1224 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFF

Therefore, the {uni } are all bounded. It follows that after taking subsequences k times there

exists a subsequence{

unk

}such that unk is a Cauchy sequence in Lp ([a,b] ;W ) . You simply

get a subsequence such that unki is a Cauchy sequence in W for each i. Then denoting this

subsequence by n,

||un−um||Lp(a,b;W ) ≤ ||un−un||Lp(a,b;W )

+ ||un−um||Lp(a,b;W )+ ||um−um||Lp(a,b;W )

≤ η

4+ ||un−um||Lp(a,b;W )+

η

4< η

provided m,n are large enough, contradicting 34.7.60.

34.8 Some Evolution InclusionsLet H be a Hilbert space and let H denote L2 (0,T ;H). Here will be an application to anevolution equation having values in H . It will always be the case that H = H ′ so this isthe simplest sort of a Gelfand triple, V = H = H ′ =V ′. First is given a maximal monotoneoperator.

Definition 34.8.1 Let D(L) ≡ {u ∈H such that u′ ∈H and u(0) = u0} . Then for u ∈D(L) ,Lu≡ u′.

Note that L is not linear.

Lemma 34.8.2 For L as just defined, L is maximal monotone L : H →H .

Proof: To show it is maximal monotone, it suffices to verify that L+ I is onto. This isby Theorem 25.7.13 on Page 881. Thus consider the equation

u′+u = f , u(0) = u0

Is there a solution? Of course there is and it equals

u(t) = e−tu0 +∫ t

0e−(t−s) f (s)ds

by the usual application of integrating factors and so forth.Then with this, the following is from Theorem 25.7.55 on Page 919. This is a well

known result found in Brezis [24].

Theorem 34.8.3 Let u0 ∈ D(φ) where φ : H → [0,∞] is proper, lower semicontinuous,and convex. Also let f ∈H be given and u0 ∈ D(φ). Then there exists a solution u to theevolution initial value problem,

u′ (t)+∂φ (u(t)) ∋ f (t) a.e.in H, u(0) = u0

This solution satisfies u(t)∈D(∂φ) for a.e. t, there exists z∈H such that z(t)∈ ∂φ (u(t))for a.e. t such that the inclusion is an equation with ∂φ (u(t)) replaced with z(t).