Approximating the cdf of the standard normal distribution

Question

Let

• $d\in\mathbb N$ with $d>1$
• $\lambda$ denote the Lebesgue measure
• $f\in C^2(\mathbb R)$ be positive and $g:=\ln f$ with $$I:=\int\left|g'\right|^2\:{\rm d}(f\lambda)=\int\frac{\left|f'\right|^2}f\:{\rm d}\lambda<\infty$$
• $\Phi$ denote the cdf of the standard normal distribution
• $\ell>0$

Now, let $$g_d(x):=\frac1{d-1}\sum_{i=2}^d\left|g'(x_i)\right|^2\;\;\;\text{for }x\in\mathbb R^d.$$

Let $x\in\mathbb R^d$ with $|g_d(x)-I|<d^{-1/8}$. Are we able to show that $$\frac d{d-1}\Phi\left(-\frac{\ell\sqrt{g_d(x)}}2\right)\xrightarrow{d\to\infty}\Phi\left(-\frac{\ell\sqrt I}2\right)\tag1?$$

Answer 1

This actually follows just from the fact that $|g_d(x) - I| < d^{-1/8}$; since $\Phi$ is continuous at $I$ (it is in fact uniformly continuous), we can interchange the limits.