# If $σ_n^2=\frac{c^2}{n-1}$, how can we show $\limsup_{n→∞}n\sup_{x\in\mathbb R}\int\mathcal N_{x,\:σ_n^2}({\rm d}y)|\varphi(y)-\varphi(x)|<\infty$?

Let

• $$c>0$$
• $$\sigma_n^2:=\frac{c^2}{n-1}$$ for $$n\in\mathbb N$$ with $$n>1$$
• $$\varphi\in C_c^\infty(\mathbb R)$$

I want to show that $$\limsup_{n\to\infty}n\sup_{x\in\mathbb R}\int\mathcal N_{x,\:\sigma_n^2}({\rm d}y)\left|\varphi(y)-\varphi(x)\right|<\infty\tag1.$$

My problem is the position of $$\left|\;\cdot\;\right|$$. Let me elaborate on what I mean: Let $$x\in\mathbb R$$ and $$Y\sim\mathcal N_{x,\:\sigma_n^2}$$ for some $$n\in\mathbb N$$ with $$n>1$$. By Taylor's theorem, $$\varphi(Y)-\varphi(x)=\varphi'(x)(Y-x)+\frac12\varphi''(Z)(Y-X)^2\tag2$$ for some random variable $$Z\in[x\wedge Y,x\vee Y]$$. Thus, $$\operatorname E\left[\varphi(Y)-\varphi(x)\right]=\varphi'(x)\underbrace{\operatorname E\left[Y-x\right]}_{=\:0}+\frac12\operatorname E\left[\varphi''(Z)(Y-X)^2\right]\tag3$$ and hence $$n\left|\operatorname E\left[\varphi(Y)-\varphi(x)\right]\right|\le\frac n2\left\|\varphi''\right\|_\infty\underbrace{\operatorname E\left[\left|Y-x\right|^2\right]}_{=\:\sigma_n^2}=\frac c2\left\|\varphi''\right\|_\infty\frac n{n-1}\tag4.$$ Since $$\frac n{n-1}\xrightarrow{n\to\infty}1$$, we're able to conclude $$\limsup_{n\to\infty}n\sup_{x\in\mathbb R}\left|\int\mathcal N_{x,\:\sigma_n^2}({\rm d}y)\left(\varphi(y)-\varphi(x)\right)\right|<\infty\tag5.$$

The crucial observation is $$\operatorname E\left[Y-x\right]=0$$ in $$(3)$$. With the position of $$\left|\;\cdot\;\right|$$ in $$(1)$$, the corresponding term is $$\operatorname E\left[\left|Y-x\right|\right]=\sqrt{\frac 2\pi}\sigma_n$$ and this is a major problem, since $$n\sigma_n\xrightarrow{n\to\infty}\infty$$. So, is the statement wrong or can we fix this?

The assertion is, in general, wrong.

Clearly,

$$\int |\varphi(y)-\varphi(x)| \mathcal{N}_{x,\sigma_n^2}(dy) = \mathbb{E}|\varphi(x+\sigma_n U)-\varphi(x)|$$

where $$U \sim N(0,1)$$. Let's consider $$x=0$$ and some function $$\varphi \in C_c^{\infty}(\mathbb{R})$$ such that $$-1 \leq \varphi \leq 1$$ and $$\varphi(x)=x$$ for $$|x| \leq 1$$. Then

\begin{align*} \mathbb{E}(|\varphi(\sigma_n U)-\varphi(0)|) = \mathbb{E}(|\varphi(\sigma_n U)|) &= \mathbb{E}(|\sigma_n U| 1_{\{|\sigma_n U| \leq 1\}}) + \mathbb{E}(|\varphi(\sigma_n U)| 1_{\{||\sigma_n U|>1\}}) \\ &\geq \mathbb{E}(|\sigma_n U| 1_{\{|\sigma_n U| \leq 1\}}) - \mathbb{P}(|\sigma_n U|>1). \tag{1} \end{align*}

Since, by Markov's inequality,$$\mathbb{P}(|\sigma_n U|>1) \leq \mathbb{E}(|\sigma_n U|^2) = \sigma_n^2 \tag{2}$$ and \begin{align*} \mathbb{E}(|\sigma_n U| 1_{\{|\sigma_n U| \leq 1\}}) &= \frac{1}{\sqrt{2\pi \sigma_n^2}} \int_{-1}^1 |y| \exp \left(- \frac{y^2}{2 \sigma_n^2} \right) \, dy \\ &= \frac{2}{\sqrt{2\pi \sigma_n^2}} \int_{0}^1 y \exp \left(- \frac{y^2}{2 \sigma_n^2} \right) \, dy \\ &=\sqrt{\frac{2}{\pi}} \sigma_n \left(1- \exp \left(- \frac{1}{2\sigma_n^2} \right) \right) \\ &\geq \delta \sigma_n \end{align*} for some $$\delta>0$$ and $$n \gg 1$$, it follows that

$$\mathbb{E}(|\varphi(\sigma_n U)-\varphi(0)|) \geq \delta \sigma_n - \sigma_n^2$$ for $$n \gg 1$$ and so

$$\limsup_{n \to \infty} n \mathbb{E}(|\varphi(\sigma_n U)-\varphi(0)|) \geq \delta \liminf_{n \to \infty} n \sigma_n - \limsup_{n \to \infty} n \sigma_n^2 = \infty.$$

• This question were the easiest road to take for the actual question I would like to ask: math.stackexchange.com/q/3208469/47771. Unfortunately, it is a bit complicated to state, but maybe you can take a look. Commented Apr 30, 2019 at 16:12