A new upper bound for the size of $s$-distance sets in boxes

Let $q,d\geq 2$ be integers. Define $$J(q,d):=\frac 1q \Big( \min_{0<x<1} \frac{1-x^q}{1-x} x^{-\frac{q-1}{d}}\Big).$$ Let \mbox{\cal G}\subseteq {\mathbb R}^n be an arbitrary subset. We denote by d(\mbox{\cal G}) the set of (non-zero) distances among points of \mbox{\cal G}: d(\mbox{$\cal G$}):=\{d( p_1, p_2):~ p_1, p_2\in \mbox{$\cal G$}, p_1\ne p_2\}. Our main result is a new upper bound for the size of $s$-distance sets in boxes. More concretely, let $A_i\subseteq \mathbb R$, $|A_i|=q\geq 2$ be subsets for each $1\leq i\leq n$. Consider the box \mbox{\cal B}:=\prod_{i=1}^n A_i\subseteq {\mathbb R}^n. Suppose that \mbox{\cal G}\subseteq \mbox{\cal B} is a set such that |d(\mbox{\cal G})|\leq s. Let $d:=\frac{n(q-1)}{s}$. Then |\mbox{$\cal G$}|\leq 2(qJ(q,d))^n. We use Tao's slice rank bounding method in our proof.