68.3. A MULTIPLE INTEGRAL 2325
which is bounded independent of k, the last step following from Lemma 64.6.4. Therefore,the Vitali convergence theorem applies in 68.2.8.
Given an h ∈ H, let sk = ∑kj=1 (h,e j)e j, the kth partial sum in the Fourier series for h.
W (sk)m =
(k
∑j=1
(h,e j)W (e j)
)m
= p(W (e1) , · · · ,W (ek))
where p is a homogeneous polynomial of degree m. Now this equals
q(H0 (W (e1)) , · · · ,H0 (W (ek)) · · ·Hm (W (e1)) , · · · ,Hm (W (ek)))
where q is a polynomial. This is because each W (e j)r is a linear combination of Hs (W (e j))
for s≤ r. Now you look at terms of this polynomial. They are all of the form cΦa for someconstant c and a∈Λ. Therefore, if X ∈ L2 (Ω) , there is a subsequence, still denoted as {sk}such that
E (W (h)n X) = limk→∞
E (W (sk)n X)
Now if X is orthogonal to each Φa, then for any h and n, there is a subsequence still denotedwith k such that
E (W (h)n X) = limk→∞
E (W (sk)n X) = 0
It follows from Lemma 64.6.4, the part about the convergence of the partial sums to eW (h)
that X is orthogonal to eW (h) for any h. Here are the details. From the lemma, for large n,∣∣∣∣∣E (eW (h)X)−E
(n
∑j=0
W (h) j
j!X
)∣∣∣∣∣< ε,
Also for large k,∣∣∣∣∣E(
n
∑j=0
W (h) j
j!X
)−E
(n
∑j=0
W (sk)j
j!X
)∣∣∣∣∣=∣∣∣∣∣E(
n
∑j=0
W (h) j
j!X
)∣∣∣∣∣< ε
Therefore, ∣∣∣E (eW (h)X)∣∣∣< 2ε
Since ε is arbitrary, this proves the desired result. By Lemma 64.6.5, X = 0 and this showsthat {Φa,a ∈ Λ} is complete.
Proposition 68.2.6 {Φa : a ∈ Λ} is a complete orthonormal set for L2 (Ω,F ,P).
68.3 A Multiple IntegralConsider trying to find ∫ 1
0
∫ 1
0dBsdBt