BANDS OF REGULAR OPERATORS
ROMULUS CRISTESCU
We consider regular operators defined on a directed vector space, with values in an (o)-complete ordered vector space. We show that some subsets of a space of regular operators are bands. In this way, we obtain some decomposition of regular operators. The terminology used is that of [1] and [2], but some definitions are also given here.
AMS 2000 Subject Classification: 47B60.
Key words: directed vector space, regular operator, band.
1. As in [3] and [5], we call space of type (R) any directed vector space which satisfies the Riesz condition ([2]).
If Z is an ordered vector space, we denote by Z+ the set of all positive elements ofZ.
Let X be a space of type (R) and Y an ordered vector space which is (o)-complete (i.e. any majorized subset ofY has a supremum). We denote by R(X, Y) the set of all regular operators which map X into Y. With respect to the usual operations and the order relation given by the cone of positive operators,R(X, Y) is a complete vector lattice ([3]).
IfE is a subset of the spaceX, we shall denote Z(E) ={U ∈ R(X, Y)|U(E) ={0}}. The next theorem generalizes a result given in [4].
Theorem 1. If 0∈E ⊂X+ and if E is a full set, then the set Z(E) is a band of the space R(X, Y).
Proof. The equality
(1) Z(E) ={U ∈ R(X, Y)| |U|(E) ={0}}
holds. Indeed, denote by E the right-hand side of (1), and let U ∈ Z(E). If a∈E then [0, a]⊂E and, since
U+(x) = supU([0, x]), ∀x∈X+,
REV. ROUMAINE MATH. PURES APPL.,52(2007),5, 555–561
we have U+(a) = 0. We also have U(a) = 0, therefore U−(a) = 0. Conse- quently|U|(a) =0, thereforeU ∈ E. Conversely, if|U|(E) =0then it follows from±U ≤ |U|that U(a) =0,∀a∈E, henceU ∈ Z(E).
By equality (1) it is easily verified thatZ(E) is a normal subspace of the space R(X, Y). On the other hand, if Uδ ↑
δ∈∆ U in the space R(X, Y) with 0≤Uδ∈ Z(E), ∀δ∈∆,then obviously U(a) =0, ∀a∈E, hence U ∈ Z(E). It results thatZ(E) is a band.
Corollary 1.If a vector subspace Gof the space Xis a directed and full set andU ∈ R(X, Y),then U can be uniquely written in the formU =U1+U2, where U1 ∈ Z(G) and U2 ∈ Z(G)⊥.
Indeed, we have Z(G) = Z(G+) and G+ is a full set. On the other hand, R(X, Y) being a complete vector lattice, any band of R(X, Y) is a component ([2]).
Corollary 2. If 0<U0 ∈ R(X, Y) and U ∈ R(X, Y) then U can be written in the form U =U1+U2, with U1, U2 ∈ R(X, Y) such that U2 ⊥U0
and U1(x) =0 if x∈X+ and U0(x) =0.
Indeed, the setE ={x∈X+ |U0(x) =0}is a full set.
Remark 1. IfA⊂X+ then the set
A={x∈X+ |0 =x∧a, ∀a∈A}
is a wedge and a full set. The set G=A−Ais a vector subspace ofX and a directed and full set.
Lemma. Let Gand H be vector subspaces of the space X. If Gand H are directed and full sets and G∩H={0},then
(2) (G+H)+=G++H+.
In particular, if a∈G, b∈H and a+b≥0,then a≥0,and b≥0. Proof. Let z ∈(G+H)+. Accordingly, z ≥0 and it can be written in the form
z=
a−a +
b−b
witha, a ∈G+ and b, b ∈H+. We have a+b ≤a+b, whence 0≤a≤a+b.
By the Riesz condition satisfied by the space X, the element a can be written in the form a = c1 +c2 with 0 ≤ c1 ≤ a and 0 ≤ c2 ≤ b. Since b ∈ H, we have c2 ∈H. But, we also have 0 ≤c2 ≤ a with a ∈ G, hence c2∈G. It results thatc2=0, therefore a =c1 ≤a.
In a similar manner, from0≤b ≤a+b we obtain b≤b.
Therefore, we have 0 ≤ a −a ∈ G and 0 ≤ b −b ∈ H, whence z ∈ G+ +H+. Consequently, (G+H)+ ⊂ G+ +H+. Since the opposite inclusion is obvious, we obtain (2).
In the sequel we shall use the following definitions ([6], [3]).
Definition 1. A vector subspaceGof a directed vector spaceZ is called adirected vector subspace if it enjous the property that ifa∈Gand ifa≤x∈ Z+, then there exists b∈G+such that a≤b≤x. A directed vector subspace ofZ which also is a full set, is called anideal ofZ.
Definition 2. We say that two vector subspacesG, Hof a directed vector space Z are a d-pair of subspaces if G∩H = {0} and G+H is a directed vector subspace of the spaceZ.
Theorem 2.Let GandHbe two ideals of the spaceXwhich are a d-pair of subspaces. If U ∈ R(X, Y)then U can be written in the form U =U1+U2
with U1, U2∈ R(X, Y) and
(3) U1(G) =U2(H) ={0}. Proof. First, we shall show that
(4) Z(G)⊥⊂ Z(H).
As we have seen in Theorem 1, we have
(5) Z(H) ={U ∈ R(X, Y) | |U|(H+) ={0}}
since Z(H) = Z(H+). Let 0 < U ∈ Z(G)⊥ and suppose that U /∈ Z(H).
By (5), let then0< x0∈H such thatU(x0)=0.By putting V0(a+b) =U(b), a∈G, b∈H,
and denoting E =G+H, we obtain a positive linear operator V0 :E → Y. By putting now
P(x) =U(x), ∀x∈X+,
we obtain a monotone sublinear operatorP :X+→Y+ for which (6) V0(x)≤P(x), ∀x∈E+.
Indeed, if 0 ≤ x =a+b with a ∈G and b ∈H, then by Lemma 1 we have a ≥ 0 (and b ≥ 0), whence we obtain (6). Consequently [5, Theorem 1.3], putting
P(x) = inf{P(v)| ±x≤v∈X+}, ∀x∈X,
there exists a positive linear operatorV :X→Y such thatV|E =V0 and V(x)≤P(x), ∀x∈X.
If x ∈ X+ then 0 ≤ V(x) ≤ P(x) = U(x), hence 0 ≤ V ≤ U. Since U ∈ Z(G)⊥, we have V ∈ Z(G)⊥ and V =0 asV (x0) =U(x0)=0. On the other hand, ifa∈G+thenV(a) =0, henceV ∈ Z(G). This is a contradiction and, consequently,U ∈ Z(H). If0=U ∈ Z(G)⊥ butU is not positive, then we take into account that|U| ∈ Z(G)⊥, hence |U| ∈ Z(H) as we have seen.
It then resultsU+, U−∈ Z(H), thereforeU ∈ Z(H).Consequently, inclusion (4) holds.
By Corollary 1 of Theorem 1, if U ∈ R(X, Y) then U can be written in the form U = U1 +U2 with U1 ∈ Z(G) and U2 ∈ Z(G)⊥, hence with U2∈ Z(H) by (4). Consequently, equalities (3) hold.
Remark 2. We can obtain an analogue of Theorem 2 for (ω)-continuous regular operators. A generalized sequence (xδ)δ∈∆ of elements of an ordered vector space Z is said [3] to (ω)-converge to an element x if there exists a generalized sequence (vλ)λ∈Λ of elements ofZ such thatvλ ↓
λ∈Λ0 and for any λ∈Λ there exists δ0 ∈∆ such that
±(xδ−x)≤vλ, ∀δ≥δ0.
The meaning of the term (ω)-continuous operator is obvious. We denote by Rω(X, Y) the space of all (ω)-continuous regular operators which mapX into Y and put
Zω(G) =Z(G)∩ Rω(X, Y)
for any subspaceGof X. IfGand H are as in Theorem 2, then we have Zω(G)⊥⊂ Zω(H)
andZω(G) is a band of the spaceR(X, Y). Consequently, if in Theorem 2 we takeU ∈ Rω(X, Y), thenU =U1+U2 withU1, U2 ∈ Rω(X, Y) satisfying (3).
2. If Z is an ordered vector space, we denote by IBo(Z) the set of all (o)-bounded subsets of Z.
If Z is an ordered vector space endowed with a vector topology τ, we shall denote by IBτ(Z) (resp. IBτ(Z+)) the set of all (τ)-bounded subset of Z (resp. of Z+).
A directed vector space Z endowed with a vector topology τ is said [2]
to haveproperty (S) if∀A∈IBτ(Z), ∃A0 ∈IBτ(Z+), A⊂A0−A0. It is known that any topological vector lattice has property (S).
In this section we shall suppose that X is a space of type (R) endowed with a vector topology τ and thatY is an (o)-complete ordered vector space.
We denote by(X, Y) the set of all linear operators mappingX intoY. An operator U : X → Y is said to be (τ o)-bounded ([1]) if U(A) ∈ IBo(Y), ∀A∈IBτ(X). We denote byMτo(X, Y) the set of all (τ o)-bounded linear operators mappingX intoY.
Theorem 3. If the space X is endowed with a locally full topology τ, then the set
N(X, Y) ={U ∈(X, Y) |U(A)∈IB0(Y), ∀A∈IBτ(X+)} is a normal subspace of the space R(X, Y).
Proof. We have
(7) N(X, Y)⊂ R(X, Y).
Indeed, ifU ∈ N(X, Y) and E ∈IBo(X+) then E∈IBτ(X+), hence U(E)∈ IBo(Y). IfA∈IBo(X) then there exists a∈X+ such that
A⊂[0, a]−[0, a]
whence it results that if U ∈ N(X, Y) then U(A) ∈ IBo(Y), hence U ∈ R(X, Y) ([3], 4.2.1, Theorem 2). Therefore, inclusion (7) holds.
It is easily verified thatN(X, Y) is a vector subspace of the spaceR(X, Y), and it is also obsious that ifU1 ∈ R(X, Y), U2 ∈ N(X, Y) and 0≤U1 ≤U2, thenU1 ∈ N(X, Y).
Let now U ∈ N(X, Y) and let us consider the positive part U+ of U in the spaceR(X, Y). Let A∈IBτ(X+). We can suppose that 0∈A and that Ais a full set (otherwise we consider the full hull ofA∪ {0}). Lety0∈Y such thatU(z)≤y0, ∀z∈A. Ifx∈Athen [0, x]⊂A, so that
0≤U+(x) = supU([0, x])≤y0,
hence U+ ∈ N(X, Y). It results that N(X, Y) is a normal subspace of the spaceR(X, Y).
Corollary. If the space X is endowed with a locally full topology with property (S),then Mτo(X, Y) is a normal subspace of the space R(X, Y).
If not only X but also the space Y is endowed with a vector topology, we denote byL(X, Y) the set of all linear operators which mapX intoY and are continuous with respect to the topologies given on X and Y.
Remark 3. If the topology in the space X is given by a norm and ifY is endowed with an (o)-continuous locally full topology ([2]), thenMτo(X, Y)⊂ L(X, Y).
Remark 4. If X is a Banach lattice, Y a complete vector lattice en- dowed with an (ω)-continuous locally convex-solid topology ([2]) andL(X, Y) is directed, then (by a result in [3] and the corollary of Theorem 3) L(X, Y) is a band of the space R(X, Y) and Mτo(X, Y) is a normal subspace of the spaceL(X, Y).
Remark 5. IfXis an Archimedean vector lattice with strong unit and the associated norm, then for any complete vector latticeZwe haveMτo(X, Z) = R(X, Z).
Definition 3. We call directed topology on a directed vector space Z, a vector topologyτ which is locally full and which satisfies the condition that if (τ)-lim
δ∈∆xδ =0in the spaceZ, then there exists a generalized sequence (cδ)δ∈∆
with values inZ such that ±xδ≤cδ, ∀δ ∈∆, and (τ)-lim
δ∈∆cδ=0.
Remark 6. Any directed topology is locally solid. On a vector lattice, a vector topology is locally solid if and only if it is a directed topology.
Remark 7. By a theorem given in [3, 5.1.4], if X is endowed with a directed locally convex topology τ, then the set Xτ∗ of (τ)-continuous linear functionals defined on X is a normal subspace of the space X# of regular functionals. In particular,Xτ∗ is a complete vector lattice.
The next theorem generalizes a result of Kawai [7] (see also [1, 6.2.8]).
Theorem 4. If X is endowed with a separated directed locally convex topology τ and Gis a vector subspace of X which also is a directed and full set, then the polar G0 of Gis a band of the space Xτ∗.
Proof. As in the case of locally convex lattices, the equality (8) G0={f ∈Xτ∗ | |f|(x) = 0, ∀x∈G+}
holds. Indeed, denoting byH the right-hand side of (8), the inclusionG0 ⊂H is based on the equality
|f|(x) = supf([−x, x]), ∀x∈X+
and the inclusionH⊂G0 is based on the fact that if x∈Gthen x=x−x with x, x ∈ G+. By (8), it is obvious that G0 is a normal subspace of the spaceXτ∗. Iffδ ↑
δ∈∆f inXτ∗ with 0≤fδ ∈G0, ∀δ∈∆,then, by Remark 7, it is immediate thatf ∈G0. It results that G0 is a band.
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Received 18 October 2006 University of Bucharest
Faculty of Mathematics and Computer Science Str. Academiei 14
010014 Bucharest, Romania
