1420 CHAPTER 43. INTERPOLATION IN BANACH SPACE

The next theorem is an example of a trace theorem. In this theorem, f ∈ Lp (0,T ;V )while f ′ ∈ Lp (0,T ;V ′) . It makes no sense to consider the initial values of f in V becauseit is not even continuous with values in V . However, because of the derivative of f it willturn out that f is continuous with values in a larger space and so it makes sense to considerinitial values of f in this other space. This other space is called a trace space.

Theorem 43.1.15 Let V and H be a Banach space and Hilbert space as described in Def-inition 43.1.14. Suppose f ∈ Lp (0,T ;V ) and f ′ ∈ Lp′ (0,T ;V ′) . Then f is a.e. equal to acontinuous function mapping [0,T ] to H. Furthermore, there exists f (0) ∈ H such that

12| f (t)|2H −

12| f (0)|2H =

∫ t

0

⟨f ′ (s) , f (s)

⟩ds, (43.1.3)

and for all t ∈ [0,T ] , ∫ t

0f ′ (s)ds ∈ H, (43.1.4)

and for a.e. t ∈ [0,T ] ,

f (t) = f (0)+∫ t

0f ′ (s)ds in H, (43.1.5)

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 (43.1.6)

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