POSITIVE OPERATORS

1.1. Basic Properties of Positive Operators 17
and
yj =
n
zij , for each j =1,...,m.
i=1
Proof. We shall use induction on m. For m = 1 the desired conclusion
follows from Theorem 1.13. Thus, assume the result to be true for some m
and all n =1, 2,... .Let
n
xi =
i=1
m+1
where the vectors xi and the yj are all positive. Since m j=1 yj ≤ n i=1 xi
holds, it follows from Theorem 1.13 that there exist vectors u1,...,un satisfying
0 ≤ ui ≤ xi for each i =1,...,n and n i=1 ui = m j=1 yj. Therefore,
from our induction hypothesis, there exists a set of positive vectors
{zij : i =1,...,n; j =1,...,m} such that:
m
n
ui = zij for i =1,...,n and yj = zij for j =1,...,m.
j=1
For each i =1,...,n put zi,m+1 = xi − ui ≥ 0 and note that the collection
of positive vectors zij : i =1,...,n; j =1,...,m+1 satisfies
xi =
m+1
j=1
j=1
zij for i =1,...,n and yj =
yj ,
i=1
n
zij for j =1,...,m+1.
Thus, the conclusion is valid for m +1 and all n =1, 2,..., and the proof is
finished.
We are now in a position to express the lattice operations of Lb(E,F)
in terms of directed sets.
Theorem 1.21. If E and F are two Riesz spaces with F Dedekind complete,
then for all S, T ∈Lb(E,F) and each x ∈ E + we have:
n (1) S(xi) ∨ T (xi): xi ∈ E + n
and xi = x ↑ [S ∨ T ](x) .
(2)
(3)
i=1
n S(xi) ∧ T (xi): xi ∈ E + and
i=1
n |T (xi)|: xi ∈ E + and
i=1
i=1
i=1
n
i=1
xi = x
n
xi = x ↑|T |(x) .
i=1
↓ [S ∧ T ](x) .