1186 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFF

Thus e1 (t) = 0 if t /∈ ∪kE1k . Also e1 (t) is right continuous because t → n1 (t) is. Thus e1

has constant values on a small interval starting at t. But what about t /∈ ∪kE1k ? Why should

it be right continuous there? If you have such a t, then as explained above, B(t) = 0.Thenletting s be arbitrary, s > t and x ∈W ,⟨

B′ (s)x,x⟩≤ kx ⟨B(s)x,x⟩

and so as above,e−Kx(s) ⟨B(s)x,x⟩ ≤ 0

Thus this case reduces to having B(s) ≡ 0 for all s ≥ t and there is nothing to prove. Youhave n1 (s) = 1 and e1 (s) = 0 for all s≥ t.

Suppose t→ ni (t) is right continuous for i≤m and that ei is also. Then define Em+1p ≡{

t :

⟨Bgp−

m

∑i=1

⟨Bgp,ei

⟩Bei,gp−

m

∑i=1

⟨Bgp,ei

⟩ei

⟩̸= 0

}∩{t : nm (t)< p}

∩nm(t)<r<p

{t :

⟨Bgr−

m

∑i=1⟨Bgr,ei⟩Bei,gr−

m

∑i=1⟨Bgr,ei⟩ei

⟩= 0

}As earlier, these sets

{Em+1

p}∞

p=1 are disjoint. As before, let nm+1 (t) = p where t ∈ Em+1p .

Then by similar reasoning to the above, for small δ , [t, t +δ ) ∈ Em+1p and nm+1 (s) = p for

s ∈ [t, t +δ ). Then from the algorithm, em+1 (t)≡

X∪pEm+1p

(t)gnm+1(t)−∑

mi=1⟨Bgnm+1(t),ei (t)

⟩ei (t)

Dm

where Dm = ⟨B(gnm+1(t)−∑

mi=1⟨Bgnm+1(t),ei (t)

⟩ei (t)

),

gnm+1(t)−∑mi=1⟨Bgnm+1(t),ei (t)

⟩ei (t)

⟩1/2

and so is right continuous. What of t /∈ ∪pEm+1p ? In this case, the process has terminated

and what is desired has been found.Then the main result in this section is the following integration by parts theorem.

Theorem 34.4.3 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 (34.4.22)

As indicated, Bu is the name of a function satisfying the above equation which satisfiesBu(t) = B(u(t)) for a.e. t. Thus Y = (Bu)′ as a weak derivative in the sense of V ′ valueddistributions. It is known that u ∈ Lp (0,T,V ) for p > 1. Then t → Bu(t) is continuousinto W ′ for t off a set of measure zero N and also there exists a continuous function t →⟨Bu,u⟩(t) such that for all t /∈ N,⟨Bu,u⟩(t) = ⟨B(u(t)) ,u(t)⟩ ,Bu(t) = B(u(t)) , and forall t,

12⟨Bu,u⟩(t) = 1

2⟨Bu0,u0⟩+

∫ t

0⟨Y (s) ,u(s)⟩ds

Note that ⟨Bu,u⟩(0) = ⟨Bu0,u0⟩.