24.5. THE SPACES Lp (Ω;X) 675

and let Φ ∈C∞c (−T,2T ) such that Φ(t) = 1 for t ∈ [0,T ]. Then you could let

θx(t)≡Φ(t) x̃(t) .

Then let {φ n} be a mollifier and define

ψnx(t)≡ φ n ∗θx(t) .

It follows ψnx is uniformly continuous because∥∥ψnx(t)−ψnx(t ′)∥∥

X

≤∫R

∣∣φ n(t ′− s

)−φ n (t− s)

∣∣∥θx(s)∥X ds

≤ C∥x∥p

(∫R

∣∣φ n(t ′− s

)−φ n (t− s)

∣∣p′ ds)1/p′

Also for x ∈C ([0,T ] ;X) , it follows from usual mollifier arguments that

∥ψnx− x∥L∞([0,T ];X)→ 0.

Here is why. For t ∈ [0,T ] ,

∥ψnx(t)− x(t)∥X ≤∫R

φ n (s)∥θx(t− s)−θx(t)∥ds

≤ Cθ

∫ 1/n

−1/nφ n (s)dsε =Cθ ε

provided n is large enough due to the compact support and consequent uniform continuityof θx.

If ||ψnx− x||L∞([0,T ];X)→ 0, then {ψnx} is a Cauchy sequence in C ([0,T ] ;X) and thisrequires that x equals a continuous function a.e. Thus C ([0,T ] ;X) consists exactly of thosefunctions, x of Lp ([0,T ] ;X) such that ∥ψnx− x∥

∞→ 0. It follows

C ([0,T ] ;X) =

∩∞n=1∪∞

m=1∩∞k=m

{x ∈ Lp ([0,T ] ;X) : ∥ψkx− x∥

∞≤ 1

n

}. (24.25)

It only remains to show

S≡{

x ∈ Lp ([0,T ] ;X) : ∥ψkx− x∥∞≤ α

}is a Borel set. Suppose then that xn ∈ S and xn → x in Lp ([0,T ] ;X). Then there exists asubsequence, still denoted by n such that xn → x pointwise a.e. as well as in Lp. Thereexists a set of measure 0 such that for all n, and t not in this set,

∥ψkxn (t)− xn (t)∥ ≡∥∥∥∥∫ 1/k

−1/kφ k (s)(θxn (t− s))ds− xn (t)

∥∥∥∥≤ α

xn (t) → x(t) .