POSITIVE OPERATORS

1.2. Extensions of Positive Operators 27
is bounded. on the other hand, if gn = fχ ( 1
n ,1), then T (gn) =lnn ∈ D holds
for each n. Therefore, D must be unbounded, a contradiction. Consequently,
in this case we have E(T )=○.
A positive operator T : G → F (where G is a vector subspace of an
ordered vector space E) issaidtohaveasmallest extension whenever
there exists some S ∈E(T ) satisfying S ≤ R for all R ∈E(T ), in which case
S is called the smallest extension of T . In other words, T has a smallest
extension if and only if min E(T ) exists in L(E,F).
It turns out that an extendable positive operator whose domain is an
ideal always has a smallest extension.
Theorem 1.30. Let E and F be two Riesz spaces with F Dedekind complete,
let A be an ideal of E, andletT : A → F be a positive operator. If E(T ) = ○,
then T has a smallest extension. Moreover, if in this case S =minE(T ),
then
S(x) =sup Ty: y ∈ A and 0 ≤ y ≤ x
holds for all x ∈ E + .
Proof. Since T has (at least) one positive extension, the formula
TA(x) =sup T (y): y ∈ A and 0 ≤ y ≤ x , x ∈ E + ,
defines a positive operator from E to F satisfying TA = T on A, andso
TA ∈E(T ). (See the proof of Theorem 1.28.)
Now if S ∈E(T ), then S = T holds on A, and hence TA = SA ≤ S.
Therefore, TA =minE(T ) holds, as desired.
For a positive operator T : E → F with F Dedekind complete, Theorem
1.30 implies that for each ideal A of E the positive operator TA is the
smallest extension of the restriction of T to A.
Among the important points of a convex set are its extreme points.
Recall that a vector e of a convex set C is said to be an extreme point of
C whenever the expression e = λx +(1−λ)y with x, y ∈ C and 0