POSITIVE OPERATORS

1.1. Basic Properties of Positive Operators 15
In addition, Tα ↓ 0 in Lb(E,F) ifandonlyifTα(x) ↓ 0 in F for each
x ∈ E + .
Proof. Fix T ∈Lb(E,F). Since T is order bounded,
sup |Ty|: |y| ≤x =sup Ty: |y| ≤x =supT [−x, x]
exists in F for each x ∈ E + , and so by Theorem 1.14 the modulus of T
exists, and moreover
|T |(x) =sup Ty: |y| ≤x .
From Theorem 1.7 we see that Lb(E,F) is a Riesz space.
Now let S, T ∈Lb(E,F) andx ∈ E + . By observing that y, z ∈ E +
satisfy y + z = x if and only if there exists some |u| ≤x with y = 1
2 (x + u)
and z = 1
2 (x − u), it follows from Theorem 1.7 that
[S ∨ T ](x) = 1
2 Sx + Tx+ |S − T |x)
Sx + Tx+sup{(S − T )u: |u| ≤x}
= 1
2
= 1
2 supSx + Su + Tx− Tu: |u| ≤x
= sup S 1
2 (x + u) + T 1
2 (x − u) : |u| ≤x
= sup S(y)+T (z): y, z ∈ E + and y + z = x .
The formula for S ∧ T can be proven in a similar manner.
Finally, we establish that Lb(E,F) is Dedekind complete. To this end,
assume that 0 ≤ Tα ↑≤ T holds in Lb(E,F). For each x ∈ E + let S(x) =
sup{Tα(x)} and note that Tα(x) ↑ S(x). From Tα(x + y) =Tα(x)+Tα(y), it
follows (by taking order limits) that the mapping S : E + → F + is additive,
and so S defines a positive operator from E to F . Clearly, Tα ↑ S holds in
Lb(E,F), proving that Lb(E,F) is a Dedekind complete Riesz space.
From the preceding discussion it follows that when E and F are Riesz
spaces with F Dedekind complete, then each order bounded operator
T : E → F satisfies
T + (x) = sup Ty: 0 ≤ y ≤ x , and
T − (x) = sup −Ty: 0 ≤ y ≤ x
for each x ∈ E + . From T = T + − T − , it follows that Lb(E,F) coincides
with the vector subspace generated by the positive operators in L(E,F). In
other words, when F is Dedekind complete we have Lr(E,F)=Lb(E,F).
Recall that a subset D of a Riesz space is said to be directed upward
(in symbols D ↑) whenever for each pair x, y ∈ D there exists some z ∈ D
with x ≤ z and y ≤ z. The symbol D ↑ x means that D is directed upward