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For $n$ points in a triangle and two fixed endpoint vertices, the longest convex chain length is the largest number of sampled points in a convex chain joining those endpoints. It is a certifiable function: a chain of length $a$ supplies a certificate of exactly $a$ point coordinates. Changing one point changes the maximum by at most one, since deleting that point from any optimal chain loses at most one vertex. For independent uniform points, its median is $O(n^{1/3})$ and its fluctuation scale about the median is at most $n^{1/6}$, as proved by convex-chain probability in a triangle and convex-chain median concentration.

The precise external concentration result we use is Talagrand's convex distance inequality: for a finite product of independent standard probability spaces, a measurable event $B$ of positive probability, and $u\ge 0$,Our coordinates are the independent uniformly sampled points of the triangle, so the hypothesis holds. Part (ii) therefore gives, for integers $a>b\ge 0$,Indeed every configuration in the first event has at least the distance from part (ii) to the second event. If one event is empty, the displayed product bound is automatic. This is the only use of the product-space theorem; the geometric separation was proved directly in part (ii). A precise source for that theorem is Section 4.1 of Talagrand's original product-space concentration paper.

Choose an integer median $m=m_n$, so $\mathbb{P}(L_n\le m)\ge 1/2$ and $\mathbb{P}(L_n\ge m)\ge 1/2$. First set $a=m+t$, $b=m$ in the product bound, and then set $a=m$, $b=m-t$. For positive integers $t$ this yields the convex-chain median concentration estimatesThe lower event is empty if $m-t<1$ in our sampling model; the upper event is empty if $m+t>n$. Thus the stated bounds remain valid for all positive integers $t$. Also $m\ge 1$ for $n\ge 1$.

By part (i), $m\le C{n}^{1/3}$ with $C=4e$. Let $h_n=\omega (n)n^{1/6}/2$, take $t_n=\lfloor h_n\rfloor +1$, and use the deterministic interval $I_n=[m-h_n,m+h_n]$. Its length is exactly $\omega (n)n^{1/6}$. As $L_n$ and $m$ are integers, falling outside this interval implies one of the two tail events with $t=t_n$. ThereforeFor positive $m,t$,Since $t_n\ge h_n$ and $m\le C{n}^{1/3}$, this exponent is at leastConsequentlyThus $L_n$ lies with high probability in the required interval. This proof works for every $\omega (n)\to \infty$, including growth faster than $n^{1/6}$; no extra restriction on $\omega$ is needed. The scale comes from the square root of the $O(n^{1/3})$ certificate size at the median.