21.8. THE RIESZ REPRESENTATION THEOREM 687

Now ∫ ∣∣∣∣r 1p′ h∣∣∣∣p′ dµ =

∫|h|p

′rdµ = ∥h∥p′

Lp′ (µ̃)< ∞.

Thus∥∥∥∥r

1p′ h∥∥∥∥

Lp′ (µ)= ∥h∥Lp′ (µ̃) =

∥∥∥Λ̃

∥∥∥ = ∥Λ∥ and represents Λ in the appropriate way. If

p = 1, then 1/p′ ≡ 0. Now consider the existence of r. Since the measure space is σ finite,there exist {Ωn} disjoint, each having positive measure and their union equals Ω. Thendefine

r (ω)≡∞

∑n=1

1n2 µ(Ωn)

−1XΩn (ω)

This proves the Lemma.

Theorem 21.8.6 (Riesz representation theorem) Let (Ω,S ,µ) be σ finite and let X ′ havethe Radon Nikodym property. Then for

Λ ∈ (Lp(Ω;X ,µ))′, p≥ 1

there exists a unique h ∈ Lq(Ω,X ′,µ), L∞(Ω,X ′,µ) if p = 1 such that

Λ f =∫

h( f )dµ.

Also ∥h∥= ∥Λ∥. (∥h∥= ∥h∥q if p > 1, ∥h∥∞ if p = 1). Here

1p+

1q= 1.

Proof: The above lemma gives the existence part of the conclusion of the theorem.Uniqueness is done as before.

Corollary 21.8.7 If X ′ is separable, then for (Ω,S ,µ) a σ finite measure space,

(Lp (Ω;X))′ ∼= Lp′ (Ω;X ′

).

Corollary 21.8.8 If X is separable and reflexive, then for (Ω,S ,µ) a σ finite measurespace,

(Lp (Ω;X))′ ∼= Lp′ (Ω;X ′

).

Corollary 21.8.9 If X is separable and reflexive and (Ω,S ,µ) is a σ finite measurespace,then if p ∈ (1,∞) , then Lp (Ω;X) is reflexive.

Proof: This is just like the scalar valued case.