Condition for a point to be extreme

Dependencies:

  1. Convex set
  2. Extreme point of a convex set

Let $S$ be a convex set. $u \in S$ is a non-extreme point of $S$ iff $\exists z \neq 0$ such that $u + z \in S$ and $u - z \in S$.

Proof

Suppose $\exists z \neq 0$ such that $u + z \in S$ and $u - z \in S$. Then $u = (1/2)(u + z) + (1/2)(u - z)$. Since $z \neq 0$, we get that $u + z \neq u - z$. Hence, $u$ is a non-extreme point of $S$.

Suppose $u$ is a non-extreme point of $S$. Then $\exists x \in S$, $\exists y \in S$, $\exists \alpha \in (0, 1)$ such that $x \neq y$ and $u = (1-\alpha)x + \alpha y$. Without loss of generality, assume $\alpha \le 1/2$. Let $w = (1-2\alpha)x + (2\alpha)y$. Then $w \in S$ since $S$ is convex. Let $z = (x - u)$. Then $z = \alpha(x - y) \neq 0$ and $u + z = x \in S$ and $u - z = 2u - x = 2(1-\alpha)x + 2\alpha y - x = w \in S$.

Dependency for:

  1. Condition for polyhedral cone to be pointed

Info:

Transitive dependencies:

  1. Group
  2. Ring
  3. Field
  4. Vector Space
  5. Convex combination and convex hull
  6. Convex set
  7. Extreme point of a convex set