24.10. WEAKLY CONVERGENT SEQUENCES 689

where we define h̄ to be the 0 extension of h̄ off [0,T ]. This is a continuous function of t.Also a.e.t is a Lebesgue point and so for a.e.t,∣∣∣∣12m

∫ t+1/m

t−1/mh̄(s)ds− h̄(t)

∣∣∣∣→ 0

∣∣h̄∗ψm (r)∣∣≡ ∣∣∣∣∫R h̄(r− s)ψm (s)ds

∣∣∣∣≤ ∥h∥L∞ ≤ ∥f∥L∞

Thus this continuous function is in L∞ (0,T,H). Letting z= zk ∈ L1 (0,T,H) be one ofthose defined above,∣∣∣∣∫ T

0

⟨h̄∗ψm (t)−f (t) ,z (t)

⟩dt∣∣∣∣≤ ∫ T

0

∣∣⟨h̄∗ψm (t)−h(t) ,z (t)⟩∣∣dt

+∫ T

0|⟨h(t)−f (t) ,z (t)⟩|dt (24.41)

for a.e. t, h̄∗ψm (t)−h(t)→ 0 and the integrand in the first integral in the above is boundedby 2∥f∥L∞ |z (t)|H so by the dominated convergence theorem, as m→ ∞, the first integralconverges to 0. As to the second, it is dominated by∫

S|⟨h(t)−f (t) ,z (t)⟩|dt ≤ 2∥f∥L∞

∫S|z (t)|dt <

2∥f∥L∞ ε

4(1+∥f∥L∞)≤ ε

2

Therefore, choosing m large enough so that the first integral on the right in 24.41 is lessthan ε

4 for each zk for k ≤M, then for each of these,

d(f,h̄∗ψm

)≤ ε

4+

M

∑k=1

2−k (ε/4)+(ε/2)1+((ε/4)+(ε/2))

=ε

4+

M

∑k=1

2−k 34

ε

34 ε +1

≤ ε

4+

3ε

4

M

∑k=1

2−k <ε

4+

3ε

4= ε

which appears to show that C ([0,T ] ,H) is weak ∗ dense in L∞ (0,T,H). However, this lastspace is obviously separable in terms of the norm topology. Let D be a countable densesubset of C ([0,T ] ,H). For f ∈ L∞ (0,T,H) let g ∈ C ([0,T ] ,H) such that d (f,g) < ε

4 .Then let h ∈ D be so close to g in C ([0,T ] ,H) that

M

∑k=1

2−k

∣∣∣⟨h−g,zk⟩L∞,L1

∣∣∣1+∣∣∣⟨h−g,zk⟩L∞,L1

∣∣∣ < ε

2

Then d (f,h) ≤ d (f,g)+ d (g,h) < ε

4 +ε

2 +ε

4 = ε It appears that D is dense in B in theweak ∗ topology. ■

24.10 Weakly Convergent SequencesThere is an interesting little result which relates to weak limits in L2 (Γ,E) for E a Banachspace. I am not sure where to put this thing but think that this would be a good place for it.It obviously generalizes to Lp spaces.