dual set of the dual set

Question

Let $X\subseteq\mathbb{R}^d$ and let $X^*$ be it's dual set i.e. $X^*=\{y\in\mathbb{R}^d| <x,y>\leq 1$ for every $x\in X\}$. How to prove that $(X^*)^*=\overline{conv(X\cup\{0\})}$? I know that $(X^*)^*$ is convex, closed and that it contains origin (being equal to $\bigcap_{x\in X^*}\mathcal{D}_0(x)^-$ where $\mathcal{D}_0$ is dual transform)

Answer 1

I came up with a solution, but I would appreciate if someone could check it for errors.

First, we prove that $X\subseteq(X^*)^*$. Let $x\in X$. From the definition of a dual set we have $<x,y>\leq 1$ for every $y\in X^*$, hence, again by the definition of dual set, we have $x\in (X^*)^*$. Since $(X^*)^*$ is closed, convex and contains origin, we have $\overline{conv(X\cup\{0\})}\subseteq (X^*)^*$.

Now we prove the reversed inclusion. Suppose there is $x_0\in(X^*)^*\setminus\overline{conv(X\cup\{0\})}$. Separation theorem for the convex sets says that there is $a\in\mathbb{R}^d$ s.t. $<a,x_0>>1$ and $<a,x>\leq 1$ for every $x\in X$. From here we have $a\in X^*$. Since $x_0\in (X^*)^*$ we have $<a,x_0>\leq 1$, by the definition of a dual set. But this is contradiction.