Multiple Ito integral viewed as an isometry in a (symmetric) tensor product space.

Question

Let $H$ be a Gaussian Hilbert space spanned by a standard Brownian motion $\{B_t\}_{t\geq 0}$, and let the isometry $I:L^2([0,\infty))\to H$ be the Ito integral.

The author of my textbook states that since $I$ is an isometry then $I^{\odot n}$ is an isometry from $L^2([0,\infty))^{\odot n}$ onto $H^{\odot n}$.

Then $L^2([0,\infty))^{\odot n}$ can be identify with the space $L^2(D_n)$ where

$$D_n:=\{(t_1,\cdots,t_n):0<t_1<\cdots<t_n<\infty\}.$$

Furthermore the symmetric tensor product space $H^{\odot n}$ is isometric to the $n$th Wiener chaos $H^{:n:}$.

Then combining these identifications the author states that there exists an isometry

$$I_n:L^2(D_n)\to H^{:n:}$$ such that

$$I_n(f_1\odot\cdots \odot f_n)=:I(f_1)\cdots I(f_n):$$

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What I don't really understand is the following:

1. Is $I^{\odot n}(f_1\odot \cdots \odot f_n)=I(f_1)\odot \cdots\odot I(f_n)$?

Answer 1

Yes, you have that $$I^{\odot n}(f_1 \odot \dots \odot f_n) = I(f_1) \odot \dots \odot I(f_n)$$ In fact, this is essentially the definition of $I^{\odot n}$.