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POSITIVE OPERATORS

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1.2. Extensions of Positive Operators 27<br />

is bounded. on the other hand, if gn = fχ ( 1<br />

n ,1), then T (gn) =lnn ∈ D holds<br />

for each n. Therefore, D must be unbounded, a contradiction. Consequently,<br />

in this case we have E(T )=○.<br />

A positive operator T : G → F (where G is a vector subspace of an<br />

ordered vector space E) issaidtohaveasmallest extension whenever<br />

there exists some S ∈E(T ) satisfying S ≤ R for all R ∈E(T ), in which case<br />

S is called the smallest extension of T . In other words, T has a smallest<br />

extension if and only if min E(T ) exists in L(E,F).<br />

It turns out that an extendable positive operator whose domain is an<br />

ideal always has a smallest extension.<br />

Theorem 1.30. Let E and F be two Riesz spaces with F Dedekind complete,<br />

let A be an ideal of E, andletT : A → F be a positive operator. If E(T ) = ○,<br />

then T has a smallest extension. Moreover, if in this case S =minE(T ),<br />

then<br />

S(x) =sup Ty: y ∈ A and 0 ≤ y ≤ x <br />

holds for all x ∈ E + .<br />

Proof. Since T has (at least) one positive extension, the formula<br />

TA(x) =sup T (y): y ∈ A and 0 ≤ y ≤ x , x ∈ E + ,<br />

defines a positive operator from E to F satisfying TA = T on A, andso<br />

TA ∈E(T ). (See the proof of Theorem 1.28.)<br />

Now if S ∈E(T ), then S = T holds on A, and hence TA = SA ≤ S.<br />

Therefore, TA =minE(T ) holds, as desired.<br />

For a positive operator T : E → F with F Dedekind complete, Theorem<br />

1.30 implies that for each ideal A of E the positive operator TA is the<br />

smallest extension of the restriction of T to A.<br />

Among the important points of a convex set are its extreme points.<br />

Recall that a vector e of a convex set C is said to be an extreme point of<br />

C whenever the expression e = λx +(1−λ)y with x, y ∈ C and 0

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