1192 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFF

valid for t ∈ D, it follows that the same formula holds for all t /∈ N. Then define ⟨Bu,u⟩(t)to equal ⟨Bu(t) ,u(t)⟩ off N and the right side for t ∈ N. Thus t→ ⟨Bu,u⟩(t) is continuousand for all t ∈ [0,T ] ,

⟨Bu,u⟩(t) = ⟨Bu0,u0⟩+2∫ t

0⟨Y (r) ,u(r)⟩dr

Also recall that t → Bu(t) was shown to be weakly continuous into W ′ on NC. Then fort,s ∈ NC,

⟨B(u(t)−u(s)) ,u(t)−u(s)⟩= ⟨Bu(t) ,u(t)⟩−2⟨Bu(t) ,u(s)⟩+ ⟨Bu(s) ,u(s)⟩

From this, it follows that t→ Bu(t) is continuous into W ′ on NC because limt→s of the rightside gives 0 and so the same is true of the left. Hence,

|⟨B(u(t)−u(s)) ,y⟩| ≤ ⟨By,y⟩1/2 ⟨B(u(t)−u(s)) ,u(t)−u(s)⟩1/2

≤ ∥B∥1/2 ⟨B(u(t)−u(s)) ,u(t)−u(s)⟩1/2 ∥y∥

so∥B(u(t)−u(s))∥W ′ ≤ ∥B∥

1/2 ⟨B(u(t)−u(s)) ,u(t)−u(s)⟩1/2

which converges to 0 as t→ s.Consider the case that t → B(u(t)) has a weak derivative, denoted as (Bu)′ (t) which

is in Lp′ (0,T ;V ′) . Then as shown above, there is a continuous function, denoted as Bu(t)which equals B(u(t)) for a.e. t and

Bu(t) = Bu(0)+∫ t

0(Bu)′ (s)ds

Then the above theorem applies. Then one obtains the following corollary.

Corollary 34.4.6 Let V ⊆W,W ′ ⊆ V ′ be separable Banach spaces, and B ∈ L (W,W ′)is nonnegative and self adjoint. Also suppose t → B(u(t)) has a weak derivative (Bu)′ ∈Lp′ (0,T ;V ′) for u ∈ Lp (0,T ;V ). Then there is a continuous function denoted as Bu(t)which equals B(u(t)) a.e. t. Say for t /∈ N. Suppose Bu(0) = Bu0, u0 ∈W. Then

Bu(t) = Bu0 +∫ t

0(Bu)′ (s)ds in V ′ (34.4.26)

Then t→ Bu(t) is in C(NC,W ′

)and also for such t,

12⟨Bu(t) ,u(t)⟩= 1

2⟨Bu0,u0⟩+

∫ t

0

⟨(Bu)′ (s) ,u(s)

⟩ds

There exists a continuous function t → ⟨Bu,u⟩(t) which equals the right side of the abovefor all t and equals ⟨Bu(t) ,u(t)⟩ off N. This also satisfies

supt∈[0,T ]

⟨Bu,u⟩(t)≤C(∥∥(Bu)′

∥∥Lp′ (0,T,V ′) ,∥u∥Lp(0,T,V )

)

1192 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFFvalid for ¢ € D, it follows that the same formula holds for all ¢ N. Then define (Bu, u) (t)to equal (Bu (t) ,u(t)) off N and the right side for t € N. Thus t — (Bu, u) (t) is continuousand for all t € [0,7],(Bu, u) (t) = (Bup, 1) +2 (Y (r) ,u(r)) drAlso recall that t + Bu(t) was shown to be weakly continuous into W’ on N°. Then fort,s ENS,(B(u(t) —u(s)) u(t) —u(s))= (Bu(t),u(t)) —2(Bu(t) ,u(s)) + (Bu(s) ,w(s))From this, it follows that t + Bu (t) is continuous into W’ on N© because lim,_,, of the rightside gives 0 and so the same is true of the left. Hence,|(B (u(t) —u(s)) ,9)| (By,y)'/? (B (u(t) —u(s)) w(t) —u(s))'/\|B\|'/? (B (w(t) —u(s)) ,w(t) —w(s))'/” [Ip]IA IANO)||B (u(t) —w(s)) [lyr < IIBI|'? (B (u(t) —u(s)) w(t) —u(s))1/?which converges toO0 ast—s. IjConsider the case that t + B(u(t)) has a weak derivative, denoted as (Bu)’ (t) whichis in L’’ (0,7;V’). Then as shown above, there is a continuous function, denoted as Bu (t)which equals B(u(t)) for a.e. t andtBu(t) = Bu(0) + [ (Bu)! (s)ds0Then the above theorem applies. Then one obtains the following corollary.Corollary 34.4.6 Let V C W,W’ CV’ be separable Banach spaces, and B € & (W,W')is nonnegative and self adjoint. Also suppose t + B(u(t)) has a weak derivative (Bu)! €L” (0,T;V’) for u € L?(0,T;V). Then there is a continuous function denoted as Bu(t)which equals B(u(t)) a.e. t. Say fort ¢ N. Suppose Bu(0) = Buo, uo € W. ThenBu(t) = Bug + [ (Bu)’ (s) ds in V' (34.4.26)Then t —> Bu(t) is in C (N©,W’) and also for such t,15 (Bu(t),u(0)) = 5 (Buo,uo) + f° ((Bu)"(),u(5))asThere exists a continuous function t + (Bu,u) (t) which equals the right side of the abovefor all t and equals (Bu(t) ,u(t)) off N. This also satisfiessup, (Buu) (0) <C(|(B4)'lpyorvy llellnory))