Elementary Malliavin Derivative question about definition.

Question

I am reading a book that defines the Malliavin derivative $D_tF$ as follows: If

1. $F = \sum_{n=0}^{\infty} I_n(f_n)$ is the Wiener Chaos expansion.

2. $F$ is in the brownian filtration and $F \in L^2(P)$.

3. $\sum_{n=0}^{\infty} n n! ||f_n||_{L^2([0,T]^n)}^2 < \infty$

Then $D_tF = \sum_{n=1}^{\infty} nI_{n-1}(f_n(.,t))$.

My question is why do we need 3 to be so strong. It seems that in the theory the $n!$ is never used. Is it defined this way in order to have it match the malliavin derivative constructed using other methods?

If it is used in the theory can you please tell me the point where it becomes important.

Thank you

Answer 1

Since the $(I_n)$ form an orthogonal family, the $\mathrm{L}^2(P)$ norm of $F$ is given by

$$\|F\|^2=\sum_{n\ge0}\|I_n(f_n)\|^2=\sum_{n\ge0}n!\|f_n\|_{\mathrm{L}^2([0,T]^n)}^2,$$

which is finite since $F\in\mathrm{L}^2(P)$. So this is where the $n!$ comes from. Condition 3. is chosen in order for $D_t F$ to be in $\mathrm{L}^2$.