POSITIVE OPERATORS

1.1. Basic Properties of Positive Operators 13
all regular operators (which is the same as the vector subspace generated by
the positive operators), then the following vector subspace inclusions hold:
Lr(E,F) ⊆Lb(E,F) ⊆L(E,F) .
Of course, Lr(E,F) andLb(E,F) with the ordering inherited from L(E,F)
are both ordered vector spaces. For brevity L(E,E), Lb(E,E), and Lr(E,E)
will be denoted by L(E), Lb(E) andLr(E), respectively.
The inclusion Lr(E,F) ⊆Lb(E,F) can be proper, as the next example
of H. P. Lotz (oral communication) shows.
Example 1.16 (Lotz). Consider the operator T : C[−1, 1] → C[−1, 1] defined
for each f ∈ C[−1, 1] by
Tf(t) =f sin 1
1
t − f sin t + t
if 0 < |t| ≤1andTf(0) = 0. Note that the uniform continuity of f, coupled
with the inequality 1
1
sin( t ) − sin(t + t ) ≤ |t|, shows that Tf is indeed
continuous at zero, and so indeed Tf ∈ C[−1, 1] for each f ∈ C[−1, 1].
Next, observe that T [−1, 1] ⊆ 2[−1, 1] holds, where 1 denotes the constant
function one on [−1, 1]. Since for every f ∈ C[−1, 1] there exists some
λ>0with|f| ≤λ1, it easily follows that T is an order bounded operator.
However, we claim that T is not a regular operator. To see this, assume
by way of contradiction that some positive operator S : C[−1, 1] → C[−1, 1]
satisfies T ≤ S. We claim that for each 0 ≤ f ∈ C[−1, 1] we have
[Sf](0) ≥ f(t) for all t ∈ [−1, 1] . (⋆)
To establish this, fix 0