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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

$(\ast)X_{i+1}(t)=x_0+\int_0^t\sigma(X_i(s))dW_s$ 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 $x_0$.

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 $x_0$.

In Bass's book on Diffusions and Elliptic Operators, the author gives a brief introduction into Malliavin Calculus. He calls a functional $F:C([0,1],\mathbb{R})$ $L^p-$smooth if there exists a sequence $F_n$ of continuous, bounded, with continuous and 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 $F$ 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 $X_{i+1}(t)=x_0+\int_0^t\sigma(X_i(s))ds$ and then 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 $X$ is $L^p$ smooth.

The idea for the iteration seems to be the following: Generalise the idea of $L^p$ functionals to include Fréchet differentiable $F:C[0,1]\rightarrow \mathbb{R}^d$. Then for suitable $\mathbb{R}^d$-valued time indexed functionals $Y_t$ one can see by approximation that the map $\int_0^t (\sigma\circ Y_s)dW_s$ will again be an $\mathbb{R}^d$ valued functional on $C[0,1]$.

By assumption $X_0(t)=x_0\in\mathbb{R}^d$ and one easily sees that for $i=0$ $X_1(t)$ is $L^p$ smooth. Now, if this holds for $i=n-1$, it is easily seen to hold for $i=n$ by the just mentioned argument. Hence, each $X_i(t)$ is $L^p$ smooth. $X_i(t)$ converges to $X(t)$ in $L^p$.

Now, how can I conclude from this that also the Fréchet derivatives of $X_i(t)$ converge to the Fréchet derivative of $X(t)$? I suppose that this is what I need the convergence of the above $DX_i(t)$ to $DX(t)$ for, but I do not understand how this helps.

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

$(\ast)X_{i+1}(t)=x_0+\int_0^t\sigma(X_i(s))dW_s$ 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 $x_0$.

In Bass's book on Diffusions and Elliptic Operators, the author gives a brief introduction into Malliavin Calculus. He calls a functional $F:C([0,1],\mathbb{R})$ $L^p-$smooth if there exists a sequence $F_n$ of continuous, bounded, with continuous and 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 $F$ 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 $X_{i+1}(t)=x_0+\int_0^t\sigma(X_i(s))ds$ and then 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 $X$ is $L^p$ smooth.

Now I do not understand what it means for $X_{i+1}(t)$ (which is in $\mathbb{R}^d)$ to be $L^p$-smooth (as this term was only defined for functionals on $C[0,1]$) (even if he means this coordinate-wise, I would still have the problem of differentiating $\sigma(X_i(t))$ which would not depend on one coordinate only) and moreover I do not see where the expression for $DX_{i+1}(t)$ comes from (however, it does look a lot like the derivative $d/dx$ of the flow $X(x,t,\omega))$. For the Frechet derivative of $X_{i+1}(t)$ I would expect him to study something like this $\lim_{\epsilon\rightarrow 0}\frac{F(W+\epsilon H)-F(W)}{\epsilon}$ where $H$ is a suitable function.

So I have two questions: 1) What does he mean by $X_{i+1}(t)$ is $L^p$ smooth? 2) How does he find the Frechet derivative of $X_{i+1}(t)$?

In Bass's book on Diffusions and Elliptic Operators, the author gives a brief introduction into Malliavin Calculus. He calls a functional $F:C([0,1],\mathbb{R})$ $L^p-$smooth if there exists a sequence $F_n$ of continuous, bounded, with continuous and 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 $F$ 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 $X_{i+1}(t)=x_0+\int_0^t\sigma(X_i(s))ds$ and then 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 $X$ is $L^p$ smooth.

The idea for the iteration seems to be the following: Generalise the idea of $L^p$ functionals to include Fréchet differentiable $F:C[0,1]\rightarrow \mathbb{R}^d$. Then for suitable $\mathbb{R}^d$-valued time indexed functionals $Y_t$ one can see by approximation that the map $\int_0^t (\sigma\circ Y_s)dW_s$ will again be an $\mathbb{R}^d$ valued functional on $C[0,1]$.

By assumption $X_0(t)=x_0\in\mathbb{R}^d$ and one easily sees that for $i=0$ $X_1(t)$ is $L^p$ smooth. Now, if this holds for $i=n-1$, it is easily seen to hold for $i=n$ by the just mentioned argument. Hence, each $X_i(t)$ is $L^p$ smooth. $X_i(t)$ converges to $X(t)$ in $L^p$.

Now, how can I conclude from this that also the Fréchet derivatives of $X_i(t)$ converge to the Fréchet derivative of $X(t)$? I suppose that this is what I need the convergence of the above $DX_i(t)$ to $DX(t)$ for, but I do not understand how this helps.