76.4. THE GENERAL CASE 2561
By the earlier convergence 76.4.30, this u is the same as the one in 76.4.30.Consider ζ . Let ψ be infinitely differentiable and equal to 0 near T and let g ∈V . Then
since Bun (0) = Bu0n,
∫ T
0⟨ζ ,ψg⟩dt = lim
n→∞
∫ T
0
⟨(Bun−B
∫ (·)
0ΦndW −Bu0n
)′,ψg
⟩dt
= − limn→∞
∫ T
0
⟨(Bun−B
∫ (·)
0ΦndW −Bu0n
),ψ ′g
⟩dt
= −∫ T
0
⟨ψ′Bg,
(u−
∫ (·)
0ΦdW −u0
)⟩dt
= −∫ T
0
⟨B(
u−∫ (·)
0ΦdW −u0
),ψ ′g
⟩dt
which shows that
ζ =
(B(
u−∫ (·)
0ΦdW −u0
))′in the sense of V ′ valued distributions. Also from the above,∫ T
0⟨ζ ,ψg⟩dt = ⟨Bu(0)−Bu0,ψ (0)g⟩
+∫ T
0
⟨(Bu−B
∫ (·)
0ΦdW −Bu0
),ψ ′g
⟩dt
= ⟨Bu(0)−Bu0,ψ (0)g⟩+∫ T
0⟨ζ ,ψg⟩dt
Hence B(u(0,ω)) = Bu0. Thus this has shown that(B(
u−∫ (·)
0ΦdW −u0
))′+ξ (·,ω) = f (·,ω) in V ′ω , Bu(0) = Bu0.
Thus integrating this, we get
Bu(t,ω)−Bu0 (ω)+∫ t
0ξ (s,ω)ds =
∫ t
0f (s,ω)ds+B
∫ t
0ΦdW (76.4.41)
Lemma 76.4.3 The above sequence does not depend on ω /∈ N. In fact, it is not necessaryto take a further subsequence.
Proof: In fact, it is not necessary to take a subsequence to get the convergences 76.4.38- 76.4.40. This is because of the pointwise convergence of 76.4.30 and Lemma 76.4.1. Ifthe original sequence did not converge, then there would be two subsequences convergingweakly to two different functions in Vω v,w which is impossible because of 76.4.30 andthis lemma since it would require v(t) = w(t) a.e.