42.9. AN EVOLUTION INCLUSION 1409

The conditions of Theorem 42.8.2 are satisfied with K1 = K2 = C. This proves L+ ∂Φ ismaximal monotone on L2 (0,T ;H) and consequently there exists a unique solution to thedifferential inclusion of the theorem.

Of course it is desirable to be able to say that [y] ∈ ∂Φ([x]) if and only if y(t) ∈∂2φ (t,x(t)) for some x ∈ [x] . To obtain this, here are two more assumptions. For all x ∈H,

t→ J1 (t)x is measurable, (42.9.70)

where J1 (t)x is the solution, y, to y(t)+∂2φ (t,y(t))∋ x, and there exists [ξ ]∈ L2 (0,T ;H)such that

[J1 (·) [ξ ]] ∈ L2 (0,T ;H) . (42.9.71)

Lemma 42.9.6 If 42.9.70 and 42.9.71 hold, and if [y] ∈ L2 (0,T ;H) , then [y] ∈ ∂Φ([x]) ifand only if there exists x ∈ [x] such that ∂2φ (t,x(t)) ̸= /0 for all t and y(t) ∈ ∂2φ (t,x(t))a.e.

Proof: First suppose y(t)∈ ∂2φ (t,x(t)) a.e. and ∂2φ (t,x(t)) ̸= /0 for all t where x∈ [x].Then for all [w] ∈ L2 (0,T ;H),

([y] , [w])L2(0,T ;H) ≡∫ T

0(y(t) ,w(t))H dt

≤∫ T

0φ (t,x(t)+w(t))dt−

∫ T

0φ (t,x(t))dt ≤Φ([x]+ [w])−Φ([x]) .

To prove the converse, define A : D(∂Φ)→P(L2 (0,T ;H)

)as follows.

[y] ∈ A [x] if and only if for some x ∈ [x] ,

∂2φ (t,x(t)) ̸= /0 for all t and y(t) ∈ ∂2φ (t,x(t)) a.e. t.

It follows A is monotone. I will show A is maximal monotone. From the first part of theproof, the graph of A is contained in the graph of ∂Φ. Since A is maximal, this will implyA = ∂Φ and prove the lemma.

It remains to show A is maximal monotone. By 42.9.70, for each x ∈ H, J1 (t)x ismeasurable. Now from 42.9.71, and using the fact that J1 (t) is a contraction,

|J1 (t)x− J1 (t)ξ (t)| ≤ |x−ξ (t)|

and so [J1 (·)x] is in L2 (0,T ;H) . Now if

s(t) =n

∑i=1

XEi (t)xi

is a simple function,

J1 (t)s(t) =n

∑i=1

XEi (t)J1 (t)x,