77.5. THE MAIN RESULT 2629

Lemma 77.5.5 For each ε > 0 there exists a solution to

(B(ω)u(·,ω))′+ εFu(·,ω)+u∗ (·,ω) = f (·,ω) in U ′, (77.5.48)Bu(0,ω) = Bu0 (ω)

u∗ (·,ω) ∈ A(u(·,ω) ,ω) (77.5.49)

this solution satisfies (t,ω)→ u(t,ω) is product measurable into V . Also (t,ω)→ u∗ (t,ω),and (t,ω)→ B(ω)u(t,ω) are product measurable into V ′ and W ′ respectively.

Next it is desired to remove the regularizing term εFu. This will involve another use ofTheorem 77.2.10. Denote by uε the solution to the above lemma. Then act on uε on bothsides. This yields

12⟨Buε ,uε⟩(T )−

12⟨Bu0,u0⟩+ ε

∫ T

0⟨Fuε ,uε⟩ds

+∫ T

0⟨u∗ε ,uε⟩ds =

∫ T

0⟨ f ,uε⟩ds (77.5.50)

Then by the coercivity assumption,

lim||u||V→∞

u∈Xr

inf{2⟨u∗,u⟩+ ⟨Bu,u⟩(T ) : u∗ ∈ A(u,ω)}||u||V

= ∞

it follows thatε ⟨Fuε ,uε⟩U ′,U +∥uε∥V ≤C (u0, f ) (77.5.51)

where the constant on the right does not depend on ε . Then

εFuε → 0 strongly in U ′

this follows because from properties of the duality map,

⟨εFuε ,v⟩ ≤ ε ⟨Fuε ,uε⟩1/r′ ⟨Fv,v⟩1/r

= ε1/r′ ⟨Fuε ,uε⟩1/r′

ε1/r ∥v∥U ≤Cε

1/r ∥v∥U

Then since A is bounded, there is a constant C independent of ε such that

∥u∗ε∥V ′ +∥∥(Buε)

′∥∥U ′ +∥uε∥V ≤C (77.5.52)

It follows there is a subsequence, still denoted with ε such that

u∗ε → u∗ weakly in V ′, (77.5.53)

(Buε)′→ (Bu)′ weakly in U ′, (77.5.54)

uε → u weakly in V . (77.5.55)