# Show that the absolute value of this function is twice differentiable except on a set of Lebesgue measure $0$

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 $$g'$$):

1. $$g(y)-g(x)-g'(x)(y-x)\ge-\frac c2|y-x|^2$$ for all $$x,y\in\mathbb R$$
2. $$f(y)\ge f(x)e^{\frac{|g'(x)|^2}{2c}}e^{-\frac c2\left|y-x-\frac{g'(x)}c\right|^2}\ge f(x)e^{-\frac c2\left|y-x-\frac{g'(x)}c\right|^2}$$ for all $$x,y\in\mathbb R$$
3. $$f(x)\le\sqrt{\frac c{2\pi}}e^{-\frac{|g'(x)|^2}{2c}}\le\sqrt{\frac c{2\pi}}$$ for all $$x\in\mathbb R$$
4. $$f$$ is Lipschitz continuous with constant $$\frac c{\sqrt{2\pi e}}$$
• Can I clarify that the answers to math.stackexchange.com/questions/3209891/… now answer your question here? Or is there something I'm missing? – Julian Newman May 11 '19 at 14:01
• @JulianNewman I'd hoped that there is a positive answer to this question, 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 $T>0$. – 0xbadf00d May 11 '19 at 15:44

Let $$h:\mathbb R\rightarrow \mathbb R$$ continuous such that the second distribution derivative $$h''$$ is continuous and bounded. We define $$\rho(x)=\vert{h(x)}\vert^{\frac{1}{2}}+\vert{h'(x)}\vert,$$ so that $$\vert{h^{(j)}(x)}\vert\le \rho(x)^{2-j}\quad \text{for 0\le j\le 1}.$$ We define as well the open set $$\Omega=\{x\in \mathbb R, \rho(x) >0\}.$$ Then there exist positive constants $$r, C$$ such that $$x\in \Omega, \vert{y-x}\vert\le r\rho(x)\Longrightarrow y\in \Omega, C^{-1}\le \frac{\rho(y)}{\rho(x)}\le C. \tag{\ast}$$ As a result, we can construct a (Calder\'on-Zygmund) partition of unity $$\mathbf 1_\Omega=\sum_{\nu\in \mathbb N}\chi_\nu(x)$$ with $$\chi_\nu\in C^\infty_c(\Omega)$$ and $$\text{supp}\chi_\nu\subset B(x_\nu, r\rho(x_\nu))$$ with each $$x_\nu\in \Omega$$, $$\forall j\in \{0,1\},\quad \sup_{\nu,x}\rho(x_\nu)^{j}\vert \chi_\nu^{(j)}(x)\vert\le C_j<+\infty.$$ You get that $$h(x)=\sum_\nu h(x) \chi_\nu(x)$$ and each function $$h_\nu=\chi_\nu h$$ has a normal form depending on which term in the definition of $$\rho$$ is dominant.
1. If $$h^{1/2}$$ is dominant, then $$h_\nu(x)$$ behaves as $$\pm\rho_\nu^2 \phi(x/\rho_\nu)$$ and thus $$\vert h_\nu\vert$$ behaves as $$\rho_\nu^2 \phi(x/\rho_\nu)$$ which is twice differentiable.
2. If $$h^{1/2}$$ is not dominant but $$\vert h'\vert$$ is dominant, then $$h_\nu(x)$$ behaves as $$\rho_\nu(x-x_\nu)$$ and thus $$\vert h_\nu\vert$$ behaves as $$\rho_\nu\vert x-x_\nu\vert$$, thus is twice differentiable except at $$x_\nu$$.
Of course, you have to prove $$(\ast)$$, the existence of a partition of unity and justify the discussion on normal forms with 1,2. My point with that answer is that it describes an example of a Calder\'on-Zygmund decomposition, a tool which looks suitable for this type of problem.
• Honestly, at least for the moment, I understand almost nothing in your answer. Just to be sure: You know that I'm interested in the result for the specific $h$ defined by $$h(z_1,y_2,\ldots,y_n):=e^{s(z_1,\:y_2,\:\ldots\:,\:y_n)}-1=\frac{f(x_1+c_nz_1)}{f(x_1)}\prod_{i=2}^n\frac{f(y_i)}{f(x_i)}-1\;\;\;\text{for }z_1,y_2,\ldots,y_n\in\mathbb R,$$ right? If that's the case: Does your answer indicate that the desired result is right or wrong? – 0xbadf00d May 12 '19 at 18:45
• The point of my answer is that considering your general function $\tilde h$, say twice differentiable, you have some normal forms via a partition of unity, that is the function behaves as some polynomial of low degree near a collection of points $x_\nu$: in the ball with center $x_\nu$ and radius $\rho(x_\nu)$, your function resembles a polynomial. Now, you may also reinforce the assumption by supposing that the function $\tilde h$ is $C^3$ with a third derivative bounded. Then you have to change the definition of $\rho$ to have $\vert \tilde h^{(j)}\vert\lesssim \rho^{3-j}$ – Bazin May 13 '19 at 9:34