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Do cones always have an extreme point?

I think that every cone has an extreme point (and only one extreme point which is the zero vector)

By definition $C \subset \mathbb R^n$ is called cone if $\forall \mathbf x \in C, \forall \alpha \ge 0:\alpha\mathbf x \in C$

Now if $C = \emptyset$ then it's trivially a cone, otherwise $\exists \mathbf x \in C$, now by the defintion of $C$ as a cone, we see that if $\alpha=0$ then $\mathbf 0=0\mathbf x=\alpha\mathbf x \in C$ which shows that every nonempty cone contains the zero vector.

On the other hand for every $\mathbf 0\ne \mathbf x \in C$ we see that $1/2\mathbf x, 3/2\mathbf x\in C$ and for $\lambda=1/2$:$$ \mathbf x=\frac{1/2\mathbf x}{2} +\frac{3/2\mathbf x}{2}$$

Which implies that there is no nonzero vector in $C$ which is also an extreme point of $C$. Now let for some $\mathbf x,\mathbf y \in C$ and $\lambda \in (0,1)$:$\mathbf 0=\lambda\mathbf x+(1-\lambda)\mathbf y$

How to conclude that $\mathbf y=\mathbf x$ and from that conclude that $\mathbf 0$ is an extreme point of $C$ and so every cone has one extreme point?

masaheb
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  • Possibly helpful: https://math.stackexchange.com/questions/1257994/why-a-convex-cone-cannot-have-more-than-one-extreme-point – Chee Han Dec 24 '21 at 17:22
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    You cannot, with the definition of cone you are using. Note e.g. the whole $\mathbb R^n$ is a cone. –  Dec 24 '21 at 17:22
  • @ Stinking Bishop, So not every cone (with my definition) has an extreme point? – masaheb Dec 24 '21 at 17:26
  • @masaheb Correct. –  Dec 24 '21 at 17:27
  • @ Stinking Bishop,As I understand you took $C=\mathbb R^n$ as a cone, but in my definition $C$ cannot be equal to $\mathbb R^n$ – masaheb Dec 24 '21 at 17:28
  • In your definition I think you are allowing $C=\mathbb{R}^n$ as a cone. For another example (in a similar vein), consider $C={(a, b) \mid a \geq 0}$, which is the right half-space. This is a cone according to your definition, but $(0,0)$ is not extremal – Nicolás Vilches Dec 24 '21 at 17:34
  • @ Nicolás Vilches, How my definition allows $C=\mathbb R^n$ while it says that $C \subset \mathbb R^n$$? Moreover under what conditions a cone always has an extreme point? – masaheb Dec 24 '21 at 17:40
  • OK, it seems that $\subset$ means strict inclusion. You should include that in the statement, since otherwise we may think it just means inclusion. – GEdgar Dec 24 '21 at 17:51

2 Answers2

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This is a (right circlular) cone in $\mathbb R^3$ and has no extreme point.

$$ \{(x,y,z) \mid x^2+y^2=z^2\} $$

image

GEdgar
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The only cones that have 0 as extreme points are the pointed cones. The proof is quite simple.

Definition: a cone $C$ is pointed if $C \cap -C = \{ 0 \}.$

For the first part, suppose that $C$ is pointed. Let $0=(1-\alpha)x + \alpha y,$ for $0<\alpha<1$ and $x,y \in C$. Hence, $x = \dfrac{\alpha}{1-\alpha}(-y).$ As $\dfrac{\alpha}{1-\alpha}>0$ and $x \in C$, also $-y \in C$. As $C$ is pointed, $y \in C$ and $-y \in C$, necessarily $y=0$. The same can be done with $x$. Thus, $x=y$. This all lets us conclude that $0$ is an extreme point of $C$.

For the second part, suppose that $0$ is an extreme point of $C$. Let $v \in C$ and $-v \in C$. Thus, $$\dfrac{1}{2} v + \left(1-\dfrac{1}{2}\right)\left(-v\right) = 0.$$ Hence, if, by absurd, $v\not=0,$ $0$ could not be an extreme point of $C$, which is not the case. Consequently, $v=0$. This proves that $C$ is pointed.

R. W. Prado
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