65.6. THE CASE THAT Q IS TRACE CLASS 2249

The remaining claim now follows from the first part of the proof.The above has discussed the integral of Φ ∈ L2

([a,T ]×Ω;L2

(Q1/2U,H

)). An obvi-

ous case to consider is when

Φ =n−1

∑k=0

ΦkX(tk,tk+1] (t)

and Φk ∈L2(Ω;L2

(Q1/2U,H

))with Φk measurable with respect to Ftk . What is

∫ t0 ΦdW?

First note that Φk ◦ J−1 ∈ L2(Ω;L2

(JQ1/2U,H

)). Let

limm→∞

Φmk →Φk ◦ J−1

in L2(Ω;L2

(JQ1/2U,H

))where Φm

k is Ftk measurable and is a simple function havingvalues in L (U1,H). Thus

Φm ≡n−1

∑k=0

Φmk X(tk,tk+1] (t)

is an elementary function and it converges to Φ ◦ J−1 in L2([a,T ]×Ω;L2

(JQ1/2U,H

)).

It follows that∫ t

aΦdW ≡ lim

m→∞

∫ t

aΦmdW ≡ lim

m→∞

n−1

∑k=0

Φmk (W (t ∧ tk+1)−W (t ∧ tk))

=n−1

∑k=0

Φk ◦ J−1 (W (t ∧ tk+1)−W (t ∧ tk)) .

Note again how it appears to depend on J but really doesn’t because there is a J in thedefinition of W .

65.6 The Case That Q Is Trace ClassIn this special case, you have a Q Wiener process with values in U and still you have

Φ ∈ L2([a,T ]×Ω;L2

(Q1/2U,H

))with Φ progressively measurable. The difference here is that in fact, Q is trace class.

Q =L

∑i=1

λ iei⊗ ei

where λ i > 0, ∑i λ i < ∞, and the ei form an orthonormal set of vectors. L is either apositive integer or ∞. Then let U0 = Q1/2U. Then Q1/2 = ∑

Li=1√

λ iei⊗ ei because thisworks, and the square root is unique. Hence Q1/2ei =

√λ iei and so an orthonormal basis

for U0 = Q1/2U is{√

λ iei}L

i=1. Now consider J = ∑Li=1√

λ i(ei⊗√

λ iei),J : U0 → U,

where the tensor product is defined in the usual way,

u⊗ v(w)≡ u(w,v)U0.