2456 CHAPTER 72. THE HARD ITO FORMULA

and both X lk and X r

k converge to X in K. Therefore, the expression

qk−1

∑j=1

∣∣X (t j+1)−X (t j)−

(M(t j+1

)−M (t j)

)∣∣2converges to 0 in probability.

In fact, the formula 72.6.12 is valid for all t ∈ [0,T ] .

Theorem 72.6.2 In Situation 72.3.1, off a set of measure zero, for every t ∈ [0,T ] ,

|X (t)|2 = |X0|2 +∫ t

0

(2⟨Y (s) , X̄ (s)⟩+ ||Z (s)||2

L2(Q1/2U,H)

)ds

+2∫ t

0R((

Z (s)◦ J−1)∗X (s))◦ JdW (s) (72.6.17)

Furthermore, for t ∈ [0,T ] , t→ X (t) is continuous as a map into H for a.e. ω . In additionto this,

E(|X (t)|2

)= E

(|X0|2

)+E

(∫ t

0

(2⟨Y (s) , X̄ (s)⟩+ ||Z (s)||2

L2(Q1/2U,H)

)ds)

(72.6.18)

Proof: Let t /∈D. For t > 0, let t (k) denote the largest point of Pk which is less than t.Suppose t (m)< t (k). Hence m≤ k. Then

X (t (m)) = X0 +∫ t(m)

0Y (s)ds+

∫ t(m)

0Z (s)dW (s) ,

a similar formula holding for X (t (k)) . Thus for t > t (m) ,

X (t)−X (t (m)) =∫ t

t(m)Y (s)ds+

∫ t

t(m)Z (s)dW (s)

which is the same sort of thing studied so far except that it starts at t (m) rather than at 0and X0 = 0. Therefore, from Lemma 72.6.1 it follows

|X (t (k))−X (t (m))|2 =∫ t(k)

t(m)

(2⟨Y (s) ,X (s)−X (t (m))⟩+ ||Z (s)||2

)ds

+2∫ t(k)

t(m)R((

Z (s)◦ J−1)∗ (X (s)−X (t (m))))◦ JdW (s) (72.6.19)

Consider that last term. It equals

2∫ t(k)

t(m)R((

Z (s)◦ J−1)∗(X (s)−X lm (s)

))◦ JdW (s) (72.6.20)

This is dominated by

2∣∣∣∣∫ t(k)

0R((

Z (s)◦ J−1)∗(X (s)−X lm (s)

))◦ JdW (s)

− 2∫ t(m)

0R((

Z (s)◦ J−1)∗(X (s)−X lm (s)

))◦ JdW (s)

∣∣∣∣