20.3. REPRESENTATION THEOREMS FOR THE DUAL SPACE OF Lp 617

Theorem 20.3.6 (Riesz representation theorem) Let (Ω,S ,µ) be σ finite and let

Λ ∈ (Lp(Ω,µ))′, p≥ 1.

Then there exists a unique h ∈ Lq(Ω,µ), L∞(Ω,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: Let {Ωn} be a sequence of disjoint elements of S having the property that

0 < µ(Ωn)< ∞, ∪∞n=1Ωn = Ω.

Define

r(x) =∞

∑n=1

1n2 XΩn(x) µ(Ωn)

−1, µ̃(E) =∫

Erdµ .

Thus ∫Ω

rdµ = µ̃(Ω) =∞

∑n=1

1n2 < ∞

so µ̃ is a finite measure. The above lemma gives the existence part of the conclusion of thetheorem. Uniqueness is done as before.

With the Riesz representation theorem, it is easy to show that

Lp(Ω), p > 1

is a reflexive Banach space. Recall Definition 17.2.14 on Page 451 for the definition.

Theorem 20.3.7 For (Ω,S ,µ) a σ finite measure space and p > 1, Lp(Ω) is reflexive.

Proof: Let δ r : (Lr(Ω))′→ Lr′(Ω) be defined for 1r +

1r′= 1 by∫

(δ rΛ)g dµ = Λg

for all g ∈ Lr(Ω). From Theorem 20.3.6 δ r is one to one, onto, continuous and linear.By the open map theorem, δ

−1r is also one to one, onto, and continuous (δ rΛ equals the

representor of Λ). Thus δ∗r is also one to one, onto, and continuous by Corollary 17.2.11.

Now observe that J = δ∗p ◦δ

−1q . To see this, let z∗ ∈ (Lq)′, y∗ ∈ (Lp)′,

δ∗p ◦δ

−1q (δ qz∗)(y∗) = (δ ∗pz∗)(y∗)

= z∗(δ py∗)

=∫(δ qz∗)(δ py∗)dµ,