2596 CHAPTER 77. STOCHASTIC INCLUSIONS

to W ′. There is also a convenient integration by parts formula, Theorem 34.4.3. For con-venience, the dependence of B on ω is often suppressed. This is not a problem because theentire approach will be to consider the situation for fixed ω .

Theorem 77.2.1 Let V ⊆W,W ′⊆V ′ be separable Banach spaces, and let Y ∈Lp′ (0,T ;V ′)and

Bu(t) = Bu0 +∫ t

0Y (s)ds in V ′, u0 ∈W,Bu(t) = B(u(t)) for a.e. t (77.2.1)

Thus Y = (Bu)′ as a weak derivative in the sense of V ′ valued distributions. It is knownthat u ∈ Lp (0,T,V ) for p > 1. Then t → Bu(t) is continuous into W ′ for t off a set ofmeasure zero N and also there exists a continuous function t→ ⟨Bu,u⟩(t) such that for allt /∈ N,⟨Bu,u⟩(t) = ⟨B(u(t)) ,u(t)⟩ ,Bu(t) = B(u(t)) , and for all t,

12⟨Bu,u⟩(t) = 1

2⟨Bu0,u0⟩+

∫ t

0⟨Y (s) ,u(s)⟩ds

Note that the formula 77.2.1 shows that Bu0 = Bu(0) . Also it shows that t→⟨Bu,u⟩(t)is continuous. To emphasize this a little more, Bu is the name of a function. Bu(t) =B(u(t)) for a.e. t and t → Bu(t) is continuous into V ′ on [0,T ] because of the integralequation.

Theorem 77.2.2 In the above corollary, the map u→ Bu(t) is continuous as a map fromX to V ′. Also if Y denotes those f ∈ Lp ([0,T ] ;V ) for which f ′ ∈ Lp ([0,T ] ;V ) , so that fhas a representative such that f (t) = f (0)+

∫ t0 f ′ (s)ds, then if ∥ f∥Y ≡ ∥ f∥Lp([0,T ];V ) +

∥ f ′∥Lp([0,T ];V ) the map f → f (t) is continuous.

Proof: First, why is u→ Bu(0) continuous? Say u,v ∈ X and say p≥ 2 first.

Bu(t)−Bv(t) = Bu(0)−Bv(0)+∫ t

0(Bu)′ (s)− (Bv)′ (s)ds

and so, (∫ T

0∥Bu(0)−Bv(0)∥p′

V ′ dt)1/p′

≤(∫ T

0∥Bu(t)−Bv(t)∥p′

V ′ dt)1/p′

+

(∫ T

0

∥∥∥∥∫ t

0(Bu)′ (s)− (Bv)′ (s)ds

∥∥∥∥p′

dt

)1/p′

and so∥Bu(0)−Bv(0)∥V ′ T

1/p′ ≤(∥B∥∥u− v∥Lp′ ([0,T ];V )

+T 1/p′ ∥∥(Bu)′− (Bv)′∥∥

Lp′ ([0,T ];V ′)

)≤C (∥B∥ ,T )∥u− v∥X