I am stuck with the following question:
Given the convex cone $$K = \{x \in \mathbb R^n : x_1 \geq x_2 \geq \cdots \geq x_n \geq 0\}$$ determine the polar set of $K$, which we denote by $K^{*}$.
I know that $K$ is a convex cone, which is easy to proof. I also know from our lecture, that $K^{*}$ is again a convex cone and
$$K^{*}=\{y\in \mathbb{R^n}: \langle x,y\rangle \leq 0\ \forall x\in K\},$$
where $\langle\cdot,\cdot\rangle$ is the inner product.
It's quite obvious, that all $y\in \mathbb{R}^n$ where $y_i\leq0\ \forall i\in[n]$ are in $K^{*}$. But there have to be some more inequalities or easier ways to determine $K^{*}$. Do you have any hints/suggestions?
Ok, with Michael Grant's hint, the solution should be the following: $Let A\in \mathbb{R^{n\times n}}$ be the upper triangular matrix consisting of $1$'s. Then $K=A\cdot \mathbb{R^{n}_+}$. A is invertiable, and the following proposition can be easily shown:
$$(MX)^* = M^{-T}X^*\hspace{1mm} where \hspace{1mm} M\in \mathbb{R^{n\times n}} \hspace{1mm} invertiable$$
So for the solution of our problem, we just have to find out $A^{-T}$ and determine what ($\mathbb{R^{n}_+})^*$ is, which is not too difficult, since for $ y\in R^{n}:{<y,x>}\leq 0 \forall \hspace{1mm} x\in \mathbb{R^{n}_+}$, so $y\in\mathbb{R^{n}_-}$.