POSITIVE OPERATORS

1.1. Basic Properties of Positive Operators 11
whenever T − S is a positive operator (i.e., whenever T (x) ≥ S(x) holds for
all x ∈ E + ) is an ordered vector space.
Definition 1.12. For an operator T : E → F between two Riesz spaces
we shall say that its modulus |T | exists (or that T possesses a modulus)
whenever
|T | := T ∨ (−T )
exists—in the sense that |T | is the supremum of the set {−T,T} in L(E,F).
In order to study the elementary properties of the modulus, we need a
decomposition property of Riesz spaces.
Theorem 1.13 (The Decomposition Property). If |x| ≤|y1+···+yn| holds
in a Riesz space, then there exist x1,...,xn satisfying x = x1 + ···+ xn and
|xi| ≤|yi| for each i =1,...,n. Moreover, if x is positive, then the xi also
can be chosen to be positive.
Proof. By using induction it is enough to establish the result when n =2.
So, let |x| ≤|y1 + y2|.
Put x1 = x ∨ (−|y1|) ∧|y1|, and observe that |x1| ≤|y1| (and that
0 ≤ x1 ≤ x holds if x is positive). Now put x2 = x − x1 and observe that
x2 = x − x ∨ (−|y1|) ∧|y1| = 0 ∧ (x + |y1|) ∨ (x −|y1|) .
On the other hand, |x| ≤|y1| + |y2| implies −|y1|−|y2| ≤x ≤|y1| + |y2|,
from which it follows that
−|y2| =(−|y2|) ∧ 0 ≤ x + |y1| ∧ 0 ≤ x2 ≤ 0 ∨ x −|y1| ≤|y2| .
Thus, |x2| ≤|y2| also holds, and the proof is finished.
An important case for the modulus to exist is described next.
Theorem 1.14. Let T : E → F be an operator between two Riesz spaces
such that sup |Ty|: |y| ≤x exists in F for each x ∈ E + . Then the
modulus of T exists and
|T |(x) =sup |Ty|: |y| ≤x
holds for all x ∈ E + .
Proof. Define S : E + → F + by S(x) =sup |Ty|: |y| ≤x for each x in
E + . Since |y| ≤x implies |±y| = |y| ≤x, it easily follows that we have
S(x) =sup Ty: |y| ≤x for each x ∈ E + . We claim that S is additive.
To see this, let u, v ∈ E + . If |y| ≤ u and |z| ≤ v, then |y + z| ≤
|y| + |z| ≤u + v, and so it follows from T (y)+T (z) =T (y + z) ≤ S(u + v)
that S(u) +S(v) ≤ S(u + v). On the other hand, if |y| ≤u + v, then by
Theorem 1.13 there exist y1 and y2 with |y1| ≤u, |y2| ≤v, andy = y1 + y2.