Extreme point
= Extreme point
{title2=$\operatorname{ext}C$}
{wiki}
= Extremal point
{synonym}
An <extreme point> of a <convex set> is a <point> that cannot be made by genuinely mixing two different <points> of the <set>. For a <convex set> $C$ in a real <vector space>, the condition on $p\in C$ is
$$
p=(1-t)y+tz,\qquad y,z\in C,\quad0<t<1
\quad\Longrightarrow\quad y=z=p.
$$
Thus $p$ lies on no open <line segment> between two distinct <points> of $C$. It is enough to exclude <midpoint> representations: if $p$ lies inside a <line segment>, two nearby <points> on opposite sides have <midpoint> $p$.<Convex geometry (BME course BMETE94AM22)>[§6, Definition 6.6, printed p. 37: extremal points of closed convex sets]{ref}
The <square> $[-1,1]^2$ has just four <extreme points>, at $(\pm1,\pm1)$. In a <convex combination> giving a corner, each <coordinate> is already at its largest or smallest allowed value, forcing both input <points> to agree with that corner. In contrast, the boundary <point> $(1,0)$ is not extreme:
$$
(1,0)=\frac12(1,-1)+\frac12(1,1).
$$
It is the <midpoint> of two other <points> on the same edge. An interior <point> also admits a small <line segment> through it.
The <unit disk> $D=\{z\in\mathbb R^2:\|z\|_2\leq1\}$ has a different shape: every <point> on its <unit circle> is extreme. For a <unit vector> $p$, the <Cauchy-Schwarz inequality> gives
$$
p\cdot z\leq\|p\|_2\|z\|_2\leq1\qquad(z\in D),
$$
and equality is possible only at $z=p$. If a <convex combination> equals $p$, applying this <linear functional> forces both ingredients to attain that equality. Both must therefore be $p$. Extremality need not look like a sharp corner.
\Image[/extreme-points-square-and-disk.png]
{title=The <square> and <unit disk> distinguish <boundary> from <extreme points>}
{description=Original planar examples. The red marks are <extreme points>: four corners for the <square>, the whole <unit circle> for the <unit disk>. Every <point> on the square's orange edge maximizes $x$, including its nonextreme <midpoint>. For the disk, the orange <supporting line> for $p\cdot z$ touches only $p$.}
These <points> help describe all allowable mixtures. Every $(x,y)\in[-1,1]^2$ is a <convex combination> of its four corners, with the weight on $(a,b)\in\{-1,1\}^2$ equal to $(1+ax)(1+by)/4$. For the <unit disk>, write $z=rp$, with $0\leq r\leq1$ and $p$ a <unit vector>; at zero choose any $p$. Then
$$
z=\frac{1+r}{2}p+\frac{1-r}{2}(-p).
$$
Two <extreme points> suffice for that mixture. The <Krein-Milman theorem> gives a much broader reconstruction: a nonempty <compact convex set> in a <locally convex space> whose <topology> is <Hausdorff> is the <closed convex hull> of its <extreme points>.
<Mark Krein> and <David Milman> published their reconstruction theorem in 1940. Milman's colleagues describe that result and its subsequent use as a method of representing <convex sets> through their <extreme points>.<David Milman (1912–1982) by Israel Gohberg and others>[David Milman (1912–1982), printed p. 4: 1940 collaboration with M. G. Krein, the reconstruction theorem and subsequent method; citation 3 refers to their original paper]{ref}