I am studying the Propagation of chaos from Sznitman. We start with an $\mathbb{R}^d$ valued process that satisfies the following SDE: $$dX_t=dB_t+\left(\int_{\mathbb{R}^d}b(X_t, y)u_t(dy)\right)dt,\;\; X_{t=0}=X_0,$$ where $u_t(dy)$ is the Law of $X_t$.
For $f\in C^2(\mathbb{R}^d)$, we obtain by applying Ito's formula $$f(X_t)=f(X_0)+\int_{0}^{t}\nabla f(X_s)\cdot dB_s+ \int_0^{t}\left(\frac{1}{2}\Delta f(X_s)+\nabla f(X_s)\cdot\int_{\mathbb{R}^d}b(X_s, y)u_s(dy)\right)ds.$$
Sznitman remarks that "after integration" this yields a weak version of the following PDE (let us call it PDE (*)) $$\partial_tu=\frac{1}{2}\Delta u-div\left(u\int b(\cdot, y)u_t(dy)\right).$$
I have very basic familiarity with SDEs and I am having trouble interpreting the last remark. My understanding is that "after integration" means taking expectations and then differentiating with respect to $t$. If I drop the $\int_0^t \nabla f(X_s)\cdot dB_s$ term then I can make some sense of the above statement.
I write everything explicitly (and interchange the integral and derivative wherever I need without justifying) and obtain
$$\int_{\mathbb{R}^d}f(x)\frac{\partial}{\partial t}u_t(dx)= \int_{\mathbb{R}^d}\frac{1}{2}\Delta f(x)u_t(dx)+\int_{\mathbb{R}^d} \left(\nabla f(x)\cdot \int_{\mathbb{R}^d}b(x, y)u_t(dy)\right)u_t(dx).$$
I can see that this is a weak version of (*). I am not sure if this is what meant by the author. And, also I am not able to handle the term $\int_0^t \nabla f(X_s)\cdot dB_s$. I can take expectation of this term and get $\int_{\Omega} \int_{0}^{t}\nabla f(X_s(\omega))\cdot dB_{s}(\omega)P(d\omega)$. I don't see why its derivative with respect to $t$ should be zero. I have an intuitive feeling that $$\int_{\Omega} \int_{0}^{t}\nabla f(X_s(\omega))\cdot dB_{s}(\omega)P(d\omega)=\int_{0}^{t}\left(\int_{\Omega} \nabla f(X_s(\omega))\cdot dB_{s}(\omega)P(d\omega)\right).$$ For every fixed $s$, the quanity in the bracket is expectation of a 'centered normal', that is, $dB_s$. And hence it is zero. This says that the intergral with respect to $t$ is zero and hence the derivative. I don't know if this intution is correct or not; but at any rate, it seems hard to formalize this idea.
Any help would be appreciated!
The term $\int_0^t \nabla f(X_s) \, dB_s$ is an Itô stochastic integral with a square integrable integrand (i.e. $\int_0^t\mathbb{E}(|\nabla f(X_s)|^2) \, \mathrm{d}s <\infty$, you will have to check this) and so is a martingale. For a martingale $M_t$, we have \begin{align} \mathbb{E}(M_t) = E(M_0) \, , \end{align} from which it follows that \begin{align} \mathbb{E}\left(\int_0^t \nabla f(X_s) \, dB_s\right) =0\, . \end{align}