78.2. MEASURABLE SOLUTIONS TO EVOLUTION INCLUSIONS 2669

Corollary 78.2.3 Let A satisfy 78.1.1-78.1.1, p> 1, and let f be measurable into V ′. Thenthere exists a solution to

Lu(ω)+w∗ (ω) = f (ω) ,Bu(0,ω) = Bu(T,ω) ,

w∗ (ω) ∈ A(u(ω) ,ω)

such that Lu,w∗,u are all measurable.

Proof: Define Λ as the restriction of L to the space {u ∈ D(L) : Bu(0) = Bu(T )} . Thisenables periodic conditions. Let

D(T )≡{

v ∈ V : v′ ∈ V and v(T ) = v(0)}, T v =−Bv′

Then consider T ∗. If u ∈ D(Λ) ,v ∈ D(T ) ,

−∫ T

0

⟨Bv′,u

⟩=−⟨Bu,v⟩ |T0 +

∫ T

0

⟨(Bu)′ ,v

⟩and so, since the boundary term vanishes, this shows that D(Λ)⊆ D(T ∗) and that T ∗ = Λ

on D(Λ).Next let u ∈ D(T ∗) . By definition, this means that

|⟨T v,u⟩| ≤Cu ∥v∥V (*)

So let v ∈C∞c ([0,T ] ;V ) .

⟨T v,u⟩=−∫ T

0

⟨Bv′,u

⟩=−

∫ T

0

⟨Bu,v′

⟩From the Riesz representation theorem, there exists a unique (Bu)′ such that the aboveequals

∫ T0⟨(Bu)′ ,v

⟩and by density of C∞

c ([0,T ] ;V ) this shows T ∗u = (Bu)′ = Lu. ThusT ∗ = L on D(T ∗) and in particular (Bu)′ ∈ V ′. It remains to consider the boundary condi-tions. For u ∈ D(T ∗) and v ∈ D(T ) ,

⟨T v,u⟩=−∫ T

0

⟨Bv′,u

⟩=−⟨Bu,v⟩ |T0 +

∫ T

0

⟨(Bu)′ ,v

⟩The boundary term is of the form

⟨Bu(0)−Bu(T ) ,v(0)⟩

If ∗ is to hold for all v ∈ D(T ) we must have Bu(0) = Bu(T ). If the difference is ξ ̸= 0,you would need to have

|⟨ξ ,v(0)⟩| ≤Cu ∥v∥Vfor all v ∈ D(T ). So pick v ∈ D(T ) such that |⟨ξ ,v(0)⟩|= δ > 0 and consider a piecewiselinear function ψn which is one at 0 and T but zero on [1/n,T − (1/n)] . Then if vn = ψnv,the left side is δ for all n but the right converges to 0.