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Consider the following set where n >= 1: $$A =\{(x_0,x_1,...,x_{2n}) \in R^{2n+1} | \sum\limits_{i=0}^{2n} c^ix_i \geq 0, \forall c \in [0,1] \}$$ Prove or Disprove whether this is a convex cone.

Prove or Disprove whether this is a pointed cone.

In order for a set C to be a convex cone, it must be a convex set and it must follow that

$$ \lambda x \in C, x \in C, \lambda \geq 0 $$

Additionally, a convex cone is pointed if the origin 0 is an extremal point of C

The 2n+1 aspect of the set is throwing me off, and I am confused by the summation. Does the summation imply that xn must be greater than 0, otherwise isn't it possible to get a point x that would sum to less than 0? Regardless, lambda*c^i should be greater than 0 since both lambda and c are greater than 0, so I think its just a matter of proving if this is a convex set or not.

A set is convex if for every combination x1, x2

$$ x_1, x_2 \in C, \lambda x_1 + (1-\lambda) x_2 = x, x \in C, 1 \geq \lambda \geq 0 $$

Like I said the R^2n+1 aspect and the x0,x1,...x2n is throwing me off on how to approach this proof

And I'm also not sure how to prove or disprove the origin is an extremal point of the set A.

73est
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3 Answers3

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To prove the set $A$ is pointed, you argue that if $x:=(x_0,x_1,\ldots,x_{2n})$ and $y:=(y_0,y_1,\ldots,y_{2n})$ are elements of $A$ such that $\alpha x+(1-\alpha)y=0$ for some $\alpha\in(0,1)$, then $x=y=0$.

So suppose $\alpha x+(1-\alpha)y=0$. For every $c\in[0,1]$ we know by definition of $A$ that $\sum c^i x_i\ge0$ and $\sum c^i y_i\ge0$. But $y=-\frac\alpha{1-\alpha}x$, so plugging in $y_i=-\frac\alpha{1-\alpha}x_i$ for every $i$ we obtain $$\sum_{i=0}^{2n}c^i x_i=0\quad\text{for all $c\in[0,1]$}.\tag1$$ From (1) you can deduce that $x$ is the zero vector, by arguing that each of its components in turn is zero. First, we have $x_0=0$ by letting $c\downarrow0$ in (1). Since the $i=0$ term is zero, we now have $$\sum_{i=1}^{2n}c^i x_i=0\quad\text{for all $c\in[0,1]$}.\tag2$$ If $c>0$ we can divide (2) by $c$ to isolate $x_1$. Now let $c\downarrow0$ to conclude $x_1=0$, and so on.

grand_chat
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This set is obviously convex. Moreover it is obviuosly convex cone. Just consider this sum. $$\sum_{i=0}^{2n}c^{i}(\alpha x_i + (1-\alpha)y_i) = \alpha \sum_{i=0}^{2n}c^{i}x_{i} + (1-\alpha) \sum_{i=0}^{2n}c^{i}y_{i} \geq 0$$. These are not series, and can easily be separated. I should think over a little about the extreme point. (I will come back:) ).

I tried this. I took n=1. So we will have that for any $c\in [0, 1]$, $\sum_{i=0}^{2}c^{i}(\frac{1}{2}x_{i}+\frac{1}{2}y_{i})=0$. (the definition of extreme point allow us to show the inexistence of the above , i.e. only for $\alpha =\frac{1}{2} $). But see what, if we take c=0, it come out that we should have $x_{0} +y_{0} =0$ , or $x_{0}= -y_{0}$. Next let's consider two different $c$-s, $c_{1} \neq c_{2}$. Now let's consider the difference which should laso be 0. So we will have the following - $$(x_{1}+y_{1})(c_{1}-c_{2})+(x_{2}+y_{2})(c^{2}_{1}- c^{2}_{2})=0$$ Having that $c^{2}_{1}- c^{2}_{2} = (c_{1}-c_{2})(c_{1}+c_{2})$, we will have. $$x_{1}+y_{1}+ (x_{2}+y_{2})(c_{1}+c_{2}) = 0$$.

We can take $c_{1}+ c_{2}$ as small as possible. So we will also have $x_{1} = - y_{1}$. From this we can have the same for $x_{2}$ and $y_{2}$. That's enough to say, that $\sum_{i=0}^{2}c^{i}x_{i} = -\sum_{i=0}^{2}c^{i}y_{i}$, so they can be equal only for the cases $\sum_{i=0}^{2}c^{i}x_{i} =0$. These can be true for every $c$ only when the sequence is exactly 0.

kolobokish
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I am not sure whether I understand the question properly, but taking $c_i = 1$ and $c_j = 0$ for $j \neq i$ you would get $x_i \geq 0$ and that your cone is just the positive orthant in the $\mathbb{R}^{2n+1}$. As already pointed out, this is a convex cone, moreover it is pointed as cone $Q$ is pointed if and only if $Q \cap -Q = \{0\}$, i.e. the intersection of $Q$ and $-Q$ is only the zero vector. This is again obviously true.

mplavala
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