Is the polar set of convex polytope also a polytope? Let $P$ be a convex polytope.
How can I prove that the polar set of $P$ (lets call it $P^*$) is polytope?
where $P^*=\{x\in\mathcal R^n:\forall v\in P, |\langle x,v\rangle|\le1\}$ .
$Thanks$
 A: If the polytope is convex, it is also necessary to suppose $0 \in int(P)$. By the way the theorem works also if initially $0 \notin int(P)$ because P can be translated around the origin.
Since P is a polytope then $P=conv\{v_i\}_{i=1}^k$. So it follows:
\begin{align}
P^*&=\{z\in \mathbb{R}^n : z^Tx\leq 1, \forall x\in P\}\\
&=\{z\in \mathbb{R}^n : z^Tv_i\leq 1, i=1,\dots,k\}
\end{align}
Thus $P^*$ is a polyhedron. We know that a bounded polyhedron is a polytope so we have to find out if this polyhedron is bounded. But since $0 \in int(P)$ then $\exists \varepsilon>0$ such that $B(0,\varepsilon)\subseteq P$, then for every $0\neq z\in P^*$ we have $\varepsilon \frac{z}{||z||}\in P$. The definition of polar yields:
\begin{equation}
\varepsilon \frac{z^T z}{||z||}\leq 1 \Leftrightarrow ||z||\leq \frac{1}{\varepsilon} 
\end{equation}
So the proof is finished and the polar of a polytope is again a polytope. But actually we can prove something more. Recalling that if $C$ is closed and convex then $C^{**}=C$.
Therefore $P^{**}$ is again a polytope and in particular a bounded polyhedron.
So we have proved something more important: a polytope is a bounded polyhedron!
