630 CHAPTER 23. REPRESENTATION THEOREMS

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

µ̃ (E)≡∫

Erdµ. (23.6)

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 23.3.4 and 23.3.1 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

)=∫

h f r−1/prdµ =∫

f(

hr1/q)

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

)=∫

h f r−1rdµ =∫

f hdµ

Thus the desired representation holds. Then in any case,|Λ( f )| ≤∥∥h̃∥∥

Lq ∥ f∥Lp so ∥Λ∥ ≤∥∥h̃∥∥

Lq . Also, as before,

∥∥h̃∥∥q

Lq(µ)=

∣∣∣∣∫Ω

h̃∣∣h̃∣∣q−2 h̃dµ

∣∣∣∣= ∣∣∣Λ(∣∣h̃∣∣q−2 h̃)∣∣∣≤ ∥Λ∥(∫

∣∣∣|h̃|q−2h̃∣∣∣p dµ

)1/p

= ∥Λ∥(∫

(∣∣h̃∣∣q/p)p)1/p

= ∥Λ∥∥h∥q/p

and so∥∥h̃∥∥

Lq(µ)≤ ∥Λ∥ . 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 23.3.6 (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.

Proof: Without loss of generality, assume µ (Ω) = ∞. By Proposition 10.13.1, either µ

is a finite measure or µ (Ω) =∞. These are the only two cases. Then let {Ωn} be a sequence

630 CHAPTER 23. REPRESENTATION THEOREMSProof: Define a new measure [, according to the ruleL(E) = | rd. (23.6)Thus pi is a finite measure on .Y. ForAe (LP (WY ALI =A (rll? (NPg)) =A( VPs)where A(g) = A(r!/?g) . Now A is in L? ((1)' becauseAw] = [a(r/re)| <tai(f, rl au)da \ 1/P= Alf [ielrmde | = LAlisllTherefore, by Theorems 23.3.4 and 23.3.1 there exists a unique / € L4 ({1) which representsA. Here q = © if p = | and satisfies 1/q+1/p = 1 otherwise. ThenA(A)=A(r VPP) = [npr verdu = | ¢ (ne'!*) agNow hr!/4 =h € L4(w) since h € L1 (ji). In case p = 1,L4 (fi) and L4 (1) are exactly thesame. In this case you haveAWN=A(r tp) = [npr indy = | prayThus the desired representation holds. Then in any case,|A(f)| < ||Al|,, |lfllz» so |All <Ne . Also, as before,~ ~ 1g-2F as . <|p 1/pVey =| fala “han| =a (\i! i) | <All (fie au)iin (f, (vit) ” =nand so lll coc) < ||A|]. It works the same for p = 1. Thus Alley = ||Al|. aA situation in which the conditions of the lemma are satisfied is the case where themeasure space is oO finite. In fact, you should show this is the only case in which theconditions of the above lemma hold.Theorem 23.3.6 (Riesz representation theorem) Let (Q,./,[) be © finite and letA € (L?(Q,))’, p > 1. Then there exists a unique h € L4(Q, UW), L*(Q, py) if p= 1 suchthat Af = [hfdp. Also \\h\| = Al]. ({\Al] = Allg FP > 1, \\hllo fp = 1). Here +7 = 1.Proof: Without loss of generality, assume pt (Q) = c. By Proposition 10.13.1, eitheris a finite measure or [1 (Q) = ce. These are the only two cases. Then let {Q,,} be a sequence