Given a duality $\left<X^*,X\right>$ over field $\mathbb R$, and any set $A \subseteq X$, the polar set of $A$ is defined as \begin{align}A^\circ = \{x^* \in X^* | \left<x^*,x\right> \leq 1\}.\end{align}
The is a famous bi-polar theorem saying that if $0 \in A$, then the bi-polar set $A^{\circ\circ}$, which is the polar set of the polar set, of a set $A$ is the closed convex hull of $A$ under weak-topology.
For a function $f: X \to \mathbb R$, its eipigraph is defined as the set of points above its graph: \begin{align}\mathrm{epi}(f) = \{(x,y) \in X \times \mathbb R | y \geq f(x)\}, \end{align} and its convex conjugate is defined as a function $f^*: X^* \to \mathbb R$, which satisfies\begin{align} f^*(x^*) = \sup_{x \in X}\left<x^*, x\right> - f(x). \end{align} There is a famous Fenchel-Moreau theorem saying that the bi-conjugate $f^{**}$ of a function $f$ is its lower semi-continous (lsc) convex hull (the largest lsc convex function that is smaller than $f$). A function is lsc iff its epigraph is a closed set in $X \times \mathbb R$, and a function is convex iff its epigraph is a convex set in $X\times \mathbb R$. Which means, taking bi-conjugate of a function $f$ is equavalent to taking the closed convex hull of $\mathrm{epi}(f)$.
Notice that, the bi-polar theorem and Fenchel-Moreau theorem both characterise a closed convex hull of a given set. More critically, they encode the same idea: a closed convex set can be (dually) described by the intersection of all half-planes that contains it.
Given this correspondence, my question is do polar sets and convex conjugates have a deeper relationship? There is a section on wikipedia saying that bipolar theorem can be proved by Fenchel-Moreau theorem. But this statement doesn't seem to hit the essence of this issue. I wonder, for example, if we can do the reverse: can we prove Fenchel-Moreau theorem by bi-polar theorem? Or if we can describe convex conjugate using the language of polar sets?