Let $f\in C^3(\mathbb R)$ with $f>0$ and $$\int f(x)\:{\rm d}x=1$$ and $g:=\ln f$. Assume $g'$ is Lipschitz continuous. Let $n\in\mathbb N$, $x_1,\ldots,x_n,y_2,\ldots,y_n\in\mathbb R$ and $$H(z_1):=\frac{f(x_1+\sigma z_1)}{f(x_1)}\prod_{i=2}^n\frac{f(y_i)}{f(x_i)}-1\;\;\;\text{for }z_1\in\mathbb R$$ for some $\sigma>0$.
Now let $$N:=\{H=0\}\cap\{H'\ne0\}.$$ Are we able to show that $$N':=\overline N\cap\{H=H'=H''=0\}$$ is empty?
Note that $$H'(z_1)=0\Leftrightarrow f'(x_1+\sigma z_1)=0$$ and $$H''(z_1)=0\Leftrightarrow f''(x_1+\sigma z_1)=0$$ for all $z_1\in\mathbb R$.
Moreover, the Lipschitz continuity (with Lipschitz constant $c$, say) of $g'$ implies the following:
- $g(y)-g(x)-g'(x)(y-x)\ge-\frac c2|y-x|^2$ 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 $\frac c{\sqrt{2\pi e}}$