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.

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    $\begingroup$ Have you seen math.stackexchange.com/q/1967361/318600? $\endgroup$ – Tachyon Oct 16 '16 at 1:45
  • $\begingroup$ @Tachyon thanks I saw it I could not understand the answer $\endgroup$ – kroner Oct 16 '16 at 1:46
  • $\begingroup$ @Tachyon The author of the answer says to consider intersection with line and I dont see how that works $\endgroup$ – kroner Oct 16 '16 at 1:53
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    $\begingroup$ How much of the answer there do you understand? $\endgroup$ – Tachyon Oct 16 '16 at 1:57
  • $\begingroup$ @Tachyon well I understand everything written I do not understand what the author intends I do not see why he takes the intersection of a line in the halfspace and why he considers intersection cases so I do not understand how to proceed with his method to obtain the desired result $\endgroup$ – kroner Oct 16 '16 at 2:01

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 $x=\frac{\lambda}{c}p + \frac{1-\lambda}{c}p'$.

  • $\begingroup$ I thank you kindly for this great answer $\endgroup$ – kroner Oct 16 '16 at 2:18

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