Show that the probability of this sequence of random variables being a root of this function tends to $0$

Let

• $$f\in C^3(\mathbb R)$$ with $$f>0$$, $$\int f(x)\:{\rm d}x=1$$ and such that $$(\ln f)'$$ is Lipschitz continuous, $$p_n(x):=\prod_{i=1}^nf(x_i)\;\;\;\text{for }x\in\mathbb R^n$$ and $$h_n^{(x)}(z):=\frac{p_n(x+z)}{p_n(x)}-1\;\;\;\text{for }x,z\in\mathbb R^n$$ for $$n\in\mathbb N$$
• $$(\sigma_n)_{n\in\mathbb N}\subseteq(0,\infty)$$ be decreasing with $$\sigma_n\xrightarrow{n\to\infty}0$$
• $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space
• $$X^{(n)},Y^{(n)}$$ be $$\mathbb R^d$$-valued random variables with $$X^{(n)}\sim p_n\lambda^n$$ ($$\lambda^n$$ denoting the Lebesgue measure on $$\mathcal B(\mathbb R^n)$$) and $$\operatorname P\left[Y^{(n)}\in B\mid X^{(n)}\right]=\mathcal N_d\left(X^{(n)},\sigma_n^2I_n\right)\;\;\;\text{almost surely for all }B\in\mathcal B(\mathbb R^n)\tag1$$ ($$I_n$$ denoting the $$n\times n$$ identity matrix) for $$n\in\mathbb N$$

Note that by $$(1)$$ there is a $$\mathbb R^d$$-valued random variable $$Z^{(n)}$$ on $$(\Omega,\mathcal A,\operatorname P)$$ independent of $$X^{(n)}$$ with $$Z^{(n)}\sim\mathcal N_d(0,\sigma_n^2I_n)$$ and $$Y^{(n)}=X^{(n)}+Z^{(n)}\tag2$$ for all $$n\in\mathbb N$$.

I want to show that $$\operatorname P\left[h_n^{\left(X^{(n)}\right)}\left(Z^{(n)}\right)=0\right]\xrightarrow{n\to\infty}0\tag3.$$

Obviously, $$h_n^{(x)}(z)=0\Leftrightarrow\prod_{i=1}^n\frac{f(x_i+z_i)}{f(x_i)}=1\;\;\;\text{for all }x,z\in\mathbb R^d.\tag4$$

We may note that the Lipschitz continuity (with Lipschitz constant $$c$$, say) of the derivative of $$g:=\ln f$$ implies the following:

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 $$\frac c{\sqrt{2\pi e}}$$

My intuition is that if you choose $$f$$ constant on a neighborhood of $$0$$, then there will be a growing probability that the coordinates of $$X^{(n)}$$ and $$Y^{(n)}$$ lay both in this neighborhood (as the variance of the Gaussian goes to zero), meaning that the probability you are interested in does not go to zero in this case.