77.6. VARIATIONAL INEQUALITIES 2637
Proof: This follows from Theorem 77.2.10. This is because there is an estimate of theright sort for the measurable functions un (·,ω) and u∗n (·,ω) and the above argument whichshows that a subsequence has a convergent subsequence which converges appropriately toa solution.
You can have K = K (ω) . There would be absolutely no change in the above the-orem. You just need to have the operator of penalization satisfy ω → P(u(ω) ,ω) =
F(
u(ω)−projK (ω) u(ω))
is measurable into V ′ provided ω → u(ω) is measurable intoV . What are the conditions on the set valued ω →K (ω) which will cause this to takeplace?
Lemma 77.6.2 Let ω →K (ω) be measurable into V . Then ω → projK (ω) u(ω) is alsomeasurable into V if ω → u(ω) is measurable.
Proof: It follows from standard results on measurable multi-functions [70] that thereis a countable collection {wn (ω)} , ω → wn (ω) being measurable and wn (ω) ∈K (ω)for each ω such that for each ω, K (ω) = ∪nwn (ω). Let
dn (ω)≡min{∥u(ω)−wk (ω)∥ ,k ≤ n}
Let u1 (ω)≡ w1 (ω) . Letu2 (ω) = w1 (ω)
on the set{ω : ∥u(ω)−w1 (ω)∥< {∥u(ω)−w2 (ω)∥}}
andu2 (ω)≡ w2 (ω) off the above set.
Thus ∥u2 (ω)−u(ω)∥= d2. Let
u3 (ω) = w1 (ω) on{
ω : ∥u(ω)−w1 (ω)∥<∥∥u(ω)−w j (ω)
∥∥ , j = 2,3
}≡ S1
u3 (ω) = w2 (ω) on S1∩{
ω : ∥u(ω)−w1 (ω)∥<∥∥u(ω)−w j (ω)
∥∥ , j = 3
}u3 (ω) = w3 (ω) on the remainder of Ω
Thus ∥u3 (ω)−u(ω)∥= d3. Continue this way, obtaining un (ω) such that
∥un (ω)−u(ω)∥= dn (ω)
and un (ω) ∈K (ω) with un measurable. Thus, in effect one picks the closest of all thewk (ω) for k ≤ n as the value of un (ω) and un is measurable and by density in K (ω) of{wn (ω)} for each ω,{un (ω)} must be a minimizing sequence for
λ (ω)≡ inf{∥u(ω)− z∥ : z ∈K (ω)}
Then it follows that un (ω)→ projK (ω) u(ω) weakly in V . Here is why: Suppose it failsto converge to projK (ω) u(ω). Since it is minimizing, it is a bounded sequence. Thus