2636 CHAPTER 77. STOCHASTIC INCLUSIONS

Then take limsupi→0 of both sides to obtain

lim supn→∞

⟨u∗n (·,ω) ,un−u⟩ ≤ 0.

Now assume the usual limit condition holds for A(·,ω). In practice, this typicallymeans A(·,ω) will be single valued, monotone and hemicontinuous because there is nocontrol on the time derivative. However, we will go ahead and assume just that the limitcondition holds. This would also take place if A(·,ω) were defined on V and maximalmonotone, for example. Then for every v ∈ V ,

lim infn→∞⟨u∗n (·,ω) ,un− v⟩ ≥ ⟨u∗ (v) ,u− v⟩

where u∗ (v) ∈ A(u,ω). In particular, this holds for v = u and so

lim infn→∞⟨u∗n (·,ω) ,un−u⟩ ≥ 0≥ lim sup

n→∞

⟨u∗n (·,ω) ,un−u⟩

showing the the limit exists. Then

⟨u∗ (v) ,u− v⟩ ≤ lim infn→∞

(⟨u∗n (·,ω) ,un− v⟩)

= lim infn→∞

(⟨u∗n (·,ω) ,un−u⟩+ ⟨u∗n (·,ω) ,u− v⟩)

= ⟨u∗,u− v⟩

and since this is true for all v ∈ V it follows that u∗ ∈ A(u(·,ω) ,ω) since otherwise,separation theorems would give a contradiction. If u∗ were not in A(u(·,ω) ,ω) therewould exist v such that for all z∗ ∈ A(u,ω) ,

⟨z∗,u− v⟩> ⟨u∗,u− v⟩

contrary to the above. Therefore, in 77.6.65 we can take the limit of both sides and concludethat for every v ∈K such that (Bv)′ ∈ V ′,Bv(0) = 0,⟨

(Bv)′ ,u− v⟩+ ⟨u∗,u− v⟩ ≤ ⟨ f (·,ω) ,u− v⟩

where u∗ ∈ A(u,ω)This has proved the first part of the following theorem which gives measurable solutions

to a variational inequality.

Theorem 77.6.1 Suppose A(·,ω) is monotone hemicontinuous bounded and single valuedand coercive as a map from V to V ′. Suppose also that for ω→ u(ω) measurable into V ,it follows that ω→ A(u(ω) ,ω) is measurable into V ′. Let K be a closed convex subset ofV containing 0 and let B ∈L (W,W ′) be self adjoint and nonnegative as above. Let therebe a regularizing sequence {ui} for each u∈K satisfying Bui (0) = 0,(Bui)

′ ∈V ′,ui ∈K ,

lim supi→∞

⟨(Bui)

′ ,ui−u⟩≤ 0

Then for each ω, there exists a solution to⟨(Bv)′ ,u− v

⟩+ ⟨A(u(·,ω) ,ω) ,u(·,ω)− v⟩ ≤ ⟨ f (·,ω) ,u− v⟩

valid for all v∈K such that (Bv)′ ∈V ′,Bv(0)= 0, and (t,ω)→ u(t,ω) , is B ([0,T ])×Fmeasurable.