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 $x_0$.