864 CHAPTER 25. NONLINEAR OPERATORS

Proof: It is given that yn → y weakly in L1 (0,T ;Y ) . It is too bad that this does notconfer pointwise convergence of some subsequence. However, what can be said is this:

weak closure of co(∪∞k=nyk) = strong closure of co(∪∞

k=nyk)

Here co signifies the convex hull. Thus something is in co(∪∞

k=nyk)

means it is of the form

vn =∞

∑k=n

cnkyk (*)

where all but finitely many of the cnk are zero and they sum to 1, each being a number

in [0,1]. Now it is given that y is in the weak closure of co(∪∞

k=nyk). In fact yn con-

verges weakly to y. Therefore, from the above observation, y is in the strong closure ofco(∪∞

k=nyk). Let vn be of the form in ∗ and let it converge in L1 (0,T ;Y ) to y. Then there

is a subsequence, still denoted as vn such that for a.e. t,

vn (t)→ y(t) in Y

Pick such a t.If y(t) /∈ F (t,x(t)) , there exist numbers k > l and y∗ ∈ Y ′ such that

⟨y∗,y(t)⟩> k > l > ⟨y∗,z⟩ for all z ∈ F (t,x(t)) .

This follows from separation theorems due to the assumption that F (t,x(t)) is a closedconvex set. Thus for all n large enough,

⟨y∗,vn (t)⟩> k > l > ⟨y∗,z⟩ for all z ∈ F (t,x(t)) . (**)

Let k− l > 2ε > 0. Consider

F (t,x(t))+By∗ (0,ε)

where the ball signifies all z ∈ Y such that

|⟨y∗,z⟩|< ε

By the weak upper semicontinuity assumption of F (t, ·) and xn (t)→ x(t), it follows thatfor k large enough,

yk (t) ∈ F (t,xk (t))⊆ F (t,x(t))+By∗ (0,ε)

Now vn is a convex combination of yk for k ≥ n and so it follows that for n large enough,

vn (t) ∈ F (t,x(t))+By∗ (0,ε)

which says that there exists zn ∈ F (t,x(t)) such that

|⟨y∗,vn (t)⟩−⟨y∗,zn⟩|< ε

However, this is a contradiction to ∗∗ because it says two things are closer than ε and alsofarther than k− l > 2ε . Thus y(t) ∈ F (t,x(t)).

It does not use that the measure space is Lebesgue measure on [0,T ] that I can see.I think it appears to work for [0,T ] replaced with Ω and t replaced with ω ∈ Ω where(Ω,F ,µ) is just some measure space.