20.4. THE DUAL SPACE OF L∞ (Ω) 621

Note also that for s simple,∣∣∣∣∫ sdν

∣∣∣∣≤ ||s||L∞ |ν |(Ω) = ||s||L∞ ||ν ||

Next the dual space of L∞ (Ω; µ) will be identified with BV (Ω; µ). First here is a simpleobservation. Let ν ∈ BV (Ω; µ) . Then define the following for f ∈ L∞ (Ω; µ) .

Tν ( f )≡∫

f dν

Lemma 20.4.5 For Tν just defined,

|Tν f | ≤ || f ||L∞ ||ν ||

Proof: As noted above, the conclusion true if f is simple. Now if f is in L∞, then itis the uniform limit of simple functions off a set of µ measure zero. Therefore, by thedefinition of the Tν ,

|Tν f |= limn→∞|Tν sn| ≤ lim inf

n→∞||sn||L∞ ||ν ||= || f ||L∞ ||ν || .

Thus each Tν is in (L∞ (Ω; µ))′ .Here is the representation theorem, due to Kantorovitch, for the dual of L∞ (Ω; µ).

Theorem 20.4.6 Let θ : BV (Ω; µ)→ (L∞ (Ω; µ))′ be given by θ (ν) ≡ Tν . Then θ is oneto one, onto and preserves norms.

Proof: It was shown in the above lemma that θ maps into (L∞ (Ω; µ))′ . It is obviousthat θ is linear. Why does it preserve norms? From the above lemma,

||θν || ≡ sup|| f ||∞≤1

|Tν f | ≤ ||ν ||

It remains to turn the inequality around. Let π (Ω) be a partition. Then

∑A∈π(Ω)

|ν (A)|= ∑A∈π(Ω)

sgn(ν (A))ν (A)≡∫

f dν

where sgn(ν (A)) is defined to be a complex number of modulus 1 such that

sgn(ν (A))ν (A) = |ν (A)|

andf (ω) = ∑

A∈π(Ω)

sgn(ν (A))XA (ω) .

Therefore, choosing π (Ω) suitably, since || f ||∞≤ 1,

||ν ||− ε = |ν |(Ω)− ε ≤ ∑A∈π(Ω)

|ν (A)|= Tν ( f )

= |Tν ( f )|= |θ (ν)( f )| ≤ ||θ (ν)|| ≤ ||ν ||