65.14. A TECHNICAL INTEGRATION BY PARTS RESULT 2265

Now let {gi} be an orthonormal basis for Q1/2U, so it follows that {Jgi} is an orthonormalbasis for JQ1/2U. Then

∑i

∣∣∣R(Z∗nX (a)−(Z ◦ J−1)∗X (a)

)(Jgi)

∣∣∣2

≡∑i

∣∣∣∣(Z∗nX (a)−(Z ◦ J−1)∗X (a) ,Jgi

)U1

∣∣∣∣2 = ∑i

∣∣(X (a) ,(Zn−Z ◦ J−1)Jgi

)H

∣∣2≤∑

i|X (a)|2H

∣∣(Zn−Z ◦ J−1)Jgi∣∣2H = |X (a)|2H

∥∥Zn−Z ◦ J−1∥∥2L2(JQ1/2U,H)

When integrated over [a,b]×Ω, it is given that this converges to 0. This has shown that

R (Z∗nX (a))→R((

Z ◦ J−1)∗X (a))

in L2(JQ1/2U,R

). In other words

R (Z∗nX (a))→(R((

Z ◦ J−1)∗X (a))◦ J)◦ J−1

It follows that (∫ t

aZ (u)dW,X (a)

)H=∫ t

aR((

Z ◦ J−1)∗X (a))◦ JdW

From localization,(∫ b∧τnp

a∧τnp

Z (u)dW,X (a)

)H

=

(∫ b

aX[0,τn

p]Z (u)dW,X (a)

)H

=∫ b

aX[0,τn

p]R((

Z ◦ J−1)∗X (a))◦ JdW

=∫ b∧τn

p

a∧τnp

R((

Z ◦ J−1)∗X (a))◦ JdW

Then it follows that, using the stopping time,

m−1

∑j=0

(∫ tnj+1∧τn

p∧t

tnj∧τn

p∧tZ (u)dW,X

(tn

j))

H

=m−1

∑j=0

∫ tnj+1∧τn

p∧t

tnj∧τn

p∧tR((

Z ◦ J−1)∗Xn(tn

j))◦ JdW

=∫ tn

m∧τnp∧t

0R((

Z ◦ J−1)∗(X ln

))◦ JdW

where X ln is the step function

X ln (t)≡

mn−1

∑k=0

X (tnk )X[tn

k ,tnk+1)

(t)