1440 CHAPTER 43. INTERPOLATION IN BANACH SPACE
provided k is large enough. This follows from 43.3.19. Since ε is arbitrary, it follows that
limk→∞
|⟨BX (t (k)) ,X (t (k))⟩−⟨BX (t) ,X (t)⟩|= 0
Then from the formula,
⟨BX (t) ,X (t)⟩= ⟨BX0,X0⟩+2∫ t
0⟨Y (u) ,X (u)⟩du
valid for t ∈ D, it follows that the same formula holds for all t. This formula impliest → ⟨BX (t) ,X (t)⟩ is continuous. Also recall that t → BX (t) was shown to be weaklycontinuous into W ′. Then
⟨B(X (t)−X (s)) ,X (t)−X (s)⟩= ⟨BX (t) ,X (t)⟩−2⟨BX (t) ,X (s)⟩+ ⟨BX (s) ,X (s)⟩
From this, it follows that t→ BX (t) is continuous into W ′ because limt→s of the right sidegives 0 and so the same is true of the left. Hence,
|⟨B(X (t)−X (s)) ,y⟩|≤ ⟨By,y⟩1/2 ⟨B(X (t)−X (s)) ,X (t)−X (s)⟩1/2
≤ ∥B∥1/2 ⟨B(X (t)−X (s)) ,X (t)−X (s)⟩1/2 ∥y∥
so∥B(X (t)−X (s))∥W ′ ≤ ∥B∥
1/2 ⟨B(X (t)−X (s)) ,X (t)−X (s)⟩1/2
which converges to 0 as t→ s.
43.4 Some Implicit InclusionsLet B∈L (W,W ′) and B satisfies ⟨Bx,x⟩ ≥ 0, ⟨Bx,y⟩= ⟨By,x⟩ for V ⊆W, W ′ ⊆V ′. WhereV is dense in the Hilbert space W . Now let
D(L)≡{
u ∈ V : (Bu)′ ∈ V ′,Bu(0) = 0}, Lu≡ (Bu)′ (43.4.20)
Then clearly D(L) is dense in V . Here V ≡ Lp ([0,T ] ,V ) where p≥ 2 for simplicity. Nowlet
D(T )≡{
u ∈ V : u′ ∈ V and u(T ) = 0}, Tu≡−B
(u′)
(43.4.21)
The idea is to show that L = T ∗ and that T is monotone. Then this will imply usingProposition 25.8.2 that L∗ is monotone. This is done by showing that G (L∗) = G (T ).
Lemma 43.4.1 T is monotone, T ∗ = L and L∗,L are both monotone.
Proof: First, why is T monotone?∫ T
0
⟨−Bu′,u
⟩dt =
∫ T
0−⟨Bu,u′
⟩dt =−⟨Bu,u⟩ |T0 +
∫ T
0
⟨(Bu)′ ,u
⟩dt
= ⟨Bu(0) ,u(0)⟩+∫ T
0
⟨Bu′,u
⟩dt