888 CHAPTER 25. NONLINEAR OPERATORS

Hencelim inf

n→∞⟨x∗n,xn− x+ t (x− y)⟩ ≥ lim inf

n→∞⟨x∗t ,xn− x+ t (x− y)⟩

and so from the above limit,

t lim infn→∞⟨x∗n,x− y⟩ ≥ t ⟨x∗t ,x− y⟩

Cancel the t.lim inf

n→∞⟨x∗n,x− y⟩= lim inf

n→∞⟨x∗n,xn− y⟩ ≥ ⟨x∗t ,x− y⟩

Now you have a fixed y and x∗t ∈ A(x+ t (y− x)) . The subspace determined by x,y is fi-nite dimensional. Also it was shown above that A is locally bounded at x and so there is asubsequence, still denoted as x∗t such that x∗t → x∗ (y) weakly. Now from the upper semi-continuity on finite dimensional spaces shown above, for every S a finite subset of X andε > 0, it follows that for all t small enough,

x∗t ∈ Ax+BS (0,ε)

Thus x∗ (y) ∈ Ax. Hence, there exists x∗ (y) ∈ Ax such that

lim infn→∞⟨x∗n,xn− y⟩ ≥ ⟨x∗ (y) ,x− y⟩

I found this in a paper by Peng. It is a very nice result.

Proposition 25.7.26 Let X and Y be reflexive Banach spaces with Y ⊆ X ′. Let 1 < p < ∞

and let q = pp−1 = p′ so 1

p +1q = 1. Let F : [0,T ]×X→P (Y ) be multivalued and satisfies.

1. F (·,x) has a measurable selection for each x ∈ X

2. F (t, ·) is maximal monotone for a.e. t ∈ [0,T ]

3. ∥y∥Y ≤ ρ1 (t)+ρ2 ∥x∥p−1X where y ∈ F (t,x) for a.e. t, and where ρ1 ∈ Lq (0,T ) and

ρ2 > 0

Let 0≤ a < b≤ T with b−a = τ . Define

Fτ x≡{

1τ

∫ ba y(t)dt : t→ y(t) is measurableand y(t) ∈ F (t,x) a.e. t ∈ (a,b)

}Then Fτ : X →P (Y ) is maximal monotone.

Proof: First note that Fτ x is convex and nonempty. To see this, say z, ẑ are in Fτ x. andlet y, ŷ be the corresponding functions. Then for λ ∈ [0,1]

λ z+(1−λ ) ẑ =1τ

∫ b

a(λy(t)+(1−λ ) ŷ(t))dt

and λy(t)+(1−λ ) ŷ(t)∈ F (t,x) because F (t, ·) is maximal monotone which implies thatthe set values are convex.