69.4. THE IMPLICIT CASE 2379

whenever k is large enough. Therefore,

Bx =∞

∑i=1⟨Bx,ei⟩Bei

in W ′. It follows that

⟨Bx,x⟩= limk→∞

⟨k

∑i=1⟨Bx,ei⟩Bei,x

⟩= lim

k→∞

k

∑i=1|⟨Bx,ei⟩|2 ≡

∞

∑i=1|⟨Bx,ei⟩|2

Theorem 69.4.2 Let V ⊆W,W ′ ⊆ V ′ be separable Banach spaces,W a separable Hilbertspace, and let Y ∈ Lp′ (0,T ;V ′)≡ K′ and

BX (t) = BX0 +∫ t

0Y (s)ds in V ′ (69.4.23)

where X0 ∈W, and it is known that X ∈ Lp (0,T,V ) ≡ K for p > 1. Then t → BX (t) is inC ([0,T ] ,W ′) and also

12⟨BX (t) ,X (t)⟩= 1

2⟨BX0,X0⟩+

∫ t

0⟨Y (s) ,X (s)⟩ds

Proof: By Lemma 65.3.1, there exists a sequence of uniform partitions{

tnk

}mnk=0 =

Pn,Pn ⊆Pn+1, of [0,T ] such that the step functions

mn−1

∑k=0

X (tnk )X(tn

k ,tnk+1]

(t) ≡ X l (t)

mn−1

∑k=0

X(tnk+1)X(tn

k ,tnk+1]

(t) ≡ X r (t)

converge to X in K and also BX l ,BX r→ BX in L2 ([0,T ] ,W ′).

Lemma 69.4.3 Let s < t. Then for X ,Y satisfying 69.4.23

⟨BX (t) ,X (t)⟩= ⟨BX (s) ,X (s)⟩

+2∫ t

s⟨Y (u) ,X (t)⟩du−⟨B(X (t)−X (s)) ,(X (t)−X (s))⟩ (69.4.24)

Proof: It follows from the following computations

B(X (t)−X (s)) =∫ t

sY (u)du

and so2∫ t

s⟨Y (u) ,X (t)⟩du−⟨B(X (t)−X (s)) ,(X (t)−X (s))⟩

= 2⟨B(X (t)−X (s)) ,X (t)⟩−⟨B(X (t)−X (s)) ,(X (t)−X (s))⟩