1442 CHAPTER 43. INTERPOLATION IN BANACH SPACE

for all (a,b∗)∈ S. That is, ⟨(y∗,−x) ,(a,b∗)⟩= 0 for all (a,b∗)∈ S. In other words ⟨y∗,a⟩=⟨x,b∗⟩ for all (a,b∗) ∈ S. Thus (τS)⊥ = τ

(S⊥). Now ττ (M) = M if M is a subspace. and(

M⊥)⊥

= M if M is a subspace. Hence

G (L∗) = (τG (L))⊥ = τ

(G (L)⊥

)= τ

((τ (G (T ))

⊥)⊥)= ττ

(G (T )⊥

)⊥= G (T )

Now it follows that, since T is monotone, it follows that L∗ is also monotone.Note that as part of this argument, we have proved that for T a densely defined linear

operator, G (T ∗∗) = G (T ).Now recall Theorem 25.8.8 on Page 924 which is listed next.

Theorem 43.4.2 Let L : D(L)⊆V →V ′ where D(L) is dense, L is monotone, L is closed,and L∗ is monotone, L a linear map. Let T : V →P (V ′) be L pseudomonotone, bounded,coercive. Then L+T is onto. Here V is a reflexive Banach space such that the norms for Vand V ′ are strictly convex.

To apply this theorem, let B be as above and V → V ≡ Lp ([0,T ] ,V ) . Letting u0 ∈ V,let

T (u)≡ A(u+u0)

where A : V →P (V ′). Suppose that T just defined is set valued pseudomonotone andcoercive. Let Lu = (Bu)′ as described above in 43.4.20. Then from Theorem 43.4.2 and iff ∈ V ′, there exists a solution u to

Lu+A(u+u0) ∋ f

Thus there exists ξ ∈ A(u+u0) such that Lu+ ξ = f in V ′. Then letting w = u+ u0, itfollows that ξ ∈ A(w) and L(w−w0)+ξ = f . Thus,

(Bw)′+ξ = f , (Bw)(0) = Bw0

Written in terms of A,(Bw)′+A(w) ∋ f in V ′, (Bw)(0) = Bu0.This proves the followingtheorem about the existence of solutions to implicit evolution inclusions.

Theorem 43.4.3 Suppose u→ A(u+u0) is set valued pseudomonotone and coercive foru0 ∈V . Also let

V ⊆W, W ′ ⊆V ′

where W is a Hilbert space, V is a reflexive Banach space dense in W. Suppose B : W →W ′

is self adjoint and nonnegative. Then there exists a solution w ∈ V to the implicit evolutionequation

(Bw)′+A(w) ∋ f in V ′, (Bw)(0) = Bu0.