1374 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE

Proof: Suppose this is not so. Then let

S≡{

t ∈ [0,T ] : f (t)− f (0)>∫ t

0g(s)ds

}and it would follow that S ̸= /0. Let a = infS. Then there exists a decreasing sequencehn→ 0 such that

f (a+hn)− f (0)>∫ a+hn

0g(s)ds (42.2.5)

First suppose a = 0. Then dividing by hn and taking the limit,

g(0)> D+ f (0)≥ g(0) ,

a contradiction. Therefore, assume a > 0. Then by continuity

f (a)− f (0)≥∫ a

0g(s)ds

If strict inequality holds, then a ̸= infS. It follows

f (a)− f (0) =∫ a

0g(s)ds

and so from 42.2.5f (a+hn)− f (a)

hn>

1hn

∫ a+hn

ag(s)ds.

Then doing limsupn→∞ to both sides,

g(a)> D+ f (a)≥ g(a)

the same sort of contradiction obtained earlier. Thus S = /0.The following is the main result.

Theorem 42.2.2 Let H be a Hilbert space and let A be a maximal monotone operator asdescribed above. Let f : [0,T ]→ H be continuous such that f ′ ∈ L2 (0,T ;H) . Then thereexists a unique solution to the evolution inclusion

y′+Ay ∋ f , y(0) = y0 ∈ D(A)

Here y′ exists a.e., y(t) ∈ D(A) a.e.,y is continuous. Also y′ ∈ L∞ (0,T ;H).

Proof: Let yλ be the solution to

y′λ+Aλ yλ = f , yλ (0) = y0 (42.2.6)

I will base the entire proof on estimating the solutions to the corresponding integral equa-tion

yλ (t)− y0 +∫ t

0Aλ yλ (s)ds =

∫ t

0f (s)ds (42.2.7)