73.7. THE ITO FORMULA 2493

due to the assumptions on Z. For {gi} an orthonormal basis of Q1/2 (U) ,

⟨BZ,Z⟩L2≡ ∑

i

(R−1BZ (gi) ,Z (gi)

)= ∑

i⟨BZ (gi) ,Z (gi)⟩

≤ ∥B∥∑i∥Z (gi)∥2

W ∈ L1 (0,T ) a.e.

This shows that for ω off a set of measure 0

limm,k→∞

⟨B(X (t (k))−X (t (m))) ,X (t (k))−X (t (m))⟩= 0

Then for x ∈W,

|⟨B(X (t (k))−X (t (m))) ,x⟩|≤ ⟨B(X (t (k))−X (t (m))) ,X (t (k))−X (t (m))⟩1/2 ⟨Bx,x⟩1/2

≤ ⟨B(X (t (k))−X (t (m))) ,X (t (k))−X (t (m))⟩1/2 ∥B∥1/2 ∥x∥W

and solim

m,k→∞

∥BX (t (k))−BX (t (m))∥W ′ = 0

Recall t was arbitrary in NCω and {t (k)} is a sequence converging to t. Then the above has

shown that {BX (t (k))}∞

k=1 is a convergent sequence in W ′. Does it converge to BX (t)? Letξ (t) ∈W ′ be what it converges to. Letting v ∈ V then, since the integral equation showsthat t→ BX (t) is continuous into V ′,

⟨ξ (t) ,v⟩= limk→∞

⟨BX (t (k)) ,v⟩= ⟨BX (t) ,v⟩ ,

and now, since V is dense in W, this implies ξ (t) = BX (t) = B(X (t)). Recall also that itwas shown earlier that BX is weakly continuous into W ′ hence the strong convergence of{BX (t (k))}∞

k=1 in W ′ implies that it converges to BX (t), this for any t ∈ NCω .

For every t ∈ D and for ω off the exceptional set of measure zero described earlier,

⟨B(X (t)) ,X (t)⟩= ⟨BX0,X0⟩+∫ t

0

(2⟨Y (s) ,X (s)⟩+ ⟨BZ,Z⟩L2

ds)

ds

+2∫ t

0

(Z ◦ J−1)∗BX ◦ JdW (73.7.39)

Does this formula hold for all t ∈ [0,T ]? Maybe not. However, it will hold for t /∈ Nω . Solet t /∈ Nω .

|⟨BX (t (k)) ,X (t (k))⟩−⟨BX (t) ,X (t)⟩|

≤ |⟨BX (t (k)) ,X (t (k))⟩−⟨BX (t) ,X (t (k))⟩|+ |⟨BX (t) ,X (t (k))⟩−⟨BX (t) ,X (t)⟩|

= |⟨B(X (t (k))−X (t)) ,X (t (k))⟩|+ |⟨B(X (t (k))−X (t)) ,X (t)⟩|