Exposed point

Remarks

Equivalently, there is a supporting hyperplane $H$ of $C$ with $H \cap C = \set{x}$. In a real topological vector space the same definition is used with continuous linear functionals.

Every exposed point is an extreme point. Suppose $x = (1 - t) y + t z$ with $y, z \in C$, $y \ne z$ and $0 < t < 1$. Then $x \ne y$ and $x \ne z$, so $f(y) < f(x)$ and $f(z) < f(x)$, and hence $f(x) = (1 - t) f(y) + t f(z) < f(x)$, a contradiction.

Example: in $[0, 1]^2$, the corner $(1, 1)$ is exposed by $f(x_1, x_2) = x_1 + x_2$, since $f \le 2$ on the square with equality only at $(1, 1)$.

Non-example: let $C$ be the stadium, the convex hull of the discs of radius $1$ centred at $(-1, 0)$ and $(1, 0)$. The point $(1, 1)$ is extreme, but the only supporting line of $C$ through it is $y = 1$, which meets $C$ in the whole segment from $(-1, 1)$ to $(1, 1)$. So no linear form is maximised on $C$ only at $(1, 1)$.