2322 CHAPTER 68. A DIFFERENT KIND OF STOCHASTIC INTEGRATION

Proof: Clearly each Hn is a closed subspace. Also, if f ∈Hn and g ∈Hm for n ̸= m,what about ( f ,g)L2(Ω)?

( f ,g)L2(Ω) = liml→∞

E

(Ml

∑k=1

alkHn

(W(

hlk

)),

Mp

∑j=1

aljHm

(W(

hlj

)))

= liml→∞

∑k, j

alkal

jE(

Hn

(W(

hlk

))Hm

(W(

hlj

)))= 0

Thus these are orthogonal subspaces. Clearly L2 (Ω) ⊇ ⊕H n. Suppose X is orthogonalto each Hn. Is X = 0? Each xn can be obtained as a linear combination of the Hk (x) fork≤ n. This is clear because the space of polynomials of degree n is of dimension n+1 and{H0 (x) ,H1 (x) , · · · ,Hn (x)} is independent on R.

This is easily seen as follows. Suppose

n

∑k=0

ckHk (x) = 0

and that not all ck = 0. Let m be the smallest index such that

m

∑k=0

ckHk (x) = 0

with cm ̸= 0. Then just differentiate both sides and obtain

m

∑k=1

ckHk−1 (x) = 0

contradicting the choice of m.Therefore, each xn is really a unique linear combination of the Hk as claimed. Say

xn =n

∑k=0

ckHk (x)

Then for |h|= 1,

W (h)n =n

∑k=0

ckHk (W (h)) ∈Hn

Hence (X ,W (h)n)L2(Ω) = 0 whenever |h| = 1. It follows that for h ∈ H arbitrary, and thefact that W is linear,

(X ,W (h)n)L2 =

(X ,

(|h|W

(h|h|

))n)= |h|n

(X ,W

(h|h|

)n)= 0

Therefore, X is perpendicular to eW (h) for every h ∈ H and so from Lemma 64.6.5, X = 0.Thus ⊕H n is dense in L2 (Ω).

Note that from Lemma 64.6.4, every polynomial in W (h) is in Lp (Ω) for all p > 1.Now what is next is really tricky.