Let $X \in \mathbb R^d$ be a large domain (a ball of radius $r$ for $r$ large should suffice)

Let $B$ be a ball of radius $1$.

Consider the ratio

$$ \frac{ \left| \left\{ x_1,\dots,x_n \in X \mid x_1,\dots,x_n \textrm{ lie in some translate of } B \right\} \right| }{|X| |B|^{n-1} n^d}$$

I can show this ratio is bounded above and below by a constant depending on $d$. Presumably, it converges to some limit as $n$ goes to $\infty$ (we can send $r$ to $\infty$ too). But how does this limit depend on $d$? Does it grow, shrink, or remain constant?

My upper bound is exponential growth and my lower bound is superexponential decline. So there is quite a lot of give between these two.

I also know that if we replace $B$ with a cube, the limit is $1$, as we may separate the different coordinates and thus reduce to the case $d=1$, and then it's very easy to check that the answer is $1$.

This question came up when I was working on lower bounds for my previous question

Upper bound proof:

If $x_1,\dots,x_n$ lie in a translate $y+B$ of $B$, then they lie in $y'+(1+\epsilon)B$ for all $y$ in a ball of radius $\epsilon$.

Hence:

$$\left| \left\{ x_1,\dots,x_n \in X \mid x_1,\dots,x_n \textrm{ lie in some translate of } B \right\} \right| \leq \frac{\int_{y \in X}\left| \left\{ x_1,\dots,x_n \in X \mid x_1,\dots,x_n \textrm{ lie in } y+ (1+\epsilon)B \right\} \right|}{ \epsilon^d |B|} \leq \frac{ \int_{y \in X} (1+\epsilon)^{nd} |B|^n } { \epsilon^d |B|} = \frac{(1 + \epsilon)^{nd}}{\epsilon^d} |X| |B|^{n-1} \leq e^d n^d |X| |B|^{n-1} $$

if $\epsilon =1/n$.

Lower bound proof

By Cauchy-Schwartz

$$ \left| \left\{ x_1,\dots,x_n \in X \mid x_1,\dots,x_n \textrm{ lie in some translate of } B \right\} \right| \leq \frac{ \left(\int_{x_1,\dots,x_n \in X} \left| \left\{ y \mid x_1, \dots, x_n \textrm{ lie in } y +B\right\}\right|\right)^2 }{ \int_{x_1,\dots,x_n \in X} \left| \left\{ y \mid x_1, \dots, x_n \textrm{ lie in } y +B\right\}\right|^2}$$

The numerator is the square of

$$\int_{y \in X} \left| \left\{ x_1,\dots,x_n \mid x_1, \dots, x_n \textrm{ lie in } y +B\right\}\right|= |X| |B|^n$$

and the denominator is

$$ \int_{y_1,y_2 \in X} \left| \left\{ x_1,\dots,x_n \mid x_1, \dots, x_n \textrm{ lie in } y_1 +B \cap y_2+B \right\}\right| = \int_{y_1,y_2 \in X} V(||y_1 -y_2||_2)^n $$ $$\approx \int_{y_1 \in X} \int_{z \in \mathbb R^d} V(||z||_2)^n = |X| \int_{z \in \mathbb R^d} V(||z||_2)^n$$

where $V(r)$ is the intersection of two copies of $B$ whose centers are a distance $r$ apart.

Now

$$ \int_{z \in \mathbb R^d} V(||z||_2)^n = n^{-d} \int_{z \in \mathbb R^d} V(||z||_2/n)^n \approx n^{-d} \int_{z \in \mathbb R^d} |B|^n e^{- \operatorname{dlog} V (0) z } = n^{-d} |B|^{n+1} d! / (\operatorname{dlog} V(0))^d$$

for a lower bound of $|X| |B|^{n-1} n^d (\operatorname{dlog} V(0))^d / d!$. Now the derivative of $V$ at $0$ is just the volume of a $d-1$-dimensional ball of radius $1$ because moving the ball $\epsilon$ along one axis and intersecting removes a slice of thickness $\epsilon$ in that direction from the ball. So the logarithmic derivative is the ratio of the volumes of the $d-1$ and $d$-dimensional balls, hence of size around $\sqrt{d}$. So the lower bound is roguhly $d^{-d/2}$.