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"?

  • 1
    $\begingroup$ See Remark 2 on page 31, where he extends the derivative to $H$-valued random variables. $\endgroup$ May 6, 2020 at 18:36
  • $\begingroup$ Thanks @NateEldredge I'll check it out. $\endgroup$
    – Chaos
    May 6, 2020 at 19:13

1 Answer 1


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.

  • $\begingroup$ Thank you very much! Nice and detailed explanation $\endgroup$
    – Chaos
    May 6, 2020 at 19:14
  • $\begingroup$ Hey Rhys sorry to bother you again, I was wondering in the penultimate step, when we have $$\langle \sum \partial_i (D^hF)\cdot e_i, h_j\rangle$$ How do we conclude the left term is the Malliavin derivative of the directional derivative? Do we assume $D^hF$ is a functional of the orthonormal vectors $e_i$'s? What I mean is that, in the definition of the Malliavin derivative we have the sum of partial derivatives of $f$ times the argument $h\in H$ of $W$. In this case we are multiplying by $e_i$, that's my doubt! Thanks in advance man! $\endgroup$
    – Chaos
    May 11, 2020 at 10:10
  • $\begingroup$ If you write $D^hF_j$ explicitly you should see it is of the right form. You have $D^hF_j = D^h[f_j(W(e_1), \dots, W(e_n))] = \sum_{k=1}^n \partial_k f_j(W(e_1), \dots, W(e_n)) \langle e_k, h \rangle$. This is a smooth function of $W(e_1), \dots, W(e_n)$. $\endgroup$ May 11, 2020 at 10:28
  • $\begingroup$ Ohh I was trying to find a representation of some kind without realizing it was already a functional of $W(e_i)$, thanks again! $\endgroup$
    – Chaos
    May 11, 2020 at 10:39

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