 Solution

We use the directed set version of the Talagrand convex distance:Here a coordinate is an entire sampled point, so $x_i\ne y_i$ means the two planar points differ. This is the distance from the upper event $A$ to the lower event $B$; its order is relevant because the convex distance need not be symmetric.

Fix $x\in A$ and choose exactly $a$ indices $I$ whose points form a convex chain with the endpoints. Such a choice exists even if $L_n(x)>a$, by taking a subchain. It is a certificate of the certifiable function $L_n$. For any $y\in B$, let $J=\{i\in I:x_i\ne y_i\}$. The $a-|J|$ unchanged selected points remain a convex chain in configuration $y$, soChoose the nonnegative weights ${\alpha }_i=a^{-1/2}$ for $i\in I$ and zero otherwise. They have squared sum one, and for every $y\in B$,Taking the infimum over $y$, then the supremum over weights, and finally the infimum over $x\in A$ provesThis proof also works for $b=0$ if zero-length chains are admitted, and gives the usual infinite separation when one of the events is empty. For the uniform sampling model, $L_n\ge 1$ almost surely when $n\ge 1$.