In a book by Bogachev-Krylov-Rockner-Shaposhnikov, I found the following statement that concludes a proof but I do not understand. I underlined in red the critical parts.
$\rho(\cdot,t)>0$ is a probability density function solving a certain parabolic PDE in a weak sense. The function $f_\epsilon$ approximates $\rho(\cdot,t)$ as $\epsilon\to0$, precisely: $$ f_\epsilon(x,t) \,:=\, \big(\rho(\cdot,t)*w_\epsilon\big)(x) \,+\, \epsilon\,\max(1,|x|)^{-d-1}$$ where $w_\epsilon(x)=\frac{1}{(2\pi\epsilon^2)^{d/2}}\,e^{-|x|^2/(2\epsilon^2)}\,$. The Sobolev space $W^{2,1}$ denotes those functions with weak derivatives up to the first order belonging to $L^2$ (notice that the indices of differentiability and integrability are reversed with respect to "usual" notation).
How can I deduce the last red statement?
The first red part is clear, since $$\nabla \sqrt{\rho(\cdot,t)} \,=\, \frac{1}{2}\,\frac{\nabla\rho(\cdot,t)}{\sqrt{\rho(\cdot,t)}}$$ hence the approximation argument shows that this weak gradient exists and belongs to $L^2(\mathbb R^d)$.
Now, since $\rho(\cdot,t)>0$ is a probability density function (by hypothesis), the second red part follows from the first red part by Cauchy-Schwarz inequality: $$\int_{\mathbb R^d} |\nabla\rho(x,t)|\,d x \,\leq\, \int_{\mathbb R^d} \frac{|\nabla\rho(x,t)|^2}{\rho(x,t)}\,d x \ \underbrace{\int_{\mathbb R^d} \rho(x,t)\,d x}_{=\,1} \ <\infty $$ for almost all $t\in(0,\tau)$, hence $\rho(\cdot,t)\in W^{1,1}(\mathbb R^d)$ for almost all $t\in(0,\tau)$.
I don't understand why the last red statement holds true. $\sqrt{\rho}\in\mathbb H^{2,2}$ means -by definition of this space- that $\partial_i\partial_j \sqrt{\rho(\cdot,t)}\in L^2(\mathbb R^d)$ for almost all $t\in(0,\tau)$ and $$\int_0^\tau \int|\partial_i\partial_j \sqrt{\rho(x,t)}|^2 d x\, dt <\infty\,.$$ Why this is the case? Why the weak second derivatives even exist?
Edit. The full reference is Theorem 7.4.1 in the book "Fokker-Planck-Kolmogorov equations" by Bogachev, Krylov, Röckner, Shaposhnikov. Here's the statement:
