# Construct $(P^*)^*$ for a polyhedron

I am trying to solve the following problem: Consider the polyhedron $$P = \{x\in \mathbb{R}^3 \mid x_1 + x_2 + x_3 \geq 1, x_j \geq 0 \; j \in [3] \}$$. The polar of a set $$S$$ is defined as $$S^* := \{y\mid y^T x \leq 1 \; \forall x\in S\}$$. Construct $$(P^*)^*$$.

What I did was to first find the extreme points and extreme rays of $$P$$, which are $$\{e_i, \; i=1, 2, 3 \}$$ for both the extreme points and extreme rays. Then I used the resolution theorem with the set of extreme points/rays I found to conclude that $$P^*$$ is the negative orthant of $$\mathbb{R}^3$$. Namely, that $$P^* = \{y\in\mathbb{R}^3 \mid y_1 \leq 0, y_2 \leq 0, y_3 \leq 0 \}$$. Then I realized that the only extreme point of $$P^*$$ is just the origin $$(0, 0, 0)$$, and the extreme rays are $$(-1, 0, 0), (0, -1, 0), (0, 0, -1)$$. Then I used the resolution theorem again to conclude that $$(P^*)^* = \{z\in\mathbb{R}^3 \mid z_1 \geq -1, z_2 \geq -1, z_3 \geq -1 \}$$.

Was there a better way to do this (assuming my answer is even correct)?

Well, but what is had for $$(P^*)^*$$ in the OP is incorrect though. [You may want to precisely state both the conditions and the conclusions of the resolution theorem.] You did conclude correctly that $$P^*$$ is indeed the set defined $$P^* = \{y \in \mathbb{R}^3; y = (y_1,y_2,y_3), {\text{ with }} y_1,y_2,y_3 \le 0\}.$$ The mistake was in calculating $$(P^*)^*$$.
In particular: You can check directly that, on the one hand, for any $$x \in \mathbb{R}^3$$ that has a negative coordinate, that there is a $$y \in P^*$$ such that $$x^Ty>1$$. Indeed, suppose $$x= (a_1,a_2,a_3)$$ is such that $$x$$ has a negative coordinate, or equivalently, there is an $$i \in \{1,2,3\}$$ such that $$a_i$$ is negative. Then let $$y$$ be any vector in $$P^*$$ such that the $$i$$-th coordinate of $$y$$ is $$\frac{2}{a_i}$$ [also a negative number], and every other coordinate of $$y$$ is $$0$$. Then $$x^Ty= \frac{2}{a_i} \times a_i = 2$$. So indeed, for any $$x \in \mathbb{R}^3$$ that has a negative coordinate, that there is a $$y \in P^*$$ such that $$x^Ty>1$$. Thus, if an $$x \in \mathbb{R}^3$$ that has a negative coordinate, then $$x$$ cannot be in $$(P^*)^*$$. So on the one hand, $$(P^*)^*$$ must be a subset of $$\{x \in \mathbb{R}^3$$; $$x = (x_1,x_2,x_3)$$, with $$x_1,x_2,x_3 \ge 0\}$$.
On the other hand, if every coordinate of $$x \in \mathbb{R}^3$$ is nonegative, then $$x^Ty \le 0$$ for all $$y \in P^*$$, so $$x$$ must be in $$(P^*)^*$$. So on the other hand, $$(P^*)^*$$ must contain $$\{x \in \mathbb{R}^3$$; $$x = (x_1,x_2,x_3)$$ with $$x_1,x_2,x_3 \ge 0\}$$.
$$(P^*)^* = \{x \in \mathbb{R}^3; x = (x_1,x_2,x_3), {\text{ with }} x_1,x_2,x_3 \ge 0\}.$$