Let
- $c>0$
- $\sigma_n^2:=\frac{c^2}{n-1}$ for $n\in\mathbb N$ with $n>1$
- $f\in C^3(\mathbb R)$ with $f>0$
- $g:=\ln f$ (and assume $g'$ is Lipschitz continuous), $$s_n(x,y):=\sum_{i=1}^n(g(y_i)-g(x_i))$$ and $$\tilde s_n(x,y):=(g(y_1)-g(x_1))+\sum_{i=2}^n\left(g'(x_i)(y_i-x_i)-\frac{\sigma_n^2}2{g'(x_i)}^2\right)$$ as well as $h_n(x,y):=\min\left(1,e^{s_n(x,\:y)}\right)$ and $\tilde h_n(x,y):=\min\left(1,e^{\tilde s_n(x,\:y)}\right)$ for $x,y\in\mathbb R^n$ and $n\in\mathbb N$
Assume $$M:=\int\frac{{f'(x)}^2}{f(x)}\:{\rm d}x<\infty$$ and let $$B_n:=\left\{x\in\mathbb R^n:\max\left(\left|\frac1{n-1}\sum_{i=2}^n{g'(x_i)}^2-M\right|,\left|-\frac1{n-1}\sum_{i=2}^ng''(x)-M\right|\right)<\frac1{n^8}\right\}$$ for $n\in\mathbb N$ with $n>1$.
We're able to show that $$\sup_{x\in B_n}\int\mathcal N_n(x,\sigma_n^2I_n)({\rm d}y)|h_n(x,y)-\tilde h_n(x,y)|\xrightarrow{n\to\infty}0.\tag1$$ Let $\varphi\in C_c^\infty(\mathbb R)$. Are we able to conclude that $$\sup_{x\in B_n}n\left|\int\mathcal N_1(x_1,\sigma_n^2)({\rm d}y_1)(\varphi(y_1)-\varphi(x_1))\int\mathcal N_{n-1}((x_2,\ldots,x_n),\sigma_n^2I_{n-1})({\rm d}(y_2,\ldots,y_n))(h_n(x,y)-\tilde h_n(x,y))\right|\xrightarrow{n\to\infty}0\tag2?$$
A clever application of Taylor's theorem might do the trick. This question is strongly related: If $σ_n^2=\frac{c^2}{n-1}$, how can we show $\limsup_{n→∞}n\sup_{x\in\mathbb R}\int\mathcal N_{x,\:σ_n^2}({\rm d}y)|\varphi(y)-\varphi(x)|<\infty$?.