The extreme points of a convex set is defined as the following,
An extreme point x of a convex set $C$, is a point st. if $x=θy+(1-θ)z$ for some $0<θ<1$ and $y,z∈C$ then $x=y=z$, i.e. $x$ is a point that can only be written as a convex combinations of itself.
The question is to prove
A hypercube $C=\{x=(x_1,…,x_n)∈{\Bbb R}^n:|x_i |≤1\}$ has $2^n$ vertexes $V=\{x:x_i=±1 {\text{ for }} i=1,…,n\}$ as its only extreme points.
I can prove $V$ are all extreme points.
For any $x∈V$, let $x=θy+(1-θ)z$ for some $y,z∈C\backslash \{x\}$. Suppose $x=(x_1,…,x_n )$,$y=(y_1,…,y_n )$,$z=(z_1,…,z_n)$. For each $i$, if we have $x_i=θy_i+(1-θ) z_i$, then when $x_i=1$, we must have $y_i=z_i=1$ since $y_i,z_i≤x_i=1$; when $x_i=-1$, we must have $y_i=z_i=-1$ since $y_i,z_i≥x_i=-1$.
My problem is how to prove other non-vertex points are not extreme points. I am wondering if we can prove without using heavy-duty theorems in convex analysis because this problem comes out quite early in the textbook. Thank you!
Note that showing that a point $x$ is not an extreme point reduces to finding some $y$ and $z$, that are not both equal to $x$, whose line segment (excluding the endpoints $y$ and $z$) contains $x$. [Make sure you understand why negating the definition gives us this characterization of what it means to not be an extreme point.] From this geometric intuition it should be easy to visualize why non-vertex points are not extreme: you can find a non-degenerate line segment that lies entirely in the cube and contains this non-vertex point.