674 CHAPTER 21. THE BOCHNER INTEGRAL

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}must be a Cauchy sequence in C ([0,T ] ;X) andthis requires that x equals a continuous function a.e. Thus C ([0,T ] ;X) consists exactly ofthose functions, 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

}. (21.5.24)

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) .

674 CHAPTER 21. THE BOCHNER INTEGRALand let ® € C2 (—T,2T) such that &(t) = 1 fort € [0,7]. Then you could letOx (t) = P(t) x(t).Then let {@,,} be a mollifier and defineWx (t) = On * Ox(t).It follows y,,x is uniformly continuous because[| Yn) — Vn (0) [Ix[lon (=s) nt =9)] 1x0) vasciy (|. lon(! 9) 9404-9 }%as)Also for x € C((0,7];X), it follows from usual mollifier arguments thatlAIAI| Wx — llr ((0,7):x) > 0.Here is why. For t € [0,7],Winx (t)—x(Ilx S [9 (6) |8x(0—s) — x (0) lasIAQa1/n/ ; ,, (5) ds€ =Cg€J—1/nprovided n is large enough due to the compact support and consequent uniform continuityof Ox.If || Wx —x||;=((0,r];x) — 9, then {y,,x} must be a Cauchy sequence in C ({0,T];X) andthis requires that x equals a continuous function a.e. Thus C ([0,7];X) consists exactly ofthose functions, x of L? ([0,7];X) such that ||y,,x —x||,, > 0. It followsC((0,7];X) =1Wat Um=t =m {1 € LP ((0,T];X) : ||Wpx—-l|.. < “. (21.5.24)It only remains to showS= {xe L? ((0,7];X) : || Wyx—xl],, < a}is a Borel set. Suppose then that x, € S and x, — x in L? ((0,7];X). Then there exists asubsequence, still denoted by n such that x, — x pointwise a.e. as well as in L?. Thereexists a set of measure 0 such that for all n, and f not in this set,1/k4 (8) (OXn (t — 5)) ds — Xn (1)<aI| Warn (t) — Xn (2) “1kXn(t) > x(t).