For each $Q \subset \Bbb R^n$, denote $Q^*:=\{z \in \Bbb R^n:z\cdot x \leq 1,\;\;\text{for all}\; x \in Q\}$.
Let $P:=\{x \in \Bbb R^n: Ax \leq b\}$, for the matrix $A$ and the vector $b$. It is clear that $P \subset P^{**}$. My question: $P = P^{**}$?
There are some particular cases that the equality happens. But I did not succeed in the general case. I would like to collect ideas here. Many thanks.
By definition, $Q^*$ is a closed convex set that contains $0$ (because it is the intersection of closed half-spaces containing $0$). So, if the equality $P=P^{**}$ holds, $P$ must be a closed convex set containing $0$.
In fact, the converse is also true: if $P$ is a closed convex set containing $0$, then $P=P^{**}$. Indeed, let $x\in \mathbb R^n\setminus P$. There is a hyperplane $H$ strictly separating $x$ from $P$. Write it as $H=\{y: \langle y,z\rangle = 1\}$, where we can choose the direction of $z$ so that $P\subseteq \{y: \langle y,z\rangle < 1\}$. Note that $z\in P^*$ by the definition of $P^*$. And since $\langle x,z\rangle >1$, we have $x\notin P^{**}$. This shows $P^{**}\subseteq P$; the reverse inclusion, as you noted, is trivial.