13.3. THE DUAL SPACE OF Lp (Ω) 327

Lemma 13.3.3 Let (Ω,S ,µ) be a measure space and suppose there exists a measur-able function, r such that r (x) > 0 for all x, there exists M such that |r (x)| < M for all x,and

∫rdµ < ∞. Then for Λ ∈ (Lp(Ω,µ))′, p ≥ 1, there exists 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). Here1p +

1q = 1.

Proof: Define a new measure µ̃ , according to the rule

µ̃ (E)≡∫

Erdµ. (13.5)

Thus µ̃ is a finite measure on S . For

Λ ∈ (Lp (µ))′ ,Λ( f ) = Λ

(r1/p

(r−1/p f

))= Λ̃

(r−1/p f

)where Λ̃(g)≡ Λ

(r1/pg

). Now Λ̃ is in Lp (µ̃)′ because

∣∣∣Λ̃(g)∣∣∣ ≡ ∣∣∣Λ(r1/pg

)∣∣∣≤ ∥Λ∥(∫Ω

∣∣∣r1/pg∣∣∣p dµ

)1/p

= ∥Λ∥

∫Ω

|g|pdµ̃︷︸︸︷

rdµ

1/p

= ∥Λ∥∥g∥Lp(µ̃)

Therefore, by Theorems 13.3.2 and 13.3.1 there exists a unique h∈ Lq (µ̃) which representsΛ̃. Here q = ∞ if p = 1 and satisfies 1/q+1/p = 1 otherwise. Then

Λ( f ) = Λ̃

(r−1/p f

)=∫

Ω

h f r−1/prdµ =∫

Ω

f(

hr1/q)

dµ

Now hr1/q ≡ h̃ ∈ Lq (µ) since h ∈ Lq (µ̃). In case p = 1,Lq (µ̃) and Lq (µ) are exactly thesame. In this case you have Λ( f ) = Λ̃

(r−1 f

)=∫

Ωh f r−1rdµ =

∫Ω

f hdµ Thus the desiredrepresentation holds. Then in any case,|Λ( f )| ≤

∥∥h̃∥∥

Lq ∥ f∥Lp so ∥Λ∥ ≤∥∥h̃∥∥

Lq . Also, asbefore,

∥∥h̃∥∥q

Lq(µ)=

∣∣∣∣∫Ω

h̃∣∣h̃∣∣q−2 h̃dµ

∣∣∣∣= ∣∣∣Λ(∣∣h̃∣∣q−2 h̃)∣∣∣≤ ∥Λ∥(∫

Ω

∣∣∣|h̃|q−2h̃∣∣∣p dµ

)1/p

= ∥Λ∥(∫

Ω

(∣∣h̃∣∣q/p)p)1/p

= ∥Λ∥∥h∥q/p

and so∥∥h̃∥∥

Lq(µ)≤ ∥Λ∥ ≤

∥∥h̃∥∥

Lq(µ). It works the same for p = 1. Thus

∥∥h̃∥∥

Lq(µ)= ∥Λ∥ . ■

A situation in which the conditions of the lemma are satisfied is the case where themeasure space is σ finite. In fact, you should show this is the only case in which theconditions of the above lemma hold.

Theorem 13.3.4 (Riesz representation theorem) Let (Ω,S ,µ) be σ finite and letΛ ∈ (Lp(Ω,µ))′, p ≥ 1. Then there exists a unique h ∈ Lq(Ω,µ), L∞(Ω,µ) if p = 1 suchthat Λ f =

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

p +1q = 1.