POSITIVE OPERATORS

1.1. Basic Properties of Positive Operators 3
consists of equivalence classes rather than functions.) It is easy to see that
under the ordering f ≤ g whenever f(x) ≤ g(x) holds for µ-almost all x ∈ X,
each Lp(µ) is a Riesz space.
There are several useful identities that are true in a Riesz space some of
which are included in the next few results.
Theorem 1.3. If x, y and z are elements in a Riesz space, then:
(1) x ∨ y = − (−x) ∧ (−y)
and x ∧ y = − (−x) ∨ (−y) (2) x + y = x ∧ y + x ∨ y.
.
(3) x +(y∨ z) =(x + y) ∨ (x + z) and x +(y∧ z) =(x + y) ∧ (x + z).
(4) α(x∨y) =(αx)∨(αy) and α(x∧y) =(αx)∧(αy) for all α ≥ 0.
Proof. (1) From x ≤ x ∨ y and y ≤ x ∨ y we get −(x ∨ y) ≤−x and
−(x ∨ y) ≤−y, andso−(x ∨ y) ≤ (−x) ∧ (−y). On the other hand, if
−x ≥ z and −y ≥ z, then −z ≥ x and −z ≥ y, and hence −z ≥ x ∨ y. Thus,
−(x ∨ y) ≥ z holds and this shows that −(x ∨ y) is the infimum of the set
{−x, −y}. That is, (−x) ∧ (−y) =−(x ∨ y). To get the identity for x ∧ y
replace x by −x and y by −y in the above proven identity.
(2) From x ∧ y ≤ y it follows that y − x ∧ y ≥ 0andsox ≤ x + y − x ∧ y.
Similarly, y ≤ x + y − x ∧ y. Consequently, we have x ∨ y ≤ x + y − x ∧ y
or x ∧ y + x ∨ y ≤ x + y. On the other hand, from y ≤ x ∨ y we see that
x + y − x ∨ y ≤ x, and similarly x + y − x ∨ y ≤ y. Thus,x + y − x ∨ y ≤ x ∧ y
so that x + y ≤ x ∧ y + x ∨ y, and the desired identity follows.
(3) Clearly, x + y ≤ x + y ∨ z and x + z ≤ x + y ∨ z, and therefore
(x + y) ∨ (x + z) ≤ x + y ∨ z. On the other hand, we have y = −x +(x + y) ≤
−x +(x + y) ∨ (x + z), and likewise z ≤−x +(x + y) ∨ (x + z), and so
y ∨ z ≤−x +(x + y) ∨ (x + z). Therefore, x + y ∨ z ≤ (x + y) ∨ (x + z)
also holds, and thus x + y ∨ z =(x + y) ∨ (x + z). The other identity can be
proven in a similar manner.
(4) Fix α>0. Clearly, (αx) ∨ (αy) ≤ α(x ∨ y). If αx ≤ z and αy ≤ z
are both true, then x ≤ 1
1
1
αz and y ≤ αz also are true, and so x ∨ y ≤ αz. This implies α(x ∨ y) ≤ z, and this shows that α(x ∨ y) is the supremum of
the set {αx, αy}. Therefore, (αx) ∨ (αy) =α(x ∨ y). The other identity can
be proven similarly.
The reader can establish in a similar manner the following general versions
of the preceding formulas in (1), (3), and (4). If A is a nonempty
subset of a Riesz space for which sup A exists, then:
(a) The infimum of the set −A := {−a: a ∈ A} exists and
inf(−A) =− sup A.