34.8. SOME EVOLUTION INCLUSIONS 1225

Proof: Define a function Φ : H → R

Φ(u)≡∫ T

0φ (u)dt

There are no measurability issues because φ is lower semicontinuous and so the compo-sition φ (u) will be appropriately measurable. Then this is clearly convex. It is properbecause Φ(u0) = φ (u0)T so u0 ∈ D(Φ). If un→ u in H , does it follow that

lim infn→∞

Φ(un)≥Φ(u)

Suppose not so Φ(u)> liminfn→∞ Φ(un) . Then choosing a subsequence such that

un→ u pointwise a.e.,

Φ(u) > lim infn→∞

Φ(un)≡ lim infn→∞

∫ T

0φ (un)dt

≥∫ T

0lim inf

n→∞φ (un)dt =

∫ T

0φ (u)dt = Φ(u)

which is a contradiction. Thus Φ is also lower semicontinuous.The constant function u≡ u0 is in D(L)∩D(Φ) . To use Theorem 25.7.55 on Page 919,

it is required to show thatΦ(Jλ u)≤Φ(u)+Cλ

In this case, the duality map is just the identity map. Hence Jλ u is the solution to

0 = (Jλ u−u)+λL(Jλ u)

Hence letting Jλ u be denoted by uλ , it follows that uλ would be the solution to

λu′λ+uλ = u, uλ (0) = u0

Using the usual integrating factor procedure, it follows that

uλ (t) = e−(1/λ )tu0 +∫ t

0e−(1/λ )(t−s) 1

λu(s)ds

Note thate−(1/λ )t +

∫ t

0

1λ

e−(1/λ )(t−s)ds = 1

Thus by Jensen’s inequality,

φ (uλ (t))≤ e−(1/λ )tφ (u0)+

∫ t

0e−(1/λ )(t−s) 1

λφ (u(s))ds

Then

Φ(uλ )≤∫ T

0e−(1/λ )t

φ (u0)dt +∫ T

0

∫ t

0e−(1/λ )(t−s) 1

λφ (u(s))dsdt