In my studies of convex optimization I came across the following result:
If $ K $ is a closed convex cone in $ \mathbb{R}^2 $, then it has a finite number of extreme rays. An extreme ray is a ray generated by an extreme vector of $K$ wich is defined as follows: $ x\in K $ is an extreme vector of $ K $ if whenever we have $ x=y+z $ for $ y,z\in K $ we have that $ y,z $ are non-negative scalar multiples of x.
I have ideas on this intuitively and visually as I know by a previous theorem for a closed convex cone we have $ K=\overline{cone(T)} $ where cone operator denotes the minimal convex cone containing T the set of all extreme vectors of the cone in question, but I have no idea how to write a rigorous proof. I suspect that it has at most two extreme vectors as I am imagining a classic pointy cone, I know it has a finite number but not sure about two exactly, could someone please help me prove there are a finite number of extreme vectors of a closed planar cones. I thank all helpers.
Building on the answer in another post, consider the three cases:
$K\cap L'=L'$:
Then $K$ is the closed half space with boundary $L$.
$K\cap L'$ is a closed half line:
Let $p$ be the endpoint of $K\cap L'$. $K$ is pointed since it does not contain a line (it was contained entirely on one side of a halfspace) and WLOG assume it is pointed at the origin. Then $K$ is the cone generated by the vector $p$ and any nonzero vector in the direction of the half line $K\cap L'$.
$K\cap L'$ is a closed line segment:
Let $p,p'$ be the endpoints of this segment. Then $p,p'$ generate $K$. For let $x\in K$. Then there is some $c\in\mathbb{R}_{>0}$ and $0\le\lambda\le 1$ such that $cx=\lambda p + (1-\lambda)p'$. (Because there is some scaling of $x$ that lands in $K\cap L'$.) But then $x=\frac{\lambda}{c}p + \frac{1-\lambda}{c}p'$.