2662 CHAPTER 78. A DIFFERENT APPROACH

for k large enough. Hence

⟨Bu,u⟩(0) = limk→∞

⟨B(tk)u(tk) ,u(tk)⟩< ε

Since ε is arbitrary, ⟨Bu,u⟩(0) = 0.Next suppose ⟨Bu,u⟩(0) = 0. Then letting v ∈ X , with v smooth,

⟨Bu(0) ,v(0)⟩= ⟨Bu,v⟩(0) = ⟨Bu,u⟩1/2 (0)⟨Bv,v⟩1/2 (0) = 0

and it follows that Bu(0) = 0.Note also that this shows that if (Bv)′ ∈ Lp′ (0,T ;V ′) as well as (Bu)′ , then there is a

continuous functiont→ ⟨B(u+ v) ,u+ v⟩(t)

which equals ⟨B(u(t)+ v(t)) ,u(t)+ v(t)⟩ for a.e.t and so, defining

⟨Bu,v⟩(t)≡ (⟨Bu,u⟩(t)+ ⟨Bv,v⟩(t)−⟨B(u+ v) ,u+ v⟩(t)) 12,

It follows that t→ ⟨Bu,v⟩(t) is continuous and equals ⟨B(u(t)) ,v(t)⟩ a.e. t.This also makes it easy to verify continuity of pointwise evaluation of Bu.Let Lu = (Bu)′ .

u ∈ D(L)≡ X ≡{

u ∈ Lp (0,T,V ) : Lu≡ (Bu)′ ∈ Lp′ (0,T,V ′)}∥u∥X ≡max

(∥u∥Lp(0,T,V ) ,∥Lu∥Lp′ (0,T,V ′)

)(78.1.11)

Since L is closed, this X is a Banach space. To see that L is closed, suppose un→ u in Vand (Bun)

′→ ξ in V ′. Is ξ = (Bu)′? Letting φ ∈C∞c ([0,T ]) and v ∈V,∫ T

0⟨ξ ,φv⟩V ′,V = lim

n→∞

∫ T

0

⟨(Bun)

′ ,φv⟩= lim

n→∞−∫ T

0

⟨Bun,φ

′v⟩

(78.1.12)

We can take a subsequence, still denoted with n such that un (t)→ u(t) pointwise a.e. Also∫ T

0

∣∣⟨Bun,φ′v⟩∣∣p ≤ ∫ T

0||un||pV dtC

(φ′,v)

and these terms on the right are uniformly bounded by the assumption that un is bounded inV . Therefore, by the Vitali convergence theorem, and using the subsequence just described,we can pass to the limit in 78.1.12.⟨∫ T

0ξ φdt,v

⟩=

⟨−∫ T

0(Bu)φ

′dt,v⟩

Since v is arbitrary, this shows that∫ T

0ξ φdt =−

∫ T

0(Bu)φ

′dt in V ′

2662 CHAPTER 78. A DIFFERENT APPROACHfor k large enough. Hence(Bu,u) (0) = lim (B (ty) u(t) u(t) <e—poo0.Since € is arbitrary, (Bu, u) (0) == 0. Then letting v € X, with v smooth,u)Next suppose (Bu, u) (0)(Bu(0) ,v(0)) = (Bu,v) (0) = (Bu,u)'/? (0) (By, v)!/7 (0) =0and it follows that Bu(0)=0.Note also that this shows that if (Bv)’ € L” (0,7;V’) as well as (Bu)’, then there is acontinuous functiont— (B(u+v),u+v) (t)which equals (B (u(t) +v(t)),u(t) +v(t)) for a.e.t and so, defining1(Bu, v) (t) = ((Bu,u) (t) + (By,v) (1) —(B(u tv) w+) (@)) 5,It follows that t + (Bu, v) (t) is continuous and equals (B(u(t)),v(t)) ae. t.This also makes it easy to verify continuity of pointwise evaluation of Bu.Let Lu = (Bu)’.u€D(L)=X= {u EL? (0,T,V) : Lu = (Bu)! € L?’ (0.7,v')}Jelly = max ([lllyro.ry) Lele ory) (78.1.11)Since L is closed, this X is a Banach space. To see that L is closed, suppose u, > u in Vand (Bun)’ > & in ¥’. Is € = (Bu)'? Letting @ € C? ({0,T]) and v EV,[ (€,0v)yry = lim " ((Bu,)! ov) = lim — " (Buy.6'v) (78.1.12)0 , no JO n-oo 0We can take a subsequence, still denoted with n such that u, (t) > u(t) pointwise a.e. AlsoT T|) \(Bu. 9)? < [ \nllf dtc (o'r)and these terms on the right are uniformly bounded by the assumption that u, is bounded inY. Therefore, by the Vitali convergence theorem, and using the subsequence just described,we can pass to the limit in 78.1.12.( [/ goat.) - (-[ (Bu) 6'at.v)Since v is arbitrary, this shows that-T T/ Soar =~ | (Bu) o'dt in V’0 0