19.3 Operator Valued Functions
Consider the case where A
separable Banach spaces.
With the operator norm ℒ
is a Banach space and so if
measurable, the Bochner integral can be defined as before. However, it is also
possible to define the Bochner integral of such operator valued functions for more
general situations. In this section,
will be a
finite measure space as
Lemma 19.3.1 Let x ∈ X and suppose A is strongly measurable. Then
is strongly measurable as a map into Y.
Proof: Since A is assumed to be strongly measurable, it is the pointwise limit of
simple functions of the form
where Akn is in ℒ
. It follows
x → A
for each s
and so, since
s → An
is a simple Y
valued function, s → A
must be strongly measurable.
Definition 19.3.2 Suppose A
for each s ∈
Ω where X,Y are separable
Banach spaces. Suppose also that for each x ∈ X,
and there exists C such that for each x ∈ X,
is defined by the following formula.
Lemma 19.3.3 The above definition is well defined. Furthermore, if 19.3.21 holds then
is measurable and if 19.3.22 holds, then
Proof: It is clear that in case s → A
is measurable for all x ∈ X
there exists a
unique Ψ ∈ℒ
This is because x →∫
is linear and continuous. Thus Ψ = ∫
definition is well defined.
Now consider the assertion about s →
the closed unit ball in Y ′
be such that D′
is countable and
Also let D be a countable dense subset of B, the unit ball of X. Then
and this is measurable because s → A
is strongly, hence weakly measurable.
Now suppose 19.3.22 holds. Then for all x,
It follows that for
This proves the lemma.
Now it is interesting to consider the case where A
s → A
is strongly measurable and A
is compact and self adjoint. Recall the
Kuratowski measurable selection theorem, Theorem
on Page 598
listed here for
Theorem 19.3.4 Let E be a compact metric space and let
be a measure space.
: E ×
Ω → ℝ has the property that x → ψ
is continuous and ω → ψ
is measurable. Then there exists a measurable function, f having values in E such
Furthermore, ω → ψ