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This question is an extension of the one I posted on ME: https://math.stackexchange.com/questions/4701500/if-alpha-nx-int-lvert-x-y-rvert-leq-1-n-lvert-x-y-rvert2-d-muy

It might be elementary for here, but I would deeply appreciate any help.

Let for each $n \in \mathbb{N}$, let $r_n : \mathbb{R}^m \to (0,\infty)$ be a sequence of smooth functions that converges to zero "pointwise" as $n \to \infty$.

Also let $\mu$ be a sufficiently regular Borel probability measure on $\mathbb{R}^m$. For concreteness, we can think of the stadard normal Gaussian measure.

Now, define \begin{equation} \alpha_n(x):=\int_{\lVert x-y \lVert \leq r_n(x)} \lVert x-y \rVert^2 d\mu(y) \end{equation} for each $n$ and let $A_n := \int_{\mathbb{R}^m} \alpha_n(x) d\mu(x)$.

Then, I suspect that for any bounded real-valued smooth function $F$ on $\mathbb{R}^m$, we have \begin{equation} \frac{1}{A_n}\int_{\mathbb{R}^m} F(x) \alpha_n(x) d\mu(x) \to \int_{\mathbb{R}^m} F(x) d\mu(x) \end{equation} as $n \to \infty$.

However, I cannot find a way to justify this guess rigorously. I tried to use convolution theorems but they do not fit in the above formula. Perhaps it is related to ergodicity?

Could anyone please help me?

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    $\begingroup$ What is the intuition behind your claim? If seems to me you are basically asking whether the sequence of measures given by $\frac{1}{A_n} \alpha_n \mu$ converges weakly to $\mu$, and I just don't see a good reason why this should be true with just pointwise convergence of $r_n$. $\endgroup$ Commented May 18, 2023 at 13:40

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$\newcommand\al\alpha\newcommand\be\beta\newcommand\R{\mathbb R}$This is not true in general. E.g., suppose that $m=1$, $\mu=N(0,1)$, and $r_n(x)=r:=1/n$ for all $n$ and $x$. Let $f$ be the standard normal pdf.

Then for each real $x$ (and $n\to\infty$) we have $$\al_n(x)\sim\int_{x-r}^{x+r}(y-x)^2\,dy\,f(x)=\frac{2r^3}3\,f(x).$$ Next, letting $X$, $Y$, and $Z$ denote independent standard normal random variables, we get $$A_n=\int\al_n f=E(X-Y)^2\,1(|X-Y|\le r) =2EZ^2\,1(|Z|\le r/\sqrt2) \\ \sim 2f(0)\int_{-r/\sqrt2}^{r/\sqrt2}dz\,z^2=f(0)\frac{r^3\sqrt2}3.$$ So, for $F:=f$ and $$L_n:=\frac1{A_n}\int F\al_n \,d\mu$$ we have $$\liminf_n L_n\ge\int f\frac{\sqrt2}{f(0)}\,f \,f=L:=\frac{2}{\sqrt{3 \pi }},$$ whereas $$R:=\int F\,d\mu=\int f^2=\frac{1}{2 \sqrt{\pi }}<L.$$ So, $L_n\not\to R$. $\quad\Box$

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    $\begingroup$ @Isaac : No -- in my example we actually have $r_n(x)=1/n$. $\endgroup$ Commented May 18, 2023 at 15:48
  • $\begingroup$ I see... So in case of $r_n(x)=1/n$, I presume it is in general $R \leq \lim_n L_n$? Perhaps I can show this? $\endgroup$
    – Isaac
    Commented May 18, 2023 at 15:54
  • $\begingroup$ @Isaac : I don't think the choice of $r_n(x)$ plays the biggest role here. Rather, there seems to be some structural defect in this setting. Note that in the above example $L_n$ is $\sim c\int f^3=L$ for some constant $c$, whereas $R=\int f^2$ -- so that $L$ and $R$ are not of the same form. $\endgroup$ Commented May 18, 2023 at 16:18
  • $\begingroup$ Oh, I considered the case when $F$ is not necessarily $f$. Then $L_n$ is of the form $c \int f^2 F$ and $R=\int Ff$...Still can we compare these two? $\endgroup$
    – Isaac
    Commented May 18, 2023 at 16:20
  • $\begingroup$ Oh, provided that $F$ is nonnegative.. $\endgroup$
    – Isaac
    Commented May 18, 2023 at 16:26

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