2568 CHAPTER 76. IMPLICIT STOCHASTIC EQUATIONS

In this case we assume

⟨(λB+A(ω))(u)− (λB+A(ω))(v) ,u− v⟩ ≥ δ ∥u− v∥2W

Then repeating the above argument with this change yields set of measure zero, still de-noted as N such that for ω /∈ N ∫ T

0∥un−un+1∥2

W ds≤ 2−n (76.4.49)

for all n large enough. Hence for such ω, un (·,ω) is Cauchy in L2 ([0,T ] ,W ) and infact un (t,ω) is a Cauchy sequence in W . Thus {un (·,ω)} converges in L2 ([0,T ] ,W ) tou(·,ω) ∈ L2 ([0,T ] ,W ) and by the above considerations involving continuous dependenceof V into W, it follows that u(·,ω) will be the same as the u from the above convergences.Now this convergence implies that in addition, for a.e. t,

limn→∞⟨Bun (t,ω)−Bu(t,ω) ,un (t,ω)−u(t,ω)⟩= 0 (76.4.50)

limn→∞

∫ T

0⟨Bun (t,ω)−Bu(t,ω) ,un (t,ω)−u(t,ω)⟩dt = 0

What is known from 76.4.35 is that for

Mn (t) =∫ t

0

(Φn ◦ J−1)∗Bun ◦ JdW

there is a continuous martingale M ∈M1T such that

limn→∞

E

(sup

t∈[0,T ]|Mn (t)−M (t)|

)= 0 (76.4.51)

Define a stopping time

τ p ≡ inf{

t : ⟨Bu,u⟩(t)+ supn⟨Bun,un⟩(t)> p

}This is a good enough stopping time because the function used to define it as a hitting timeis lower semicontinuous.

Lemma 76.4.8∫ T

0(Φn ◦ J−1

)∗Bun ◦ JdW →∫ T

0(Φ◦ J−1

)∗Bu◦ JdW in probability. Alsothere is a futher subsequence and set of measure zero such that off this set,

limn→∞

(sup

t∈[0,T ]

∣∣∣∣∫ t

0

(Φ◦ J−1)∗Bu◦ JdW −

∫ t

0

(Φn ◦ J−1)∗Bun ◦ JdW

∣∣∣∣)

= 0.

In particular, what is needed here is valid,

limn→∞

∫ T

0

(Φn ◦ J−1)∗Bun ◦ JdW =

∫ T

0

(Φ◦ J−1)∗Bu◦ JdW