Ratio of integrals with increasing dimension over Euclidean balls

Question

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.

Answer 1

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<r}\,dx, \end{equation*} 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<r}=\int dx\, f_n(x)|B_n(r)|=|B_n(r)|, \end{equation*} 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*}