64.6. WIENER PROCESSES, ANOTHER APPROACH 2223
It follows from the above that
Je j =L
∑k=1
1√λ j
δ i j︷ ︸︸ ︷(√λ je j,
√λ kek
)Q1/2H
√λ kek = e j
Then
L
∑i=1
∣∣∣J√λ iei
∣∣∣2H
=L
∑i=1
∣∣∣∣∣ L
∑k=1
(√λ iei,
√λ kek
)Q1/2H
√λ kek
∣∣∣∣∣2
H
=L
∑i=1
∣∣∣√λ iei
∣∣∣2H=
L
∑i=1
λ i < ∞
Thus it is clear that J is Hilbert Schmidt. Is JJ∗ = Q? For y ∈ Q1/2H,x ∈ H,
(J∗x,y)Q1/2H ≡ (x,J (y))H =
(x,
L
∑k=1
(y,√
λ kek
)Q1/2H
√λ kek
)H
=L
∑k=1
(x,√
λ kek
)H
(y,√
λ kek
)Q1/2H
Thus for y ∈ H,x ∈ H,
(J∗x,J∗y)Q1/2H =L
∑k=1
(x,√
λ kek
)H
(J∗y,
√λ kek
)Q1/2H
≡L
∑k=1
(x,√
λ kek
)H
(y,√
λ kJek
)H
=L
∑k=1
λ k (x,ek)H (y,ek)H = (Qx,y)
and so (JJ∗x,y) = (Qx,y) showing that JJ∗ = Q. This shows the following.
Proposition 64.6.9 Let Q ∈ L (H,H) where H is a real separable Hilbert space and(Qx,x) ≥ 0 and is trace class. Then there exists a one to one Hilbert Schmidt map J :Q1/2H → H such that JJ∗ = Q. Then the Q Wiener process is W (t) = ∑
∞k=1 ψk (t)Jgk
where {gk} is a complete orthonormal basis for the Hilbert space Q1/2H.
Note that in case H is Rp and Q is any symmetric p× p matrix, having nonnegativeeigenvalues, this is automatically trace class and so the above conclusion holds. In partic-ular, the covariance condition says in this case that
E ((ei,W(t)−W(s))(e j,W(t)−W(s)))
= E ((Wi (t)−Wi (s))(Wj (t)−Wj (s))) = (Qei,e j) = Qi j
This is a p dimensional Wiener process.