Dual cone of $\mathbb{R}^{n}$

Question

Let $E$ be a Banach space and $P\subset E$ be a cone. A nonempty subset P of a Banach space E is called a cone if P is convex, closed, and satisfies the conditions: (i) $λx\in P$ for all $x\in P$ and all real positive number $λ$. (ii) $x,−x\in P$ imply $x = θ.$

Let P be a cone of a real Banach space E. The set $P=\left\{ψ \in E^{*} | ψ(x) ≥ 0, \forall x\in P\right\}$ is called the dual cone of P, where $E^{*}$ is the topological dual of $E$.

Let P = {x = (x_1, x_2, . . . , x_n) ∈ R^n | x_i ≥ 0, i = 1, 2, . . . , n}. How can I show that $P^{*}=P$.

Answer 1

First, you can show that $P \subset P^*.$ This is straight forward, since for any $y \in P$ and $ x \in P$ the product $$ \langle y, x \rangle = \sum_{i = 1}^n y_i x_i \ge 0 $$ since all terms involved are positive. On the other hand take a $y \in \mathbb{R}^n$ with one negative component, say $y_{i_0} < 0.$ Then you know that the basis vector $e_{i_0} \in P$ where $e_{i_0}$ has all entries zero but the $i_0-$th one, which is $1$. Then you can see that $$ \langle y,e_{i_0} \rangle = y_{i_0} < 0 $$

So that $y$ cannot belong to $P^*.$ This proves that $P = P^*$, at least under the assumption of choosing two dual bases for $\mathbb{R}^n$ and it's dual.