Kenneth Kuttler

626 CHAPTER 23. REPRESENTATION THEOREMS

Corollary 23.2.9 Suppose (Ω,S ) is a measure space and µ is a finite nonnegativemeasure on S . Then for h ∈ L1 (µ) , define a complex measure, λ by λ (E) ≡

∫E hdµ.

Then |λ |(E) =∫

E |h|dµ. Furthermore, |h|= gh where gd |λ | is the polar decomposition ofλ , defined by λ (E) =

∫E gd |λ |.

Proof: From Corollary 23.2.8, there exists g such that |g| = 1, |λ | a.e. and for allE ∈S

λ (E) =∫

Egd |λ | , λ (E)≡

∫E

hdµ, so∫

Ehdµ =

∫E

gd |λ | (23.1)

Since |g| = 1, there is a sequence sn of simple functions converging pointwise to g with|sn| ≤ 1. (Approximate the positive and negative parts of the real and imaginary parts ofḡ with an increasing sequence of simple functions. Then assemble these to get sn. SeeTheorem 10.7.6 on Page 286.) Then from 23.1,

∫E gsnd |λ | =

∫E snhdµ. Passing to the

limit using the dominated convergence theorem,∫

E d |λ |=∫

E ghdµ. It follows gh≥ 0 a.e.and |g|= 1 a.e. Therefore, |h|= |gh|= gh. It follows from the above, that

|λ |(E) =∫

Ed |λ |=

∫E

ghdµ =∫

E|h|dµ ■

Formally: If dλ = hdµ, and dλ = gd |λ | , |g| = 1, then d |λ | = |h|dµ and so you mightexpect to have dλ = g |h|dµ so g |h| = h and so |h| = ḡh. That which should be true is.Emphasizing the most significant part of this, if dλ = hdµ, then d |λ |= |h|dµ .

23.3 Representation for the Dual Space of Lp

Recall the concept of the dual space of a Banach space in the chapter on Banach spacestarting on Page 533. The next topic deals with the dual space of Lp for p ≥ 1 in the casewhere the measure space is σ finite or finite. In what follows q = ∞ if p = 1 and otherwise,1p +

1q = 1. In what follows, |·| is the usual norm on C.

Theorem 23.3.1 (Riesz representation theorem) Let ∞ > p > 1 and let (Ω,S ,µ)be a finite measure space. If Λ ∈ (Lp(Ω))′, then there exists a unique h ∈ Lq(Ω) such that

Λ f =∫

Ω

h f dµ .

This function satisfies ∥h∥q = ∥Λ∥ where ∥Λ∥ is the operator norm of Λ.

Proof: (Uniqueness) If h1 and h2 both represent Λ, consider

f = |h1−h2|q−2(h1−h2),

where h denotes complex conjugation. By Holder’s inequality, it is easy to see that f ∈Lp(Ω). Thus 0 = Λ f −Λ f =

∫h1|h1− h2|q−2(h1− h2)− h2|h1− h2|q−2(h1− h2)dµ =∫

|h1 − h2|qdµ . Therefore h1 = h2 and this proves uniqueness in every case regarlesswhether µ is finite.

Now let λ (E) = Λ(XE). Since this is a finite measure space, XE is an element ofLp (Ω) and so it makes sense to write Λ(XE). In fact λ is a complex measure having finitetotal variation. First I show that it is a measure. Then by Theorem 23.2.6, λ has finite totalvariation.

626 CHAPTER 23. REPRESENTATION THEOREMSCorollary 23.2.9 Suppose (Q,-/) is a measure space and y is a finite nonnegativemeasure on .Y. Then for h € L'(), define a complex measure, 4 by A(E) = frhdu.Then |A| (E) = fy |h| du. Furthermore, |h| = gh where gd |A| is the polar decomposition ofA, defined by 0 (E) = frgd|A|.Proof: From Corollary 23.2.8, there exists g such that |g| = 1,|A| a.e. and for allECSA(E)= | eda), A(E) = | hd, so [nau = [aia (23.1)Since |g| = 1, there is a sequence s, of simple functions converging pointwise to g with|s,| < 1. (Approximate the positive and negative parts of the real and imaginary parts ofg with an increasing sequence of simple functions. Then assemble these to get s,. SeeTheorem 10.7.6 on Page 286.) Then from 23.1, fp gsnd|A| = Jp snhd. Passing to thelimit using the dominated convergence theorem, [;,d|A| = fi, ghd. It follows gh > 0 ae.and |g| = 1 a.e. Therefore, |h| = |gh| = gh. It follows from the above, thatA\(E)= fdia\= [ ghau = [olduFormally: If dA = hdp, and dA = gd|A|,|g| = 1, then d|A| = |h|du and so you mightexpect to have dA = g|h|du so g|h| =h and so |h| = gh. That which should be true is.Emphasizing the most significant part of this, if dA = hdp, then d|A| = |h|du.23.3 Representation for the Dual Space of L’Recall the concept of the dual space of a Banach space in the chapter on Banach spacestarting on Page 533. The next topic deals with the dual space of L? for p > | in the casewhere the measure space is o finite or finite. In what follows q = © if p = | and otherwise,; + ; = 1. In what follows, |-| is the usual norm on C.Theorem 23.3.1 (Riesz representation theorem) Let «> p > 1 and let (Q,.7%, UL)be a finite measure space. If A € (L?(Q))’, then there exists a unique h € L1(Q) such thatAf= | hfdy.QThis function satisfies ||h\|q = \|A|| where \|Al|| is the operator norm of A.Proof: (Uniqueness) If h; and h2 both represent A, considerf= |hi—ha|* (iy ~ In),where h denotes complex conjugation. By Holder’s inequality, it is easy to see that f €L?(Q). Thus 0= Af —Af = fhi\hy —ho|* (Ay — hy) — holy — ho|?? (ty — ho) dp =f\f1 —A2\%dp. Therefore h, = hy and this proves uniqueness in every case regarlesswhether is finite.Now let A(E) = A(z). Since this is a finite measure space, 2 is an element ofL? (Q) and so it makes sense to write A (2%). In fact A is a complex measure having finitetotal variation. First I show that it is a measure. Then by Theorem 23.2.6, A has finite totalvariation.