On proposition $1.3.1$ of Nualart's book "The Malliavin Calculus and
Related Topics" it's stated the following. Let $H$ be a real separable
Hilbert space, and let $W=\{W(h),h\in H\}$ be a Gaussian isonormal
process.Let $u$ be a process of the form $$u=\sum_{j=1}^n F_j h_j,$$ where the
$F_j$ are smooth random variables (in the Malliavin calculus sense)
and the $h_j$ are elements on the Hilbert space $H$.We have the following commutativity relationship between the Malliavin
derivative and the divergence operators.$$D^h(\delta(u)):=\langle D\delta(u),h\rangle_H=\langle u,
h\rangle_H+\delta (D^h u) (\star)$$
My attempt:
Using the integration by parts formula I can see that
$$\delta(u)= \sum_{i=1}^n F_i W(h_i)-\sum_{i=1}^n \langle DF_i,h_i\rangle_H,$$
hence
$$D^h(\delta(u))=D^h\left(\sum_{i=1}^n F_i W(h_i)-\sum_{i=1}^n \langle DF_i,h_i\rangle_H\right)$$
The linearity of the Malliavin derivative allows us to write:
$$=\sum_i D^h(F_i W(h_i))-\sum_i D^h\langle DF_i,h_i\rangle_H$$
For the first term we can use the product rule and we obtain
$$=\langle h,u\rangle_H+ \sum_i (D^hF_i)W(h_i)-\color{red}{D^h\langle DF_i,h_i\rangle_H}.$$
At this point the author writes the red term as:
$$D^h\langle DF_i,h_i\rangle_H=\langle D(D^hF_i),h_i\rangle_H$$
(1) Why does this hold?
The definitions tells me:
$$D^h\langle DF_i,h_i\rangle= \langle D \langle DF_i,h_i\rangle, h\rangle=\langle D(D^{h_i}F_i),h\rangle.$$
(2) The Malliavin derivative is an operator that takes a real-valued random variable and returns us an $H$-valued random variable. Why on $\star$ we have $D^h u:= \langle Du, h\rangle$? $u$ is not a real-valued random variable, how to take the Malliavin derivative of this "stochastic process"?
Best Answer
Let me start by answering the second point. Nate Eldredge points to the right place in Nualart's book in a comment. Here I will just recreate the detail from there that if $G$ is a smooth $H$-valued random variable i.e. $$G = \sum_{j=1}^n G_j v_j$$ where the $G_j$ are smooth random variables and $v_j \in H$ then e.g. $$DG = \sum_{j=1}^n DG_j \otimes v_j$$ is valued in $H \otimes H$. One then checks in much the same way as the real-valued case that this defines a closable operator etc.
Now I turn to your first question.
Since $F_j$ is smooth, we can write $F_j = f_j(W(e_1), \dots, W(e_n))$ for $\{e_i: i = 1, \dots, n\}$ orthonormal in $H$ and $f_j$ smooth. Then \begin{align} D^h \langle DF_j, h_j \rangle =& D^h \left (\sum_{i=1}^n \partial_if_j(W(e_1),\dots,W(e_n)) \langle e_i, h_j \rangle \right) \\ =& \sum_{k=1}^n \sum_{i=1}^n \partial_k \partial_i f_j(W(e_1), \dots, W(e_n)) \langle e_i, h_j \rangle \langle e_k, h \rangle \\ =& \sum_{i=1}^n \sum_{k=1}^n \partial_i\partial_k f_j(W(e_1), \dots, W(e_n))\langle e_k, h \rangle\langle e_i, h_j \rangle \\ =& \sum_{i=1}^n \partial_i D^h F_j \langle e_i, h_j \rangle \\ =& \langle DD^h F_j, h_j \rangle \end{align} as desired.