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POSITIVE OPERATORS

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1.1. Basic Properties of Positive Operators 19<br />

In particular, we have n m i=1 j=1 zij = x. On the other hand, using the<br />

lattice identity x ∨ y = 1<br />

2 (x + y + |x − y|), we see that<br />

Similarly,<br />

n<br />

S(xi) ∨ T (xi)<br />

i=1<br />

= 1<br />

2<br />

= 1<br />

2<br />

≤ 1<br />

2<br />

=<br />

n <br />

S(xi)+T (xi)+|S(xi) − T (xi)| <br />

i=1<br />

n m S(zij)+<br />

i=1<br />

j=1<br />

m m<br />

<br />

T (zij)+ S(zij) − T (zij) <br />

<br />

<br />

j=1<br />

j=1<br />

n m <br />

S(zij)+T (zij)+|S(zij) − T (zij)| <br />

i=1<br />

n<br />

i=1 j=1<br />

j=1<br />

m<br />

S(zij) ∨ T (zij) .<br />

m<br />

S(yj) ∨ T (yj) ≤<br />

j=1<br />

holds, and so D is directed upward.<br />

n<br />

i=1 j=1<br />

m<br />

S(zij) ∨ T (zij)<br />

(2) Use (1) in conjunction with the identity T ∧ S = − (−S) ∨ (−T ) .<br />

(3) Use (1) and the identity |T | = T ∨ (−T ).<br />

The next result presents an interesting local approximation property of<br />

positive operators.<br />

Theorem 1.22. Let T : E → F be a positive operator between two Riesz<br />

spaces with F Dedekind σ-complete. Then for each x ∈ E + there exists a<br />

positive operator S : E → F such that:<br />

(1) 0 ≤ S ≤ T .<br />

(2) S(x) =T (x).<br />

(3) S(y) =0for all y ⊥ x.<br />

Proof. Let x ∈ E + be fixed and define S : E + → F + by<br />

S(y) =sup T (y ∧ nx): n =1, 2,... .<br />

(The supremum exists since F is Dedekind σ-complete and the sequence<br />

{T (y ∧ nx)} is bounded above in F by Ty.) We claim that S is additive.<br />

To see this, let y, z ∈ E + .From(y + z) ∧ nx ≤ y ∧ nx + z ∧ nx we get<br />

T (y + z) ∧ nx ≤ T (y ∧ nx)+T (z ∧ nx) ≤ S(y)+S(z) ,

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