# If $\alpha_n(x):=\int_{\lVert x-y\rVert \leq 1/n } \lVert x-y\rVert^2 d\mu(y)$, how should I understand limit of normalized $\alpha_n$?

The question is as in the title.

Let $$\alpha_n(x):=\int_{\lVert x-y\rVert \leq 1/n } \lVert x-y\rVert^2 d\mu(y)$$ where $$x,y \in \mathbb{R}^m$$ and $$\mu$$ is sufficiently regular Borel proability measure; for concreteness, we can think of the standard normal Gaussian measure.

Also let $$A_n:=\int_{\mathbb{R}^m} \alpha_n(x) d\mu(x)$$.

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

However, I cannot justify this rigorously or know what kind of concept it is. Perhaps it is related to ergodicity?

Could anyone please provide any information?

• It feels true. But does not seems easy to justify. May 18, 2023 at 5:33

I think you should consider $$\frac{1}{A_n} \int_{\mathbb R^m} F(x) \alpha_n(x) dx$$ instead of the integral with respect to $$d\mu(x)$$. The intuitive reason for this is that you are integrating with respect to $$\mu$$ twice on the LHS of the convergence (once in $$\alpha_n$$, once in the actual LHS), but only once on the RHS.
I will nevertheless try to adress both the question as posted, and the question with my suggested edit. In what follows, $$\mu$$ is any Borel probability measure $$\mu$$ (no further regularity required).
The original question. Using $$[A]$$ for the indicator function of the proposition $$A$$ that is $$1$$ if and only if $$A$$ is true and $$0$$ otherwise, $$\int F(x) \alpha_n(x) d\mu(x) = \int \int F(x) \|x-y\|^2 [\|x-y\| \leq n^{-1}] d\mu(x) d\mu(y) .$$ Similarly, $$A_n$$ is the same expression with $$F \equiv 1$$. Assuming that $$\mu$$ has a smooth density $$p$$ with respect to $$dx$$ and writing $$\omega_m = \int_{\|x\|\leq 1} \|x\|^2 dx$$, asymptotically $$\int F(x) \alpha_n(x) d\mu(x) \sim n^{-2-m} \omega_m \int F(x) p(x)^2 dx$$ so that $$\frac{1}{A_n} \int F(x) \alpha_n(x) d\mu(x) \to \int F(x) \frac{p^2(x)}{\int p^2} dx ,$$ which is not the convergence you are looking for.
The modified question. We now integrate the LHS with respect to $$dx$$ and not $$d\mu(x)$$. Define $$\rho(x) = \begin{cases} \frac{\|x\|^2}{\int_{\|x\|\leq 1} \|x\|^2 dx} &\text{ if } \|x\|\leq 1, \\ 0 &\text{ otherwise.} \end{cases}$$ Note $$\int \rho = 1$$. Then $$\alpha_n(x) = \left( \int_{\|x\|\leq 1} \|x\|^2 dx\right) \int n^2 \rho(n(x-y)) d\mu(y) = \left( \int_{\|x\|\leq 1} \|x\|^2 dx\right) n^{2-m} \rho_n * \mu(x) ,$$ where $$\rho_n(t) = n^m \rho(nt)$$ is such that $$\int \rho_n = 1$$. It is pretty standard that $$(\rho_n \star \mu)(x) dx \to \mu$$ in the weak topology as $$n\to\infty$$, i.e. for every continuous and bounded $$F$$ (again, no further regularity required) $$\int (\rho_n \star \mu)(x) F(x) dx \to \int F(x) d\mu(x) .$$ To use your notation with the modified $$A_n = \int \alpha_n(x) dx$$ (instead of an integral against $$d\mu(x)$$): $$\frac{1}{A_n} \int \alpha_n(x) F(x) dx \to \int F(x) d\mu(x) .$$
Proof of the "pretty standard" convergence. Assume first that $$F$$ is continuous with compact support. We use that $$\rho(-x) = \rho(x)$$: \begin{align} \int (\rho_n * \mu)(x) F(x) dx &= \int \int \rho_n(x-t) F(x) dx d\mu(t) = \int (\rho_n * F)(t) d\mu(t) . \end{align} The exchange of the integrals is okay because the function $$(t,x) \mapsto \rho_n(x-t) F(x)$$ is integrable under $$dxd\mu(t)$$, thus Fubini’s theorem applies. Then use convergence of convolutions and approximation of unity to prove that $$\rho_n*F \to F$$ uniformly enough for the convergence to hold. This proves the convergence $$\rho_n * \mu(x) dx \to \mu$$ in the vague topology, which is equivalent to the convergence in the weak topology, see Corollary 10.5 in Relation between vague convergence and weak convergence. (The results I link also hold on $$\mathbb R^m$$ and not only $$\mathbb R$$.)