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.
Best Answer
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.