Supporting disks and activity below the diameter

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The later long-rank gap theorem uses the linear core surplus proved here to obtain adjacent gaps $O_{K,k}(\Delta/n^2)$ at rich fixed ranks, and growing odd girth for fixed top-rank unions. These are further necessary conditions, not an unrestricted conclusion.

All counts are unordered. For an occurring distance $r$, let

$$S=\{x\in A:\text{some }y\in A\text{ has }|x-y|>r\}, \qquad T=A\setminus S.$$

Write $H=G_r(A)$, $s=|S|$, and $h=e(H[S])$. The key point is that every radius-$r$ disk centered at a point of $T$ contains all of $A$.

Exposed points and circular leaves

Every $r$-active point of $T$ is exposed in $\operatorname{conv}T$: if $|x-y|=r$, the disk centered at $y$ contains $T$ and its tangent at $x$ supports $T$ uniquely. If $x$ has an $r$-neighbor in $T$, the same disk contains all of $A$, so $x$ is exposed in $\operatorname{conv}A$.

For any $y\in A$, its $r$-neighbors in $T$ occupy an arc of angle at most $\pi/3$. Indeed, their mutual distances are at most $r$; the angular localization argument in star_diameter applies after scaling. Every neighbor strictly between the two extremes has global $r$-degree one. The unit-circle arc lemma applies because each extreme belongs to $T$, and hence is within $r$ of every point of $A$. Thus each $y\in S$ has at most two nonleaf neighbors in $T$.

Partition $T$ into $C,F,P,Z$: $C$ consists of points incident with an $r$-edge inside $T$; $F$ consists of global degree-one points whose neighbor is in $S$; $P$ consists of points with no neighbor in $T$ and global degree at least two; and $Z$ consists of isolated points. The preceding argument gives

$$e_H(S,T)\le |F|+2s.$$

The graph on $C$ is a diameter graph, so $e(H[T])\le |C|$. Consequently the following elementary bound needs no stronger description of that graph:

$$m_A(r)\le n+h+s-|P|-|Z|. \tag{1}$$

In particular, richness implies $|P|+|Z|\le h+s-1$, while $T\setminus Z$ is strictly convex. At the second-largest distance, $S$ is the set of diameter-active vertices. Its points lie on a constant-width boundary, so the boundary theorem gives $h\le s$. Thus a rich second distance supplies a convex subset of size at least $n-3s+1$.

The sharper inactivity estimate

Let $D$ be the diameter, $d=m_A(D)$, and let $r$ be the second-largest distance. Vesztergombi's theorem states

$$m_A(r)\le n+2d. \tag{2}$$

This was checked on p. 380 of the full 1996 paper.

Suppose $m_A(r)>n$, and delete all $z$ points isolated in $G_r(A)$. The retained set must still contain a diameter pair: otherwise its $r$-graph would be a diameter graph with more edges than vertices. Applying (2) to the retained set, whose second-largest distance is still $r$, gives

$$n+1\le m_A(r)\le n-z+2d,$$

so $z\le2d-1$. Deleting the diameter-active vertices as well leaves a strictly convex subset of size at least

$$n-|S|-2d+1\ge n-4d+1. \tag{3}$$

Thus bounded diameter multiplicity and a rich second distance force all but a bounded number of points into convex position.

Scope and the unresolved step

These disk and counting arguments were checked by the author and are author-recorded. The source inequality is used as stated; no stronger inequality such as $m_A(r)\le n+d$ is assumed. That proposed strengthening is false: Vesztergombi's even-ring construction has $d=n/4$ and $m_A(r)=3n/2$.

Neither (1) nor (3) supplies a constant-width boundary or bounded individual rank omissions for the resulting convex subset. Under bounded rarity, (3) only gives a convex core with $q\le |C|/2+O(d+R(A))$. The required arbitrary-convex inverse at bounded additive excess remains unproved. No induction on diameter multiplicity or unrestricted asymptotic conclusion follows.

A circle bound for the active support of a long distance

Let $k\ge1$ be an integer and $r>0$. Suppose at most $k-1$ distinct distances exceed $r$, and let $X=\{q\in A:\deg_{G_r(A)}q>0\}$. Every circle of radius $R>r$, with arbitrary center, contains at most $4k-2$ points of $X$.

To prove this, apply the balanced-circle lemma to its $N\ge2$ points of $X$, selecting $q$, and choose an $r$-neighbor $w$ of $q$. The center $o$ of the chosen circle differs from $w$, since $R>r$. Writing $\rho=|o-w|$, its intersection with the closed radius-$r$ disk about $w$ satisfies

$$\cos\theta\ge\frac{R^2+\rho^2-r^2}{2R\rho}>0.$$

This is an arc shorter than a semicircle, or a singleton, with endpoint $q$. One semicircle from $q$, excluding $q$, is strictly outside the disk and contains at least $\lfloor(N-1)/2\rfloor$ selected points. Their distances to $w$ have at most $k-1$ possible values, each with at most two circle intersections. Thus $N\le4k-2$. The cases $N\le1$ are immediate.

In particular, every larger-distance graph induced on $X$ has maximum degree at most $4k-2$. Unlike the original core lemma, this assertion requires no vertex peeling.

Linear surplus in a rich fixed-long-rank core

Suppose $R(A)\le K$ along an unbounded sequence. Choose a rich smallest-distance rank $s\le K+1$, and let $r_k$ be any rich fixed largest-distance rank. Such a $k$ is necessarily at least two. The finitely many possible choices of $s$ allow all asymptotic estimates below to be uniform, or one may pass to a subsequence with $s$ fixed.

Write $B_k=L_1\cup\cdots\cup L_k$. The short-edge inverse gives only $o(n)$ edges of the chosen short class touching $B_k$. Its degree bound $6s$ therefore implies

$$n-|B_k|\ge n/(3s)-o(n).$$

The $r_k$-graph is supported on $B_k$. If its 2-core $H$ has $v$ vertices, peeling removes at most $|B_k|-v$ edges. Consequently

$$e(H)-v\ge m_A(r_k)-|B_k|\ge n/(3s)-o(n). \tag{4}$$

The 2-core bound gives $\Delta(H)\le4k-2$. Since $\sum_{x\in H}(\deg_Hx-2)=2(e(H)-v)$, at least $n/[6s(k-1)]-o(n)$ core vertices have degree at least three.

There is also a linear matching of pairs longer than $r_k$ inside this core. Let $J$ be the graph of those longer pairs, and let $\mu$ be its maximum matching size. The $2\mu$ endpoints $U$ of a maximum matching cover $J$. Thus the remaining points have diameter at most $r_k$, so $e(H-U)\le v-|U|$. At most $(4k-2)|U|$ edges of $H$ touch $U$, yielding

$$e(H)\le v+2(4k-3)\mu,\qquad \mu\ge n/[6s(4k-3)]-o(n).$$

If $k$ is the first rich largest rank, all the $k-1$ larger classes are rare, and at least one contains a linear matching. These conclusions use the coexistence of a rich short class with a rich long class, not just the long class's core-degree bound.

For a rich second-largest distance, the checked bound $m_A(r_2)\le |B_2|+2m_A(\Delta)$, obtained by applying (2) on $B_2$, also gives $m_A(\Delta)\ge n/(6s)-o(n)$. In particular, a hypothetical unbounded $R=1$ family has at least $(1/6-o(1))n$ diameter pairs, since its minimum distance is rich and one may use $s=1$.

The author checked the active-circle geometry, peeling deficit, matching deletion, and unordered-pair factors; the argument is author-recorded. Linear branching and a linear diameter matching do not imply circular concentration: the even-order extremal deformations already exhibit those features without five concyclic points. No new connection to the interior short-distance palette follows from (4).