Let

- $f\in C^3(\mathbb R)$ with $f>0$ and $$\int f(x)\:{\rm d}x=1\tag1$$
- $g:=\ln f$ and assume that $g'=\frac{f'}f$ is Lipschitz continuous (note that this implies that $f'(x)\xrightarrow{|x|\to\infty}0$)
- $n\in\mathbb N$, $x\in\mathbb R^n$, $$s(z_1,y_2,\ldots,y_n):=g(x_1+c_nz_1)-g(x_1)+\sum_{i=2}^n(g(y_i)-g(x_i))\;\;\;\text{for }z_1,y_2,\ldots,y_n\in\mathbb R$$ for some $c_n>0$ and $$h(z_1,y_2,\ldots,y_n):=e^{s(z_1,\:y_2,\:\ldots\:,\:y_n)}-1=\frac{f(x_1+c_nz_1)}{f(x_1)}\prod_{i=2}^n\frac{f(y_i)}{f(x_i)}-1\;\;\;\text{for }z_1,y_2,\ldots,y_n\in\mathbb R$$

For an arbitrary differentiable $\tilde h:\mathbb R\to\mathbb R$, we're able to show that $$N:=\left\{a\in\mathbb R:\tilde h(a)=0\text{ and }\tilde h'(a)\ne0\right\}$$ is countable and $|\tilde h|$ is differentiable on $\mathbb R\setminus N$ with $$|\tilde h|'(x)=\begin{cases}\displaystyle\frac{\tilde h(a)}{\left|\tilde h(a)\right|}\tilde h'(a)&\text{, if }\tilde h(a)\ne0\\0&\text{, if }\tilde h(a)=0\text{ and }\tilde h'(a)=0\end{cases}\tag2$$ for all $a\in\mathbb R$. So, an analogous result can be shown for the partial derivative of $h$ with respect to $z_1$. Are we even able to show that there is a Borel measurable $N'\subseteq\mathbb R^n$ with Lebesgue measure $0$ such that $|h|$ is twice partially differentiable with respect to $z_1$ on $\mathbb R^n\setminus N'$?

Feel free to assume that $c_n$ is decreasing in $n$ with $c_n\xrightarrow{n\to\infty}0$ and that $n$ is as large as you like.

**EDIT**: Let me note that the assumptions yield the following properties of $f$, which might be useful (let $c$ denote the Lipschitz constant of $g'$):

- $g(y)-g(x)-g'(x)(y-x)\ge-\frac c2|y-x|^2$ for all $x,y\in\mathbb R$
- $f(y)\ge f(x)e^{\frac{|g'(x)|^2}{2c}}e^{-\frac c2\left|y-x-\frac{g'(x)}c\right|^2}\ge f(x)e^{-\frac c2\left|y-x-\frac{g'(x)}c\right|^2}$ for all $x,y\in\mathbb R$
- $f(x)\le\sqrt{\frac c{2\pi}}e^{-\frac{|g'(x)|^2}{2c}}\le\sqrt{\frac c{2\pi}}$ for all $x\in\mathbb R$
- $f$ is Lipschitz continuous with constant $\frac c{\sqrt{2\pi e}}$

thisquestion, since the setting here is more concrete. If it's not possible for all $x\in\mathbb R^d$, it would be enough for me if it can be shown for a suitable rich class $F_d\in\mathcal B(\mathbb R^d)$ of $x$ such that for the process $X^{(d)}$ from this question it holds $\operatorname P\left[X^{(d)}_t\in F_d\text{ for all }t\in[0,T]\right]\xrightarrow{d\to\infty}1$ for all $T>0$. $\endgroup$ – 0xbadf00d May 11 '19 at 15:44