25.7. MAXIMAL MONOTONE OPERATORS 889

Next is a claim that Fτ x is closed and also has the property that if zn ∈ Fτ xn and ifxn→ x strongly in X and zn→ z weakly in Y, then z ∈ Fτ x. Let yn (t) ∈ F (t,xn) a.e. suchthat zn = 1

τ

∫ ba yn (t)dt. These xn are bounded and so by the assumed estimate, it follows

that the yn are bounded in Lq (0,T ;Y ) . Therefore, there is a subsequence, still denoted withn such that yn → ŷ weakly in Lq (0,T ;Y ). Now this means that ŷ is in the weak closureof the convex hull of {yk : k ≥ n}. However, this is the same as the strong closure becauseconvex and closed is the same as convex and weakly closed. Therefore, there are functions

limn→∞

∞

∑k=n

cnkyk = ŷ strongly in Lq (0,T ;Y ) where

∞

∑k=n

cnk = 1,cn

k ≥ 0,

and only finitely many are nonzero. Thus a subsequence still denoted with subscript n alsoconverges to ŷ(t) for each t off a set of measure zero. The function F (t, ·) is maximal mono-tone and defined on X and so it is pseudomonotone by Theorem 25.7.25. The estimate alsoshows that it is bounded. Therefore, as shown in the section on set valued pseudomonotoneoperators, x→ F (t,x) is upper semicontinuous from strong to weak topology. Thus, forlarge n depending on t, all of the F (t,xn) are contained in F (t,x)+BS (0,r/2) where S isa finite subset of points of X .

B≡ BS (0,r/2)≡{

w∗ : |w∗ (x)|< r2

for all x ∈ S}

Thus, for a fixed t not in the exceptional set, off which the above pointwise convergencetakes place, ŷ(t) ∈ F (t,x)+D where

D≡ {w∗ : |w∗ (x)| ≤ r for all x ∈ S}

Since S,r are arbitrary, separation theorems imply that ŷ(t) ∈ F (t,x) for t off a set ofmeasure zero: If not, there would exist u ∈ X such that ŷ(t)(u)> l > l−δ > p(u) for allp ∈ F (t,x) . But then you could take r = δ/2 and Bu (0,δ/2) and find that ŷ(t) = p+w∗

where p ∈ F (t,x) and |w∗ (u)| ≤ δ . Hence p(u) = ŷ(t)(u)−w∗ (u) > l− δ > p(u) anobvious contradiction. Is z ∈ Fτ x? Certainly so if z = 1

τ

∫ ba ŷ(t)dt. Letting φ ∈ X ,

⟨z,φ⟩ = limn→∞⟨zn,φ⟩= lim

n→∞

⟨1τ

∫ b

ayn (t)dt,φ

⟩= lim

n→∞

⟨1τ

∫ b

a

∞

∑k=n

cnkyk (t)dt,φ

⟩

=

⟨1τ

∫ b

aŷ(t)dt,φ

⟩Since φ is arbitrary, it follows that z = 1

τ

∫ ba ŷ(t)dt and so z ∈ Fτ x.

Is Fτ monotone? Say z, ẑ are in Fτ (x) ,Fτ (x̂) respectively. Consider

⟨z− ẑ,x− x̂⟩=⟨

1τ

∫ b

ay(t)− ŷ(t)dt,x− x̂

⟩=

1τ

∫ b

a⟨y(t)− ŷ(t) ,x− x̂⟩dt

but y(t) ∈ F (t,x) similar for ŷ and so the above is ≥ 0. Thus Fτ is indeed monotone.This has also shown that Fτ satisfies the necessary modified hemicontinuity condition of