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In Bass's book on Diffusions and Elliptic Operators, the author gives a brief introduction into Malliavin Calculus. He calls a functional $F:C([0,1],\mathbb{R})\rightarrow \mathbb{R}$ $L^p-$smooth if there exists a sequence $F_n$ of continuous, bounded functionals, with continuous, bounded Fréchet derivatives such that

a) $F_n\rightarrow F$ in $L^p(\mu)$, where $\mu$ is the Wiener measure and

b) $D_WF_n(H)\rightarrow D_WF(H)$ in $L^p(\mu)$, where $D_WF(H)$ denotes the Fréchet derivative of $F$ at $W$ in direction $H=\int_0^\cdot h_s ds$ for suitable $h$, i.e., $\mathbb{E}[\int_0^1 h_s^2 ds]<\infty$.

In Theorem (2.3) on page 196 he shows that if $X$ is the solution to the SDE $dX_t=\sigma(X_t)dW_t$ ($\sigma$ is $mxd$ matrix, $X_0=x_0\in\mathbb{R}^d$, $b,\sigma$ satisfy a linear growth condition and are smooth), then $F(W):=f(X_1)$ ($f$ smooth) is $L^p$-smooth.

In the proof he uses the following iteration, starting with $X_0(t)=x_0$,

$(\ast) X_{i+1}(t)=x_0+\int_0^t\sigma(X_i(s))dW_s$ and says $DX_{i+1}(t)=I+\int_0^t\sigma'(X_i(s))DX_i(s)dW_s.$

He then goes on to show that $X_i$ and $DX_i$ converge in $L^p$ to the solutions $X$ and $DX$ of the respective SDEs without $i$-dependence. Then concludes that $X_t$ is $L^p$ smooth.

My problem is with $(\ast)$ and especially with the meaning of $DX_i$. Up until here the derivative $D_WF(H)$ only made sense in direction $H$. So if I assume for the moment that $DX_i$ is is the linear map that at $W$ maps $H$ to the directional derivative $D_WX_i(H)$, I would expect

$D_WX_i(H)=H+\int_0^t \sigma'(X_i(s))D_WX_i(s)H dW_s$.

However, even for the simple case $i=0$, I do not get this equality, as

$X_1(t)=x_0+\int_0^t\sigma'(X_0(s))dW_s=x_0+\sigma'(x_0)W_t$.

Therefore, the directional derivative would be

$\lim_{\epsilon\rightarrow 0}\frac{X_1(t)(\omega+\epsilon H)-X_1(t)(\omega)}{\epsilon}=\lim_{\epsilon\rightarrow 0}\frac{\sigma'(x_0)(\epsilon H_t}{\epsilon}=\sigma'(x_0)H_t$.

But $DX_1H(t)=H_t+\int_0^t \sigma'(x_0)DX_0(s)dW_s=H_t$

because I would expect $DX_0(s)=Dx_0=0$, as $x_0$ maps paths to the the constant value $x_0$.

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  • $\begingroup$ Does anyone have an idea? I know that Nualart does something similar in his monograph but it is only much later in the book that he covers this application. As I am very familiar with Bass's book it would be very helpful to understand his exposition of Malliavin Calculus and in particular its use in proofs of smooth densities. I would very much appreciate any ideas and comments. $\endgroup$ Commented Jan 30, 2016 at 23:04

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As Bass and Nualart mention (Theorem 2.2.1 in "MC and related topics"), they are talking about the matrix derivative DX, not the Malliavin derivative $D_{h}X$. That's what Bass means by "where we use the notation of (I.10.4).".

To the get Malliavin Lp-smootheness, he uses the approximation result (2.2) Lemma by the Lp smooth solutions built in the proof for both the X and DX sdes.

He doesn't explicitly work with Frechet derivatives but mostly leaves it as exercise at the step "it is easy to see that... is Frechet differentiable". Indeed, since we know that $Y$ is Lp-smooth, we can just use the product rule version for Malliavin.

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