Ornstein-Uhlenbeck operator and divergence operator

Question

So I'm still struggling with Malliavin calculus, and this time about the divergence operator. We are working in the classical Wiener space $(W,H,\mu)$ where $W$ is the Wiener space ($\mathcal{C}([0,1])$), $\mu$ the Wiener measure and $H$ is the Cameron Martin space.

Let us denote $P_t$ the Ornstein-Uhlenbeck operator s.t. : $$P_t f(x)=\int_W f\left(e^{-t}x+\sqrt{1-e^{-2t}}y\right)d\mu(y)$$

And $\nabla$,$\delta$ the Malliavin derivative and the divergence operator (adjoint of $\nabla$).

I want to show that $\boxed{P_t\delta u=e^{-t}\delta(P_t u)}$ ($u\in\mathbb{D}_{p,1}(H)$)

I already showed that $\nabla P_t\phi=e^{-t}P_t\nabla\phi$ for $\phi\in\mathbb{D}_{p,1}$ (which is pretty straightforward). I imagine that for divergence it should also be straightforward but for some reason I'm stucked :

Let $\phi\in\mathbb{D}_{q,1}$ (q conjuguate of p). We start from the left hand side :

\begin{align*} \mathbb{E}[\delta(P_tu)\phi]&=\mathbb{E}[\langle P_tu,\nabla\phi\rangle_H]\\ &= \mathbb{E}\left[\int_W \langle u(e^{-t}x+\sqrt{1-e^{-2t}}y),\nabla \phi\rangle_H d\mu(y)\right] \\ &=\int_W \mathbb{E}\left[\langle u(e^{-t}x+\sqrt{1-e^{-2t}}y),\nabla \phi\rangle_H\right] d\mu(y) \end{align*}

But from here I'd like to do some variable change, ie : $\mathbb{E}[\langle u(z),\nabla \psi(z)\rangle_H]$ to be able to get $\delta u$ with the duality identity, but I don't know how to do this properly without assuming something similar than what I'm trying to prove...

I think my problem is that I'm not used to duality reasoning. So if there is a general scheme to process this kind of identity, please feel free to mention it !

Answer 1

So since I was missing something and didn't get any answer, I tried to start all over from Nualart's book. The point that I missed was the invariance of the product Wiener measure to rotations.

Now with this fact we have that $\mathbb{E}[\langle F,P_t y\rangle]=\mathbb{E}[\langle P_tF,u\rangle]$ \begin{align*} \mathbb{E}[\delta(P_tu)]&=\mathbb{E}[\langle P_t u,\nabla \phi\rangle_H]\\ &=\mathbb{E}[\langle u,P_t\nabla\phi\rangle_H]\\ &=\mathbb{E}[\langle u,e^t\nabla P_t\phi\rangle_H]\\ &=e^t\mathbb{E}[\delta(u)P_t\phi] \end{align*}

Where we used the dual result at the third equality. This completes the proof.