688 CHAPTER 24. THE BOCHNER INTEGRAL

Lemma 24.9.1 Let B = B(0,L) be a closed ball in L∞ (0,T,H) . Then B is a Polishspace with respect to the weak ∗ topology. The closure is taken with respect to the usualtopology.

Proof: Let {zk}∞

k=1 = X be a dense countable subspace in L1 (0,T,H) . You start witha dense countable set and then consider all finite linear combinations having coefficients inQ. Then the metric on B is

d (f,g)≡∞

∑k=1

2−k

∣∣∣⟨f −g,zk⟩L∞,L1

∣∣∣1+∣∣∣⟨f −g,zk⟩L∞,L1

∣∣∣Is B complete? Suppose you have a Cauchy sequence {fn} . This happens if and only if{⟨fn,zk⟩}∞

n=1 is a Cauchy sequence for each k. Therefore, there exists

ξ (zk) = limn→∞⟨fn,zk⟩ .

Then for a,b ∈Q, and z,w ∈ X

ξ (az+bw) = limn→∞⟨fn,az+bw⟩= lim

n→∞a⟨fn,z⟩+b⟨fn,w⟩= aξ (z)+bξ (w)

showing that ξ is linear on X a dense subspace of L1 (0,T,H). Is ξ bounded on this densesubspace with bound L? For z ∈ X ,

|ξ (z)| ≡ limn→∞|⟨fn,z⟩| ≤ lim sup

n→∞

∥fn∥L∞ ∥z∥L1 ≤ L∥z∥L1

Hence ξ is also bounded on this dense subset of L1 (0,T,H) . Therefore, there is a uniquebounded linear extension of ξ to all of L1 (0,T,H) still denoted as ξ such that its normin L1 (0,T,H)′ is no larger than L. It follows from the Riesz representation theorem thatthere exists a unique f ∈ L∞ (0,T,H) such that for all w ∈ L1 (0,T,H) , ξ (w) = ⟨f,w⟩and ∥f∥ ≤ L. This f is the limit of the Cauchy sequence {fn} in B. Thus B is complete.

Is B separable? Let f ∈ B. Let ε > 0 be given. Choose M such that ∑∞k=M+1 2−k < ε

4 .Then the finite set {z1, · · · ,zM} is uniformly integrable. There exists δ > 0 such that ifm(S) < δ , then

∫S |zk|H dm <

(ε

4(1+∥f∥L∞ )

)Then there is a sequence of simple functions

{sn} which converge uniformly to f off a set of measure zero, N, ∥sn∥L∞ ≤ ∥f∥L∞ . Byregularity of the measure, there exists a continuous function with compact support hn suchthat sn = hn off a set of measure no more than δ/4n and also ∥hn∥L∞ ≤ ∥f∥L∞ . Then offa set of measure no more than 1

3 δ , hn (r)→ f (r). Now by Eggorov’s theorem and outerregularity, one can enlarge this exceptional set to obtain an open set S of measure no morethan δ/2 such that the convergence is uniform off this exceptional set. Thus f equals theuniform limit of continuous functions on SC. Define

h(r)≡

 limn→∞hn (r) = f (r) on SC

0 on S\N0 on N

Then ∥h∥L∞ ≤ ∥f∥L∞ . Now consider h̄∗ψm (r) where ψr is approximate identity.

ψm (t) =12

mX[−1/m,1/m] (t) , h̄∗ψm (t)

=12

m∫ 1/m

−1/mh̄(t− s)ds =

12

m∫ t+1/m

t−1/mh̄(s)ds