618 CHAPTER 20. REPRESENTATION THEOREMS

J(δ qz∗)(y∗) = y∗(δ qz∗)

=∫(δ py∗)(δ qz∗)dµ .

Therefore δ∗p ◦δ

−1q = J on δ q(Lq)′ = Lp. But the two δ maps are onto and so J is also onto.

20.4 The Dual Space Of L∞ (Ω)

What about the dual space of L∞ (Ω)? This will involve the following Lemma. Also recallthe notion of total variation defined in Definition 20.2.2.

Lemma 20.4.1 Let (Ω,F ) be a measure space. Denote by BV (Ω) the space of finitelyadditive complex measures ν such that |ν |(Ω)<∞. Then defining ||ν || ≡ |ν |(Ω) , it followsthat BV (Ω) is a Banach space.

Proof: It is obvious that BV (Ω) is a vector space with the obvious conventions involv-ing scalar multiplication. Why is ||·|| a norm? All the axioms are obvious except for thetriangle inequality. However, this is not too hard either.

||µ +ν || ≡ |µ +ν |(Ω) = supπ(Ω)

{∑

A∈π(Ω)

|µ (A)+ν (A)|}

≤ supπ(Ω)

{∑

A∈π(Ω)

|µ (A)|}+ sup

π(Ω)

{∑

A∈π(Ω)

|ν (A)|}

≡ |µ|(Ω)+ |ν |(Ω) = ||ν ||+ ||µ|| .

Suppose now that {νn} is a Cauchy sequence. For each E ∈F ,

|νn (E)−νm (E)| ≤ ||νn−νm||

and so the sequence of complex numbers νn (E) converges. That to which it converges iscalled ν (E) . Then it is obvious that ν (E) is finitely additive. Why is |ν | finite? Since ||·||is a norm, it follows that there exists a constant C such that for all n,

|νn|(Ω)<C

Let π (Ω) be any partition. Then

∑A∈π(Ω)

|ν (A)|= limn→∞

∑A∈π(Ω)

|νn (A)| ≤C.

Hence ν ∈ BV (Ω). Let ε > 0 be given and let N be such that if n,m > N, then

||νn−νm||< ε/2.