Upper bound on intersection of $\ell_p$-norm balls in $\mathbb R^n$, with different centers and the same radius

Question

Let $x$ and $x'$ be points in $\mathbb R^n$ (for a large positive integer $n$) and let $d(x,x') := \sup_{1 \le j \le n} |x_j-x'_j|$ be their $\ell_\infty$-norm distance apart. Given $r \ge 0$, let $B(x;r) := \{z \in \mathbb R^n \mid d(x,z) \le r\}$ be the closed ball of radius $r$ centered at $x$.

Question. What are good upper and lower bounds on the volume $v_n(x,x';r)$ of intersection of $B(x;r)$ and $B(x';r)$ as a function of $d(x,x')$, $r$, and $n$ ?

Note. By a paper of M. Gromov, we know that $v_n(x,x';r)$ is a decreasing function of $d(x,x')$.

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Motivation. A lower bound on $v_n(x,x';r)$ would let me compute an upper bound for the total-variation distance $T_n(x,x';r)$ between the uniform distribution on $B(x;r)$ and the uniform distribution on $B(x';r)$. Indeed, by definition, one can easily obtain the formula $$ T_n(x;x';r) = 1-\frac{v_n(x,x';r)}{(2r)^n}. $$

Answer 1

Disclaimer. Sorry for self-answering, but the problem turns out to be really easy.

So, we can write $B(x;r) := \{z \in \mathbb R^n \mid x_j - r \le z_j \le x_j + r\;\forall j\}$. Thus, we have the equivalance $$ z \in B(x;r) \cap B(x';r) \iff z_j \in I_j\;\forall j, $$ where $I_j: [c_j^-,c_j^+]$, with $c_j^- := \max(x_j,x'_j)-r$ and $c_j^+ := \min(x_j,x_j')+r$. That is, $B(x;r) \cap B(x';r)=I_1 \times \ldots \times I_n$, a cuboid. Note that $I_j$ is empty if $x_j \ge x'_j + 2r$ or $x_j' \ge x_j + r$; else it has length equal to $\min(x_j,x_j')-\max(x_j,x_j')+2r = 2r - |x_j-x_j'|$. Thus $$ \begin{split} v_n(x,x';r) := \text{vol}(B(x;r)\cap B(x';r)) &= \text{vol}(I_1\times \ldots \times I_n)= \Pi_{j=1}^n\text{length}(I_j)\\ &= \begin{cases}0,&\mbox{ if }d(x,x') \ge 2r,\\\Pi_{j=1}^n(2r-|x_j-x_j'|),&\mbox{else.}\end{cases} \end{split} $$

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Thus, if $d(x,x') \ge 2r$ and we let $u_j := |x_j-x_j'|/(2r)$, then we immediately get $$ \begin{split} \frac{v_n(x,x';r)}{(2r)^n} &\ge \Pi_{j=1}^n\left(1-u_j\right) \ge e^{-\sum_{j=1}^n\frac{u_j}{1-u_j}}=e^{-\sum_{j=1}^n(1+u_j+u_j^2+\ldots)}\\ &=e^{-\sum_{p=1}^\infty\|u\|_p^p}=e^{-\sum_{p=1}^\infty\left(\frac{\|x-x'\|_p}{2r}\right)^p} \ge 1 -\sum_{p=1}^\infty\left(\frac{\|x-x'\|_p}{2r}\right)^p, \end{split} $$ which in turn yields

$$ \begin{split} 0 \le T_n(x,x';r) &:= 1-\frac{v_n(x,x';r)}{(2r)^n} \\ &\le \begin{cases}1-e^{-\sum_{p=1}^\infty\left(\frac{\|x-x'\|_p}{2r}\right)^p} \le \sum_{p=1}^\infty\left(\frac{\|x-x'\|_p}{2r}\right)^p,&\mbox{if }\|x-x'\|_\infty \le 2r,\\1,&\mbox{ else.} \end{cases} \end{split} $$