658 CHAPTER 21. THE BOCHNER INTEGRAL

Therefore, ∫Ω

yn (s)dµ ∈ D(A) ,∫

Ω

yn (s)dµ →∫

Ω

x(s)dµ in X ,

and since yn is a simple function and A is linear,

A∫

Ω

yn (s)dµ =∫

Ω

Ayn (s)dµ →∫

Ω

Ax(s)dµ in Y.

It follows, since A is a closed operator, that∫

Ωx(s)dµ ∈ D(A) and

A∫

Ω

x(s)dµ =∫

Ω

Ax(s)dµ.

Here is another version of this theorem which has different hypotheses.

Theorem 21.2.12 Let X and Y be separable Banach spaces and let A : D(A)⊆ X → Y bea closed operator. Also let (Ω,F ,µ) be a σ finite measure space and let x : Ω→ X beBochner integrable such that x(s) ∈D(A) for all s. Also suppose Ax is Bochner integrable.Then ∫

Axdµ = A∫

xdµ

and∫

xdµ ∈ D(A).

Proof: Consider the graph of A,

G(A)≡ {(x,Ax) : x ∈ D(A)} ⊆ X×Y.

Then since A is closed, G(A) is a closed separable Banach space with the norm ∥(x,y)∥ ≡max(∥x∥ ,∥y∥) . Therefore, for g∗ ∈ G(A)′ , one can apply the Hahn Banach theorem andobtain (x∗,y∗) ∈ (X×Y )′ such that g∗ (x,Ax) = (x∗ (x) ,y∗ (Ax)) . Now it follows from theassumptions that s→ (x∗ (x(s)) ,y∗ (Ax(s))) is measurable with values in G(A) . It is alsoseparably valued because this is true of G(A) . By the Pettis theorem, s→ (x(s) ,A(x(s)))must be strongly measurable. Also

∫∥x(s)∥+ ∥A(x(s))∥dµ < ∞ by assumption and so

there exists a sequence of simple functions having values in G(A) ,{(xn (s) ,Axn (s))}whichconverges to (x(s) ,A(s)) pointwise such that

∫∥(xn,Axn)− (x,Ax)∥dµ → 0 in G(A) .

Now for simple functions is it routine to verify that∫(xn,Axn)dµ =

(∫xndµ,

∫Axndµ

)=

(∫xndµ,A

∫xndµ

)Also ∥∥∥∥∫ xndµ−

∫xdµ

∥∥∥∥ ≤∫∥xn− x∥dµ

≤∫∥(xn,Axn)− (x,Ax)∥dµ