42.7. AN EVOLUTION INCLUSION 1401

≥ |ξ − xλ |2−Cξ y |ξ − xλ |where Cξ y depends on ξ and y but is independent of λ because of the assumption thatξ ∈D(A)∩D(Φ) and Lemma 42.1.3 which gives a bound on |Aλ ξ | in terms |y| for y ∈ Ax.Therefore, there exist constants, C1 and C2, depending on ξ and y but not on λ such that

Φ(ξ )≥Φ(xλ )+ |xλ |2−C1 |xλ |−C2.

Since Φ≥ 0, this shows that |xλ | is bounded independent of λ .

2(

Φ(ξ )+C2 +C2

12

)≥Φ(xλ )+ |xλ |2 .

This shows |xλ | is bounded independent of λ . Therefore, by 42.6.44, |Aλ xλ | is boundedindependent of λ . By Theorem 42.6.1 this shows there exists x ∈ D(∂Φ)∩D(A) such that

y ∈ Ax+∂Φ(x)+ x

and so A+∂Φ is maximal monotone since y ∈ H was arbitrary.

42.7 An Evolution InclusionIn this section is a theorem on existence and uniqueness for the initial value problem

x′+∂φ (x) ∋ f , x(0) = x0.

Suppose φ is a mapping from H to [0,∞] which satisfy the following axioms.

φ is convex and lower semicontinuous, and proper, (42.7.45)

Lemma 42.7.1 For x ∈ L2 (0,T ;H) , t→ φ (x) is measurable.

Proof: This follows because φ is Borel measurable and so φ ◦ x is also measurable.Now define the following function Φ, on the Hilbert space, L2 (0,T ;H) .

Φ(x)≡{ ∫ T

0 φ (x(t))dt if x(t) ∈ D for a.e. t+∞ otherwise

(42.7.46)

Lemma 42.7.2 Φ is convex, nonnegative, and lower semicontinuous on L2 (0,T ;H) .

Proof: Since φ is nonnegative and convex, it follows that Φ is also nonnegative andconvex. It remains to verify lower semicontinuity. Suppose, xn→ x in L2 (0,T ;H) and let

λ = lim infn→∞

Φ(xn) .

Is λ ≥ Φ(x)? Then is suffices to assume λ < ∞. Suppose not. Then λ < Φ(x) . Taking asubsequence, we can have λ = limn→∞ Φ(xn) and we can take a further subsequence forwhich convergence of xn to x is pointwise a.e. Then

λ < Φ(x)≡∫ T

0φ (x(t))dt ≤

∫ T

0lim inf

n→∞φ (xn (t))dt

≤ lim infn→∞

∫ T

0φ (xn (t))dt = lim inf

n→∞Φ(xn) = λ