lewis in [6] we introduce the notion of integrable function with respect to m. Definition 1.1. A measurable function fis integrable with respect to mif (i) fis integrable with respect to the scalar measure m, x∗, for all x∗∈X∗. (ii) For every A∈ΣthereexistsanelementAfdm ∈Xsuch that A fdm,x∗=A fdm, x∗. It is well-known that this definition is equivalent to the one given by Bartle, Dunford and Schwartz in [2]. The space L1(m) is the Banach lattice (of classes) of integrable functions with respect to mequipped with the norm fm:= sup x∗∈BX∗|f|d|m, x∗|,f∈L1(m). where |m, x∗| denotes the variation of the measure m, x∗In fact, L1(m)isa K¨othe function space over any measure of the type μ=|m, x∗| that satisfies the Rybakov Theorem (see [7, 1.b.17] for the definition of K¨othe function space and [4, IX.2] for the Rybakov Theorem). The order in L1(m) is the usual μ-almost everywhere order. Note that L1(m) reduces to the ordinary space of Lebesgue integrable functions if the measure mis scalar. We will use standard notation of Riesz spaces and Banach lattices which can be found in [1] and [8]. We recall that a lattice is a partially ordered set (X, ≤)
such that every subset consisting of two elements has supremum and infimum. We denote by x∨yand x∧ythe supremum and the infimum of {x, y}, respectively. On the other hand, (X, ≤) is said to be an ordered vector space if Xis a vector space and ≤is a partial order compatible with the algebraic structure of X, i.e., (i) x≤yimplies x+z≤y+zfor every x, y, z ∈X,and (ii) x≥0 implies αx ≥0 for each real number α≥0andforeveryx∈X. An ordered vector space that is also a lattice is called a Riesz space.If(X, ≤) is a Riesz space, we write C+ Xfor its positive cone,thatis,theset{x∈X:x≥0}. Given x∈Xthe positive part x+,thenegative part x−, and the absolute value |x| of xare respectively defined by x+:= x∨0, x−:= (−x)∨0, |x|:= x∨(−x)and they verify x=x+−x−and |x|=x++x−. Anormin·on a Riesz space Xis a lattice norm if |x|≤|y|implies x≤yfor all x, y ∈X. A Riesz space equipped with a lattice norm is called a normed Riesz space and a complete normed Riesz space is called a Banach lattice. We also recall that a quasi-norm on a vector space Xis any map · :X−→ R+verifying the following properties: (i) x= 0 if and only if x=0. (ii) αx=|α|x, for all α∈Rand x∈X. (iii) There exists M≥1 such that x+y≤M(x+y), for all x, y ∈X. Note that if the constant Mis equal to 1 then ·is in fact a norm. To finish this introductory section and for the purpose of completeness we prove the following result. Lemma 1.2. If Xis a Banach lattice and m:Σ−→ Xis a positive vector measure, then fm=|f|dm,∀f∈L1(m). Proof. Since mis positive and the integration map Im:L1(m)→Xis continuous, it is also positive. Therefore, for every f∈L1(m) the element |f|dm belongs to C+ Xand then fm≥sup x∗∈BX∗ |f|dm, x∗= |f|dm =sup x∗∈BX∗ |f|dm, |x∗| =sup x∗∈BX∗|f|d|m, |x∗|| ≥ fm, since X∗is also a Banach lattice and thus x∗=|x∗| for every x∗∈X∗. 2. The first representation theorem Although the positivity of the measure mprovides the alternative formula for the norm of L1(m) given in Lemma 1.2, the converse is not true: there are non-positive measures such that formula (1) also gives the norm · m. Let us show this with an example.
Example 2.1.Consider the Lebesgue measure space ([0,1],Σ,μ) and the vector measure τ:Σ→2given by τ(A):= ∞ n=1 (−1)n 2nμ(An∩A)en,A∈Σ, where {An:n=1,2,...}is a non-trivial measurable partition of [0,1] and {en: n=1,2,...}is the canonical basis of 2. A direct calculation shows that, although the measure is clearly not positive, |·|dτ=· τ. The purpose of this section is to describe the class of measures nyielding a quasi-norm |·|dnequivalent to · n. To this end, we introduce the notion of cone-open transformation of a measure. Definition 2.2. Let Xbe a normed Riesz space and Ya normed space. A linear and continuous operator S:X−→ Yis called a cone-open operator (resp. a cone-isometry) if there exists K≥0 such that S(x)≥Kx,∀x∈C+ X(resp. S(x)=x,∀x∈C+ X). We also introduce the dual notion in the following sense: let Xand Ybe two normed spaces and assume that Xis also a Riesz space. A linear operator S:X−→ Yis said to be cone-continuous if there exists Q≥0 such that S(x)≤Qx,∀x∈C+ X. However, whenever Xis a normed Riesz space, it is easy to see that if Sis cone-continuous then Sis in fact continuous (and hence both concepts coincide). To see this, take an element xof X. Then S(x)=S(x+−x−)≤S(x+)+S(x−)≤2Q·max{x+,x−}≤2Q·x, since |x+|≤|x|,|x−|≤|x|and ·is a lattice norm. Definition 2.3. Given a vector measure m:Σ−→ Xand a cone-open operator S:X−→ Y, we will say that n:= S◦mis a cone-open transformation of the measure m. Proposition 2.4. Let Xbe a Banach lattice and Ya Banach space. If m:Σ−→ X is a vector measure and S:X−→ Yis a linear and continuous operator, then n:= S◦mis also a vector measure, L1(m)⊂L1(n),andAfdn =S(Afdm), ∀f∈L1(m),∀A∈Σ. Moreover, if Sis cone-open (resp. cone-isometry) and mis positive then n and mare equivalent vector measures. Thus, |·|dnis a quasi-norm (resp. norm) on L1(m). Proof. The first part of the proposition is well-known. Given f∈L1(m)and y∗∈Y∗, it is clear that m, S∗(y∗)=n, y∗. Since fis integrable with respect to m, S∗(y∗)we have that fis integrable with respect to n, y∗for all y∗∈Y∗, and from
SA fdm,y∗=A fdm,S∗(y∗)=A fdm, S∗(y∗)=A fdn, y∗, for every y∗∈Y∗, we conclude that f∈L1(n) with Afdn =S(Afdm). Note that m(A) = 0 always implies that n(A) = 0 and, if Sis cone-open and mis positive, the converse is also true. Moreover, in this case, it is clear that |f|dn= 0 if and only if f=0,|αf|dn=|α| |f|dnand, since there exist K, Q ≥0 such that Kx≤S(x)≤Qxfor all x∈C+ X, we deduce that |f+g|dn = S|f+g|dm ≤Q |f+g|dm ≤Q |f|dm + |g|dm ≤Q K S|f|dm + S|g|dm ≤Q K |f|dn + |g|dn . Therefore |·|dmis a quasi-norm (and in fact a norm if and only if Q=K, that is, if Sis a cone-isometry). Proposition 2.4 guarantees that every cone-open transformation of a positive measure yield a quasi-norm |·|dnon L1(m). It is a natural matter to study the relation of these quasi-norms with the canonical norm on L1(n). To finish this section we characterize vector measures for which the function f→ |f|dm gives an equivalent expression for the norm of L1(m). Proposition 2.5. Let m:Σ−→ Xbe a positive vector measure, S:X−→ Ybe a cone-open operator and n:= S◦m.Then: (i) The quasi-norm |·|dnis equivalent to the norm |·|dmon L1(m). (ii) ·nand ·mare equivalent norms on L1(m). In particular, L1(n)=L1(m). Proof. Since Sis continuous and cone-open, there exist K, Q ≥0 such that Kx≤S(x)≤Qx,∀x∈C+ X. Applying these inequalities to x=|f|dm ∈ C+ X,foreachf∈L1(m) it follows that |·|dn and |·|dm are equivalent on L1(m). To prove (ii), we will work with the following equivalent norms of L1 (see [3]): · n∼sup A∈Σ A ·dn and · m∼sup A∈Σ A ·dm Thus, there exist convenient constants K,K,Q ,Q such that fn≤Qsup A∈Σ A fdn ≤Qsup A∈Σ S(A fdm) ≤Q sup A∈Σ A fdm ≤Qfm
and fm≤Ksup A∈Σ A fdm ≤Ksup A∈Σ A f+dm +sup A∈Σ A f−dm ≤ ≤K sup A∈Σ A f+dn +sup A∈Σ A f−dn ≤2K sup A∈Σ A fdn ≤2Kfn which yield the equivalence between both norms. Using these results, we will prove that the cone-open transformations of positive measures are precisely the measures nfor which the map |·|dnis a quasi-norm describing the topology of L1(n), that is Theorem 2.6. Let Ybe a normed space and n:Σ−→ Ybe a vector measure. The following statements are equivalent: (i) The function |·|dnis a quasi-norm equivalent to (resp. a norm coinciding with) the norm · non L1(n). (ii) There exist a Banach lattice X, a positive vector measure m:Σ−→ Xand a cone-open operator (resp. cone-isometry) S:X−→ Ysuch that n=S◦m. Moreover, in this case, L1(n)and L1(m)are isomorphic Banach lattices. Proof. (i) ⇒(ii) Setting X:= L1(n), m(A):=χA,∀A∈ΣandS(f):=fdn, ∀f∈Xwe have a positive vector measure m:Σ−→ Xand a cone-open operator (resp. cone-isometry) S:X−→ Ysuch that n=S◦m. (ii) ⇒(i) By proposition 2.5 and lemma 1.2 we conclude that ·n∼ ·m= |·|dm∼|·|dnon L1(m)=L1(n). (The equivalence is an equality if Sis an isometry.) 3. The second representation theorem In this section we study if it is possible to define an order on the image of the vector measure min such a way that if the expression |·|dmis a quasi-norm for L1(m), then mcan be considered, in a sense, as a positive vector measure. This happens for instance in Example 2.1, where τis positive whenever the new order (λi)∞ i=1 ≤(ηi)∞ i=1 iff λi≤ηifor i=2,4,... and λi≥ηifor i=1,3,5,... is considered in 2. Therefore, a natural question arises: is this in general true?, i.e. is it always possible to define a new order on the space such that mis positive with respect to this order? We will see that cone-open transformations of positive vector measures are closely related to positive vector measures since cone-open operators induce a natural order on its range which is partially compatible with the normed space structure in the following precise sense:
Proposition 3.1. Let Xbe a Banach lattice, let S:X−→ Ybe a cone-open operator and let Zbe the range rg(S)of S. The relation z1≤Sz2⇔∃x∈C+ X: z2−z1=S(x)defines an order in Zunder which Zis an ordered vector space. Moreover, there exists M≥1such that if z1,z 2∈Z,z1≥S0,z2≥S0and z1≤Sz2then z1≤Mz2(and M=1if and only if Sis an cone-isometry). Proof. Since z−z=0=S(0),∀z∈Z, the relation ≤Sis reflexive. Given z1≤Sz2 and z2≤Sz3there exist x1,x 2∈C+ Xsuch that z2−z1=S(x1)andz3−z2=S(x2). Thus, we have z3−z1=S(x1+x2) with x1+x2∈C+ X,thatis,z1≤Sz3, and consequently the relation ≤Sis transitive. To prove that ≤Sis also antisymmetric, assume that z1≤Sz2and z2≤Sz1. Thus, there exist x1,x 2∈C+ X such that z2−z1=S(x1)andz1−z2=S(x2) which implies that S(x1+x2)=0. Since Sis cone-open we conclude that x1=x2= 0 and hence z1=z2. The compatibility of this order with the vector space structure of Zfollows directly from the definition of the order and the linearity of S. Finally, given z1,z 2∈Z,z1≥S0, z2≥S0andz1≤Sz2, there exist x1,x 2,x 3∈C+ Xsuch that z1=S(x1), z2=S(x2)andz2−z1=S(x3). In addition, there exist K, Q ≥0 such that Kx≤S(x)≤Qx,∀x∈C+ X. Therefore z2=S(x3+x1)≥Kx3+x1≥Kx1≥K QS(x1)=K Qz1 and thus M:= Q Kverifies the required property. Definition 3.2. Given M≥1, a normed space Xwith norm ·is called an M- normed Riesz space if there exists an order ≤such that Xis a Riesz space and x1≤Mx2holds for every x1,x 2∈Xwith x1≥0, x2≥0andx1≤x2. Definition 3.3. A vector measure n:Σ−→ Yon a normed space Yis called pseudo-positive if there exists an order ≤in Ysuch that Yis an M-normed Riesz space and nis positive for this order. Definition 3.4. Let Xbe a normed Riesz space and let Ybe a normed space. Let S:X−→ Ybe a cone-open operator and let ≤Sbe the order induced by Sin Z:= rg(S). We will say that Sis lattice generating if for all z1,z 2∈Zthere exist x1,x 2∈Xsuch that (i) z1=S(x1), z2=S(x2), and (ii) ∀z∈Z:z≥Sz1,z≥Sz2,∃h∈X:S(h)=z, h ≥x1∨x2. Example 3.5.A in a sense canonical example of a lattice generating cone-open transformation is the integration map f→ Iμ(f)=fdμ with respect to a probability measure μ,Iμ:L1(μ)→R. To see this is enough to consider, for every element r∈R, the function r·χΩ. Taking these functions as the associated elements x1,x 2∈L1(μ)toz1=r1,z 2=r2,r1,r 2∈Rin Definition 3.4, it is clear that (ii) in this definition is satisfied for h:= max{r1,r 2}·χΩ. Other simple example is given by the identity map S=Id :X→Xin any Banach lattice X. In this case, S(x1)=x1,S(x2)=x2and clearly (ii) is satisfied for h:= x1∨x2.
Corollary 3.6. The range of every cone-open and lattice generating operator is an M-normed Riesz space. Therefore every cone-open and lattice generating transformation of a positive vector measure is a pseudo-positive vector measure. Proof. It is sufficient to prove that z1∨Sz2is well defined by S(x1∨x2) for every z1,z 2∈rg(S) independently of x1,x 2satisfying (i) and (ii) of Definition 3.4. Let us check it. Let x1,x 2and x 1,x 2two couples associated to the elements z1and z2satisfying (i) and (ii) above. Then S(x1∨x2)=S(x 1∨x 2). To see this, let us define z=S(x1∨x2)andz=S(x 1∨x 2), and let us prove that z=z. Since clearly S(x 1∨x 2)≥Sz1=S(x 1)=S(x1)andS(x 1∨x 2)≥S z2=S(x 2)=S(x2), we have that there is an element hsuch that S(h)=z and h≥x1∨x2. Therefore, the positive element r:= h−x1∨x2satisfies that S(h)−S(x1∨x2)=S(r), and then z≥Sz. The same argument can be given for proving that z≥Sz, that implies that z=zas a consequence of the fact that Sis cone-open and then ≤Sis an order relation by Proposition 3.1. In the opposite direction we prove in Theorem 3.7 that pseudo-positive measures are precisely this particular type of cone-open transformations of positive vector measures, that is: Theorem 3.7. If n:Σ−→ Yis a pseudo-positive vector measure then ncan be written as a cone-open and lattice generating transformation of a positive vector measure (into a normed Riesz space). This result can be deduced from the following facts: (i) If Yis an M-normed Riesz space with norm ·then |||y||| := ||(|y|)||,∀y∈Y defines a quasi-norm on Ysuch that ||y|| ≤ 2M|||y||| for all y∈Y. (ii) If Yis an M-normed Riesz space with norm ·then |||y|||0:= inf{|||z||| :|y|≤|z|},∀y∈Y is a lattice norm on Ywhich is equivalent to ||| · |||. In fact, |||y|||0≤ |||y||| ≤ M|||y|||0for all y∈Y (iii) If Yis an M-normed Riesz space with norm · then the identity map Id :(Y,||| · |||0)−→ (Y,·) is a (continuous) cone-open operator. (iv) If Yis an M-normed Riesz space, Xis a normed space, T:Y−→ Xis a linear and cone-continuous operator and n ynis a convergent series of positive terms yn∈Y, then n T(yn) is also a convergent series in Xand T∞ n=1 yn= ∞ n=1 T(yn)
(v) If n:Σ−→ Yis a pseudo-positive vector measure and T:Y−→ Xis cone-continuous then ˜n:= T◦nis a vector measure into X. Proof. (i) It is clear that |||y||| = 0 if and only if y= 0 and that |||αy||| =|α| |||y||| for all α∈Rand for all y∈Y. Moreover, given x, y ∈Ywe have |||x+y||| =(|x+y|)≤M((|x|)+(|y|))=M(|||x||| +|||y|||) and since y+≤|y|,y −≤|y|it follows that y+≤M|||y|||,y−≤M|||y||| and hence y=y+−y−≤y++y−≤2M|||y|||. (ii) Let us check that ||| · |||0is a norm. Only the triangle inequality is not obvious. Now take >0 and consider two elements x, y ∈Y. Then there are zx,z y∈Ysuch that |x|≤|zx|and |y|≤|zy|,and|||zx||| ≤ |||x|||0+ 2and |||zy||| ≤ |||y|||0+ 2. Then |x+y|≤|zx|+|zy|,andso |||x+y|||0≤ ||| |zx|+|zy| ||| =|zx|+|zy| ≤ |||zx||| +|||zy||| ≤ |||x|||0+|||y|||0+. Moreover, if |x|≤|y|then for any z∈Ysuch that |y|≤|z|in particular we have |x|≤|z|and hence |||x|||0≤ |||z|||.Thus|||x|||0≤inf{|||z||| :|y|≤|z|} = |||y|||0, which proves that ||| · |||0is in fact a lattice norm. The equivalence between ||| · ||| and ||| · |||0is easy to check. On the one hand it is evident that |||y|||0≤ |||y||| for all y∈Y, and on the other hand given any z∈Ywith |y|≤|z|it follows that |||y||| ≤ M|||z||| and hence |||y||| ≤ M|||y|||0 for all y∈Y. (iii) Id :(Y,||| · |||0)−→ (Y,·) is a cone-open operator since for all y∈Y, Id(y)=y≤2M|||y||| ≤ 2M2|||y|||0 and for all y∈Ywith y≥0, Id(y)=y=|||y||| ≥ |||y|||0. (iv) Since Tis cone-continuous, there exists Q≥0 such that T(y)≤Qy, for all y≥0. Thus, given n∈Nwe have T∞ k=1 yk− n k=1 T(yk) = ∞ k=n+1 yk which converges to 0 when ngoes to infinity. (v) ˜nis countably additive since for all pairwise disjoint Ai∈Σ, n(Ai)≥0, ˜n∞ i=1 Ai=T◦n∞ i=1 Ai=T∞ i=1 n(Ai)= ∞ i=1 T◦n(Ai)= ∞ i=1 ˜n(Ai).