68.4. THE SKOROKHOD INTEGRAL 2341

If {h1, · · · ,hn} is independent, then λ has a normal density function and λ ≪ mn and soF2 (x) = 0 for a.e. x. Since F is smooth, this means that F = 0 everywhere. Hence DkF = 0and so DF = 0. Thus the case where the hi are independent is easy.

Next suppose without loss of generality that a basis for

span(h1, · · · ,hn)

is{h1, · · · ,hr}

where r < n. Say hk = ∑ri=1 ck

i hi for k > r. Then

0 = F

(W (h1) , · · ·W (hr) ,W

(r

∑i=1

cr+1i hi

)· · ·W

(r

∑i=1

cni hi

))

= F

(W (h1) , · · ·W (hr) ,

r

∑j=1

cr+1j W (h j) · · ·

r

∑j=1

cnjW (h j)

)≡ G(W (h1) , · · ·W (hr))

and so in terms of {h1, · · · ,hr} ,

DF =r

∑i=1

(DiF)hi +n

∑i=r+1

(DiF)

=hi︷ ︸︸ ︷r

∑j=1

cijh j

=r

∑j=1

(D jF)h j +r

∑j=1

n

∑i=r+1

(DiF)cijh j

=r

∑j=1

(D jF +

n

∑i=r+1

(DiF)cij

)h j

Now it was just shown that G(x) is identically 0 and so D jG = 0, j ≤ r. So what is D jG?From the above, it equals

D jF +n

∑i=r+1

(DiF)cij = 0

Hence DF = 0. Now if F (W (h1) , · · · ,W (hn)) = G(W (k1) , · · · ,W (km)) , then F−G = 0and so from what was just shown, D(F−G) = DF−DG = 0. Thus the derivative is welldefined.

Lemma 68.4.3 Let P denote the set of all polynomials in W (h) for h ∈ H. Then P isdense in Lp (Ω).

Proof: Let g ∈ Lp′ (Ω) and suppose that for every f ∈ D,∫

Ωg f dP = 0. Does it follow

that g = 0? If so, then by the Riesz representation theorem, P is dense in Lp (Ω). From