Extreme point ($\mathrm{ext}C$)

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 $p\in C$ isThus $p$ 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 $p$.^[ref]

The square $[-1,1{]}^2$ has just four extreme points, at $(\pm 1,\pm 1)$. 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: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 $D=\{z\in {\mathbb{R}}^2:\Vert z{\Vert }_2\le 1\}$ has a different shape: every point on its unit circle is extreme. For a unit vector $p$, the Cauchy-Schwarz inequality givesand 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.

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 $0\le r\le 1$ and $p$ a unit vector; at zero choose any $p$. ThenTwo 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.^[ref]

An exposed point is a point uniquely selected by maximizing a linear functional. In ${\mathbb{R}}^d$, a point $p$ of a convex set $C$ is exposed if some vector $c$ satisfiesGeometrically, a supporting hyperplane touches the convex set only at $p$. Every exposed point is an extreme point: a convex combination that attains the maximum can put positive weight only on maximizers, and there is only one.^[ref]

The unit disk calculation in extreme point exposes every point of the unit circle. For a corner $(a,b)$ of the square, $c=(a,b)$ exposes it: $ax+by$ attains two only at $(a,b)$. However, maximizing $x$ exposes no single point on the right edge, since all of them have the same objective value. A linear functional may have several optimal points, even when some of them are extreme.

The distinction appears at a smooth join between a straight segment and a curved boundary. Consider the convex hullThe upper tangent from $P$ meets the unit circle at $T=(1/2,\sqrt{3}/2)$. The supporting line isEvery point of the unit disk has $T\cdot z\le 1$, with equality only at $T$. Hence the convex face of $C$ where equality holds is exactly the line segment $TP$.

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.

For the real l-infinity sequence space, every coordinate of an extreme point must be an endpoint of $[-1,1]$; any slack coordinate permits opposite perturbations. For the real absolutely summable sequence space, a unit vector with two nonzero coordinates permits a transfer of mass between them, so only signed coordinate vectors are extreme. For the real l2 sequence space, the parallelogram law makes every unit vector extreme. In all three cases interior points of the unit ball are nonextreme.

For compact Hausdorff $K$, the extreme points of the real unit ball in the space of continuous functions on a compact space are precisely its continuous sign-valued functions. A point with $|f|<1$ permits a nonzero continuous bump perturbation supported where there is uniform slack, while pointwise equality at an endpoint of $[-1,1]$ forces both functions in a midpoint decomposition to agree. For the Cantor set, clopen indicators give the perturbations directly.

Every real continuous function of supremum norm at most one on the Cantor set is a uniform limit of convex combinations of continuous sign functions. Approximate on a finite Cantor cylinder partition by values $a_j\in [-1,1]$. The sign vector $s$ has weight ${\prod }_j(1+s_j{a}_j)/2$, whose coordinate means are $a_j$. This proves a closed convex hull identity without weak compactness.

For real scalars the extremes are the almost-everywhere sign-valued functions. A positive-measure region uniformly inside $[-1,1]$ permits opposite nontrivial bounded perturbations, proving necessity. Equality at either endpoint forces both members of a convex decomposition to agree, proving sufficiency. For complex scalars the same criterion uses unit-modulus functions and the strict convexity of the complex unit disk.