= Extreme point that is not an exposed point
The distinction appears at a smooth join between a straight segment and a curved boundary. Consider the <convex hull>
$$
C=\operatorname{conv}(D\cup\{P\}),\qquad D=\{z:\|z\|_2\leq1\},\qquad P=(2,0).
$$
The upper tangent from $P$ meets the <unit circle> at $T=(1/2,\sqrt3/2)$. The <supporting line> is
$$
T\cdot z=1,\qquad T\cdot P=T\cdot T=1.
$$
Every <point> of the <unit disk> has $T\cdot z\leq1$, with equality only at $T$. Hence the <convex face> of $C$ where equality holds is exactly the <line segment> $TP$.
\Image[/extreme-point-not-exposed-rounded-hull.png]
{title=An <extreme point> can fail to be an <exposed point>}
{description=Original analytic <convex hull> of the <unit disk> and $P=(2,0)$. The tangent contact $T=(1/2,\sqrt3/2)$ is extreme, but its only <supporting line> contains the entire <line segment> $TP$. A different marked <point> on the curved arc is exposed by its own tangent <line>.}
The <point> $T$ is extreme. If a nontrivial <convex combination> of two <points> of $C$ equals $T$, both must belong to the <convex face> $TP$. Since $T$ is an endpoint of that <line segment>, both must equal $T$.
Yet $T$ is not exposed. A nonzero <linear functional> maximizing at $T$ on $C$ must also maximize there on the included <unit disk>. Its <vector> must be a positive multiple of $T$, by the equality condition in the <Cauchy-Schwarz inequality>. It then gives $P$ the same value as $T$. No <linear functional> can select $T$ alone. Being an <extreme point> is a property of possible mixtures; being an <exposed point> adds the ability to distinguish it by one linear objective.
Back to article page
