636 CHAPTER 23. REPRESENTATION THEOREMS

and so ∥ν∥ ≤ ∥Λ∥ showing that ν ∈ BV (Ω; µ). Also from 23.7, if s = ∑nk=1 ckXEk is a

simple function,

∫sdν =

n

∑k=1

ckν (Ek) =n

∑k=1

ckΛ(XEk

)= Λ

(n

∑k=1

ckXEk

)= Λ(s)

Then letting f ∈ L∞ (Ω; µ) , there exists a sequence of simple functions converging to funiformly off a set of µ measure zero and so passing to a limit in the above with s replacedwith sn it follows that Λ( f ) =

∫f dν and so θ is onto. ■

23.6 Non σ Finite CaseIt turns out that for p> 1, you don’t have to assume the measure space is σ finite. The Rieszrepresentation theorem holds always. The proof involves the notion of uniform convexity.First recall Clarkson’s inequalities. These fundamental inequalities were used to verify thatLp (Ω) is uniformly convex. More precisely, the unit ball in Lp (Ω) is uniformly convex.

Lemma 23.6.1 Let 2≤ p. Then∥∥∥∥ f +g2

∥∥∥∥p

Lp+

∥∥∥∥ f −g2

∥∥∥∥p

Lp≤ 1

2(∥ f∥p

Lp +∥g∥pLp)

Let 1 < p < 2. then for 1/p+1/q = 1,∥∥∥∥ f +g2

∥∥∥∥q

Lp+

∥∥∥∥ f −g2

∥∥∥∥q

Lp≤(

12∥ f∥p

Lp +12∥g∥p

Lp

)q/p

Recall the following definition of uniform convexity.

Definition 23.6.2 A Banach space, X, is said to be uniformly convex if whenever∥xn∥ ≤ 1 and

∥∥ xn+xm2

∥∥→ 1 as n,m→∞, then {xn} is a Cauchy sequence and xn→ x where∥x∥ = 1. More precisely, for every ε > 0, there is a δ > 0 such that if ∥x+ y∥ > 2−δ for∥x∥ ,∥y∥ both 1, then ∥x− y∥< ε .

Observe that Clarkson’s inequalities imply Lp is uniformly convex for all p > 1. Con-sider the harder case where 1 < p. The other case is similar. Say ∥ f∥ = ∥g∥ = 1 and

∥ f +g∥Lp > 2−δ . Then from the second inequality(

2−δ

2

)q+∥∥∥ f−g

2

∥∥∥q

Lp≤ 1and so

∥ f −g∥qLp ≤ 2q

(1−(

2−δ

2

)q)< ε

provided δ is small enough.Uniformly convex spaces have a very nice property which is described in the following

lemma. Roughly, this property is that any element of the dual space achieves its norm atsome point of the closed unit ball.

Lemma 23.6.3 Let X be uniformly convex and let φ ∈ X ′. Then there exists x ∈ X suchthat ∥x∥= 1, φ (x) = ∥φ∥ .

636 CHAPTER 23. REPRESENTATION THEOREMSand so ||v|| < ||A|| showing that v € BV (Q;y). Also from 23.7, if s = Y~_) ce 2x, is asimple function,[sav = ¥ cv (Ex) = ¥ oA (Zin) =A e 2%) = A(s)k=1 k=1k=1Then letting f € L*(Q;U), there exists a sequence of simple functions converging to funiformly off a set of 6 measure zero and so passing to a limit in the above with s replacedwith s,, it follows that A(f) = f fdv and so @ is onto.23.6 Nono Finite CaseIt turns out that for p > 1, you don’t have to assume the measure space is o finite. The Rieszrepresentation theorem holds always. The proof involves the notion of uniform convexity.First recall Clarkson’s inequalities. These fundamental inequalities were used to verify thatLP (Q) is uniformly convex. More precisely, the unit ball in L? (Q) is uniformly convex.Lemma 23.6.1 Let 2 < p. Then|S) s;f+g2 2PL(lfllz» + lisllz»)NI]Pp LPLet 1 < p <2. then for 1/p+1/q=1,q q(53f+82 21 P 1 P q/P< (SIMs + 5 lelLP LPRecall the following definition of uniform convexity.Definition 23.6.2 A Banach space, X, is said to be uniformly convex if whenever\|xn|| <1 and |||] + 1 as n,m —> &, then {xy} is a Cauchy sequence and xy — x where||x|| = 1. More precisely, for every € > 0, there is a 6 > 0 such that if \|x+-y|| > 2—6 forI|x|| ,|ly|] Both J, then ||x—yl| <e.Observe that Clarkson’s inequalities imply L? is uniformly convex for all p > 1. Con-sider the harder case where 1 < p. The other case is similar. Say ||f|| = ||g|| = 1 and_ iy (2-8)4 4 || fe ||4\|f + ll» > 2—6. Then from the second inequality (5° } +]/45 ji < land soP2-6\!Ital s2*(1- (25°) J <eprovided 6 is small enough.Uniformly convex spaces have a very nice property which is described in the followinglemma. Roughly, this property is that any element of the dual space achieves its norm atsome point of the closed unit ball.Lemma 23.6.3 Let X be uniformly convex and let @ € X'. Then there exists x € X suchthat |\x|| = 1, 9 (x) = |||].