Separation exponent at least 2.10

For every real number $\varepsilon>0$ and every natural number $S\ge 2$, there exist a natural number $n$ and a Boolean function $f:\{0,1\}^{\{0,\ldots,n-1\}}\to\{0,1\}$ such that $S\le s(f)$ and $s(f)^{21/10-\varepsilon}\le \operatorname{bs}(f)$, where the natural numbers $s(f)$ and $\operatorname{bs}(f)$ are regarded as real numbers in the latter inequality and the exponentiation is real exponentiation. For an input $x$ and a set $B\subseteq\{0,\ldots,n-1\}$, let $x^B$ be the input obtained by complementing precisely the coordinates in $B$. Then $s(f)$ is the maximum, over all inputs $x$, of the number of coordinates $i$ for which $f(x^{\{i\}})\ne f(x)$; and $\operatorname{bs}(f)$ is the maximum, over all inputs $x$ and all families $\mathcal B$ of pairwise disjoint nonempty subsets of $\{0,\ldots,n-1\}$ satisfying $f(x^B)\ne f(x)$ for every $B\in\mathcal B$, of $|\mathcal B|$. The empty family is allowed in this maximum and contributes zero. The witnesses $n$ and $f$ may depend on both $\varepsilon$ and $S$. Although $n$ ranges over all natural numbers, including zero, $n=0$ gives both sensitivities equal to zero and cannot satisfy $S\le s(f)$; the lower bound also excludes constant functions and ensures that the base of the real power is positive. There is no upper bound on $\varepsilon$, so the assertion includes zero and negative exponents when $\varepsilon\ge21/10$.