Complete equality condition for the Kahn–Saks inequality

Let $P=(X,\prec)$ be a finite poset with $n=|X|$, and let $x,y\in X$ be distinct. Denote by $\operatorname{F}(k)$ the number of linear extensions $L\in\operatorname{\mathcal{E}}(P)$ such that $L(y)-L(x)=k$. A pair $(x,y)$ satisfies the $k$-midway property when the conditions specified in the paper hold, and the dual $k$-midway property is defined by reversing the order. Suppose that $k\in\{2,\ldots,n-2\}$ and $\operatorname{F}(k)>0$. Complete equality condition for the Kahn–Saks inequality. The following are equivalent: (a) $\operatorname{F}(k)=\operatorname{F}(k+1)=\operatorname{F}(k-1)$; (b) there is an element $z\in\{x,y\}$ such that, for every $L\in\operatorname{\mathcal{E}}(P)$ with $L(y)-L(x)=k$, there are elements $u,v\in X$ incomparable with $z$ and satisfying $L(u)+1=L(z)=L(v)-1$; and (c) $(x,y)$ satisfies either the $k$-midway property or the dual $k$-midway property. This gives a characterization of equality in the Kahn–Saks inequality under the stated hypotheses; the surrounding discussion presents it as a partial result toward complete equality conditions for general posets.