# Ratio of integrals with increasing dimension over Euclidean balls

Let $$f_n(x)\geq0$$ be any sequence of nonnegative $$L^1(\mathbb{R}^n)$$ functions such that $$\int_{\mathbb{R}^{n}}f_n(x)dx=1$$ where $$dx$$ is the Lebesgue measure on $$\mathbb{R}^n$$. For any $$a>1,\epsilon>0$$, does there exist a sequence $$x_n\in\mathbb{R}^n$$ such that $$\lim_{n\to+\infty}a^n\frac{\int_{\|x-x_n\|^2\leq n^{1-\epsilon}}f_n(x)dx}{\int_{\|x-x_n\|^2\leq n}f_n(x)dx}=0$$?

If it does not hold, is there any counter example of the sequence $$f_n$$? What additional conditions do we need on the sequence $$f_n$$? A similar question is to show that $$\lim_{n\to+\infty}a^n\frac{\int_{\|x-x_n\|^2\leq n^{1-\epsilon}}\exp(-\|x_n-x\|^2)f_n(x)dx}{\int_{\|x-x_n\|^2\leq n}\exp(-\|x_n-x\|^2)f_n(x)dx}=0$$ ？This can be implied by the first display. Therefore, this is a weaker statement.

Such a sequence $$(x_n)$$ always exists. Indeed, the displayed ratio expression under the limit sign equals $$R_n(x_n)$$, where $$\begin{equation*} R_n(y):=\frac{g_{n,r_n}(y)}{g_{n,s_n}(y)},\quad r_n:=n^{(1-\epsilon)/2}, \quad s_n:=n^{1/2}, \end{equation*}$$ $$\begin{equation*} g_{n,r}(y):=\int f_n(x)1_{|x-y|_n and $$|\cdot|_n$$ is the Euclidean norm in $$\mathbb R^n$$. Note that for all real $$r>0$$ $$\begin{equation*} \int g_{n,r}(y)\, dy=\int dx\, f_n(x)\int dy\,1_{|x-y|_n the volume of any ball in $$\mathbb R^n$$ of radius $$r$$, whence $$\begin{equation*} \inf_y \frac{g_{n,r_n}(y)}{g_{n,s_n}(y)} \le\frac{\int g_{n,r_n}(y)\, dy}{\int g_{n,s_n}(y)\, dy}=\Big(\frac{r_n}{s_n}\Big)^n=n^{-\epsilon n/2}. \tag{1} \end{equation*}$$ So, for some sequence $$(x_n)$$ (depending only on $$\epsilon$$ but not on $$a$$) with $$x_n\in\mathbb R^n$$ we have $$\begin{equation*} R_n(x_n)=\frac{g_{n,r_n}(x_n)}{g_{n,s_n}(x_n)}<\inf_y \frac{g_{n,r_n}(y)}{g_{n,s_n}(y)}+n^{-n} \le n^{-\epsilon n/2}+n^{-n} =o(1/a^n) \end{equation*}$$ for all real $$a>0$$, so that $$a^nR_n(x_n)\to0$$, as desired.
Details on the inequality in (1): Let $$\begin{equation*} c:=\inf_y \frac{g_{n,r_n}(y)}{g_{n,s_n}(y)}. \end{equation*}$$ Then $$g_{n,r_n}(y)\ge c g_{n,s_n}(y)$$ for all $$y$$, whence $$\int g_{n,r_n}(y)\, dy\ge c\int g_{n,s_n}(y)\,dy$$ and $$\begin{equation*} \frac{\int g_{n,r_n}(y)\, dy}{\int g_{n,s_n}(y)\,dy}\ge c=\inf_y \frac{g_{n,r_n}(y)}{g_{n,s_n}(y)}. \end{equation*}$$
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