Left and right halves of convex curve

Adopting complex notations, let's consider the open half-planes $H_0:=\{z\in \mathbb C: \Re z>0 \}$ and $H_\theta:=e^{i\theta}H_0$ for $\theta\in[0, 2\pi]$. For $p\in S$, we are interested in the cardinality $\nu(p,\theta):=\big| (p+H_\theta) \cap S\big|.$ When $\theta$ varies from $0$ to $2\pi$, this cardinality jumps by $+1$ or $-1$ each time the boundary line $\partial(p+H_\theta)=p+ ie^{i\theta}\mathbb R$ hits a point of $S$ lying on the left, resp. on the right to $p$. Let $\{q_1,\dots,q_m\}$ be the points of $S$ on the left to $p$, listed in counterclockwise cyclic order, that is corresponding to angles $0<\theta_1<\dots<\theta_m<\pi$ (meaning $q_k\in \partial(p+H_{\theta_k})$. Thus for $\theta\in[\theta_k,\theta_{k+1}]$ we have $\nu(p,\theta) \le\nu(p,\theta_k)+1$. To exploit the above property, it is convenient to state the following Lemma (a discrete version of the Intermediate Value Theorem; see below for the simple proof).

Lemma. Let $f$ be an integer valued function defined on an interval of integers $[a,b]\subset\mathbb Z$ such that $f(a)\le f(b)$ and $f(k+1)\le f(k)+1$ for every $a\le k<b$. Then $f([a,b])$ is an interval of integers.

As a consequence one has the following facts:

1. Every point $p$ on the right half of $S$ ends at least one convex chain;

2. Two convex chains share no edge;

3. Every point $p$ on the left half of $S$ ends no convex chain;

that is, the convex chains are a partition of the halving edges, and are exactly indexed by their ending points, which are exactly the $n/2$ points of the right half of $S$.

The arguments to prove these facts are similar. Here is the proof of the first one (I'll write down the others too, if needed).

Proof of Claim 1. Let $p$ be on the right half of $S$, so $\nu(p,\theta_1)\le \nu(p,0)\le\frac {n-2}2 $ and $\nu(p,\theta_m)\ge \nu(p,\pi)-1\ge\frac{n-2}2 $. By the Lemma this implies that for at least one index $1\le k\le m$ one has $\nu(p,\theta_k)=\frac {n-2}2 $, that is, $q_kp$ is halving. Let $k_*$ be the minimum of these indices, so $\nu(p,\theta_{k_*})=\frac {n-2}2 $ and again by the Lemma, $\nu(p,\theta_k)<\nu(p,\theta_{k_*})$ for all $1\le k<k_*$. This implies that in fact no line $\partial(p+H_\theta)$ can be halving for $0\le \theta\le \theta_{k_*}$, that is $q_{k_*}p$ has no next.

Proof of the Lemma: Given an integer $ f(a)\le v\le f(b)$, let $u$ be the maximum of the non-empty set $\{k\in[a,b]: f(k)\le v\}$. Then $u<b$ and $v<f(u+1)\le f(u)+1\le v+1$ whence $f(u)=v$.