76.7. OTHER EXAMPLES, INCLUSIONS 2589

=1h

∫ t

t−h⟨B(u−q)−B(v−q) ,u− v⟩ds

+1

2h

∫ t−h

0⟨B(u−q)−B(v−q) ,(u− v)⟩ds

− 12h

∫ t−h

h⟨B(u−q)−B(v−q) ,u− v⟩ds

− 12h

∫ t

t−h⟨B(u−q)−B(v−q) ,u− v⟩ds

=12h

∫ t

t−h⟨B(u−q)−B(v−q) ,u− v⟩ds

+1

2h

∫ h

0⟨B(u−q)−B(v−q) ,(u− v)⟩ds

which is nonnegative as can be seen by replacing u− v with (u−q)− (v−q) and usingmonotonicity of B.

Now pass to a limit in 76.7.68 as h→ 0 to get the desired inequality. Therefore, from76.7.67,

lim supn→∞

∫ T

0⟨A(t,un)+h(t,ω)Jn (un) ,un−u⟩dt ≤ 0

From the above strong convergence, the left side equals

lim supn→∞

∫ T

0⟨A(t,un) ,un−u⟩dt ≤ 0

It follows that for all v ∈ Vω ,∫ T

0⟨A(t,u) ,u− v⟩dt

≤ lim infn→∞

∫ T

0⟨A(t,un) ,un− v⟩dt

= lim supn→∞

[∫ T

0⟨A(t,un) ,un−u⟩dt +

∫ T

0⟨A(t,un) ,u− v⟩dt

]≤

∫ T

0⟨ξ ,u− v⟩dt

Since v is arbitrary, A(·,u) = ξ ∈ V ′ω . Passing to the limit in the integral equation yields

Bu(t)−Bu0 +∫ t

0A(s,u)ds+

∫ t

0h(s,ω)ζ (s,ω)ds =

∫ t

0ΦdW

What is h(s,ω)ζ (s,ω)?∫ T

0⟨h(s,ω)Jn (un (s)) ,v(s)−un (s)⟩ds≤

∫ T

0h(s,ω)(φ n (v)−φ n (un))ds