This is a numerical I am trying to solve related to polarity.
Given $$G = \{x \in \Bbb{R}^d: \langle x,c_i \rangle \le 1\text{ for }i = 1,\ldots, n\},$$ where $c_i \ne 0$ for $i = 1, \ldots, n$, be a non-empty set. Prove that $(G^\circ)^\circ$ is $\operatorname{conv}(0, c_1, . . . , c_n)$.
What I have tried: I am trying to take the polar of $G$ i.e. $(G^\circ)$ will be given by $\{y \in \Bbb{R}^d: \langle y,x \rangle \le 1\text{ where }x \in G\}$. Should I again take the polar of $(G^\circ)$? I read somewhere that $(G^\circ)^\circ$ = $G$. So every element $d \in (G^\circ)^\circ$ can be written as $d= t_1c_1 +...+t_nc_n$ for $t \in [0,1]$ and $\sum_{n}t_n =1$. Now I do not know how to proceed. I am stuck over here. Could you please help me with this?
Thanks in advance.