2366 CHAPTER 69. GELFAND TRIPLES

and for all t ∈ [0,T ] , ∫ t

0f ′ (s)ds ∈ H, (69.2.11)

and for a.e. t ∈ [0,T ] ,

f (t) = f (0)+∫ t

0f ′ (s)ds in H, (69.2.12)

Here f ′ is being taken in the sense of V ′ valued distributions and 1p +

1p′ = 1 and p≥ 2.

Proof: Let Ψ ∈C∞c (−T,2T ) satisfy Ψ(t) = 1 if t ∈ [−T/2,3T/2] and Ψ(t) ≥ 0. For

t ∈ R, define

f̂ (t)≡{

f (t)Ψ(t) if t ∈ [−T,2T ]0 if t /∈ [−T,2T ]

and

fn (t)≡∫ 1/n

−1/nf̂ (t− s)φ n (s)ds (69.2.13)

where φ n is a mollifier having support in (−1/n,1/n) . Then by Minkowski’s inequality∣∣∣∣∣∣ fn− f̂∣∣∣∣∣∣

Lp(R;V )=

(∫R

∣∣∣∣∣∣∣∣ f̂ (t)−∫ 1/n

−1/nf̂ (t− s)φ n (s)ds

∣∣∣∣∣∣∣∣pV

dt)1/p

=

(∫R

∣∣∣∣∣∣∣∣∫ 1/n

−1/n

(f̂ (t)− f̂ (t− s)

)φ n (s)ds

∣∣∣∣∣∣∣∣pV

dt)1/p

≤(∫

R

(∫ 1/n

−1/n

∣∣∣∣∣∣ f̂ (t)− f̂ (t− s)∣∣∣∣∣∣

Vφ n (s)ds

)p

dt)1/p

≤∫ 1/n

−1/nφ n (s)

(∫R

∣∣∣∣∣∣ f̂ (t)− f̂ (t− s)∣∣∣∣∣∣p

Vdt)1/p

ds

≤∫ 1/n

−1/nφ n (s)εds = ε

provided n is large enough. This follows from continuity of translation in Lp with Lebesguemeasure. Since ε > 0 is arbitrary, it follows fn → f̂ in Lp (R;V ) . Similarly, fn → f inL2 (R;H). This follows because p ≥ 2 and the norm in V and norm in H are related by|x|H ≤C ||x||V for some constant, C. Now

f̂ (t) =

Ψ(t) f (t) if t ∈ [0,T ] ,Ψ(t) f (2T − t) if t ∈ [T,2T ] ,Ψ(t) f (−t) if t ∈ [0,T ] ,0 if t /∈ [−T,2T ] .

An easy modification of the argument of Lemma 69.2.11 yields

f̂ ′ (t) =

Ψ′ (t) f (t)+Ψ(y) f ′ (t) if t ∈ [0,T ] ,Ψ′ (t) f (2T − t)−Ψ(t) f ′ (2T − t) if t ∈ [T,2T ] ,Ψ′ (t) f (−t)−Ψ(t) f ′ (−t) if t ∈ [−T,0] ,0 if t /∈ [−T,2T ] .

.