POSITIVE OPERATORS

1.1. Basic Properties of Positive Operators 21
The direct sum Σ⊕Ei (or more formally Σi∈I ⊕Ei) is the vector subspace
of ΠEi consisting of all vectors x = {xi} for which xi = 0 holds for all
but a finite number of indices i. With the pointwise algebraic and lattice
operations Σ ⊕ Ei is a Riesz subspace of ΠEi (and hence a Riesz space in
its own right). Note that if, in addition, each Ei is Dedekind complete, then
ΠEi and Σ ⊕ Ei are likewise both Dedekind complete Riesz spaces.
It is not difficult to see that every operator T :Σ⊕ Ei → Σ ⊕ Fj between
two direct sums of families of Riesz spaces can be represented by a matrix
T =[Tji], where Tji: Ei → Fj are operators defined appropriately. Sometimes
it pays to know that the algebraic and lattice operations represented
by matrices are the pointwise ones. The next result (whose easy proof is left
for the reader) clarifies the situation.
Theorem 1.24. Let {Ei : i ∈ I} and {Fj : j ∈ J} be two families of Riesz
spaces with each Fj Dedekind complete. If S =[Sji] and T =[Tji] are order
bounded operators from Σ ⊕ Ei to Σ ⊕ Fj, then
(1) S + T =[Sji + Tji] and λS =[λSji], and
(2) S ∨ T =[Sji ∨ Tji] and S ∧ T =[Sji ∧ Tji]
hold in Lb(Σ ⊕ Ei, Σ ⊕ Fj).
Exercises
1. Let E be an Archimedean Riesz space and let A ⊆ R be nonempty and
bounded above. Show that for each x ∈ E + the supremum of the set
Ax := {αx: α ∈ A} exists and sup(Ax) =(supA)x.
2. Show that in a Riesz space x ⊥ y implies
(a) αx ⊥ βy for all α, β ∈ R, and
(b) |x + y| = |x| + |y|.
Use the conclusion in (b) to establish that if in a Riesz space the
nonzero vectors x1,...,xn are pairwise disjoint, then x1,...,xn are linearly
independent. [Hint: If|x|∧|y| = 0, then
|x + y| ≥
|x|−|y| = |x|∨|y|−|x|∧|y|
= |x|∨|y| + |x|∧|y| = |x| + |y| ≥|x + y| .]
3. In this exercise we ask you to complete the missing details in Example
1.11. Let G be the lexicographic plane. (That is, we consider
G = R 2 as a Riesz space under the lexicographic ordering
(x1,x2) ≥ (y1,y2) whenever either x1 >y1 or else x1 = y1 and x2 ≥ y2.)
Also, let φ: R → R be an additive function that is not linear (i.e., not of
the form φ(x) =cx).
Show that the mapping T : R + → G + defined by
T (x) =(x, φ(x))