In the geometrical version of the Hahn-Banach Theorem where $A$ is an open convex set and $B$ is any convex set, with $A$ and $B$ disjoint, we get that there is a continuous linear functional $f$ and a $s\in \mathbb R$ such that $f(a)<s\leq f(b)$ for all $a\in A$, $b\in B$.
If we look at singleton $B=\{ b\}$, then we get that there is $f$ such that $f(a) < f(b)$ for all $a\in A$ and this is true for all $b\notin A$. Let us write $f_b$ such a continuous linear functional that separates $A$ and $\{b\}$, then \begin{align*} A\subseteq \bigcap_{b\in A^c} f_b^{-1}(]-\infty, f_b(b)[) \end{align*} And also if $b\notin A$, then $b\notin f_b^{-1}(]-\infty,f_b(b)[)$ therefore $A=\bigcap_{b\in A^c} f_b^{-1}(]-\infty, f_b(b)[)$. From there since $f_b^{-1}(]\infty,f_b(b)[)\supseteq f_b^{-1}(f_b(A))$ we get that \begin{align*} A\subseteq \bigcap_{f\in E^\ast} f^{-1}(f(A)) \subseteq \bigcap_{b\in A^c} f_b^{-1}(]-\infty,f_b(b)[)=A \end{align*} Therefore whenever $A$ is open and convex $A=\bigcap_{f\in E^\ast} f^{-1}(f(A))$, this is also true whenever $A$ is closed and convex (easier to prove using the fact that a closed convex set is the intersection of all containing half space).
Conversely if $A$ is a convex set such that $A=\bigcap_{f\in E^\ast} f^{-1}(f(A))$ and $B$ is any convex set disjoint from $A$, then we can show that there is a continuous linear functional $f$ such that $\sup f(A)\leq \inf f(B)$ as well as $\inf f(A) < sup f(B)$.
Going a bit further, it turns out that the set of convex sets $A$ that are such that $A=\bigcap_{f\in E^\ast} f^{-1}(f(A))$ behaves nicely under several operations, meaning that it is closed under those operations and they also have a appealing form. Those operation are (well there is probably more but some that I could find) :
- Closure, in which case $\overline{A} = \bigcap_{f\in E^\ast} f^{-1}\left(\overline{f(A)}\right)$
- Interior, in which case $\mathrm{int} A=\bigcap_{f\in E^\ast} f^{-1}\left(\mathrm{int}{f(A)}\right)$
I have intuition that they also behave nicely under the following operations :
- Scalar Augmentation (there may be a better name for this) : For $A$ of the form mentioned above and $\Gamma$ a convex subset of $\mathbb R$, then define $\Gamma A\triangleq \{ \lambda a: \lambda\in Gamma, a\in A \}$, and then $\Gamma A=\bigcap_{f\in E^\ast} f^{-1}(\Gamma f(A))$
- Conification, in which case $\mathrm{cone} A=\bigcap_{f\in E^\ast} f^{-1}\left(\mathrm{cone}{f(A)}\right)$
- Affine hull, in which case $\mathrm{aff} A=\bigcap_{f\in E^\ast} f^{-1}\left(\mathrm{aff}{f(A)}\right)$
- Relative interior, in which case $\mathrm{ri} A=\bigcap_{f\in E^\ast} f^{-1}\left(\mathrm{ri}{f(A)}\right)$
I don't really know how to prove those but here is what I have so far. $\mathrm{cone}(A)=]0,\infty[ A$, $\mathrm{span}(A)=\mathbb R A$ (because $A$ is convex) so $\mathrm{cone}$ and $\mathrm{span}$ both follows from the scalar augmentation part. The relative interior also follows from $\mathrm{cone}$ and $\mathrm{span}$, indeed if we assume that $x\in\mathrm{ri}A$, then $\mathrm{cone}(A-x)=\mathrm{span}(A-x)$ so for any $f\in E^\ast$, \begin{align*} \mathrm{cone}(f(A)-f(x))&=\mathrm{cone}(f(A-x))\\ &=f(\mathrm{cone}(A-x))\\ &=f(\mathrm{span}(A-x))\\ &=\mathrm{span}(f(A-x))\\ &=\mathrm{span}(f(A)-f(x)) \end{align*} therefore $f(x)$ is in $\mathrm{ri}(f(A))$, i.e. $x\in f^{-1}(\mathrm{ri}(f(A)))$ and as this is true for any $f\in E^\ast$ and $x\in \mathrm{ri}(A)$ we get $\mathrm{ri}(A)\subseteq \bigcap_{f\in E^\ast}f^{-1}(\mathrm{ri}(f(A)))$. Suppose now that $x\in A\setminus \mathrm{ri}(A)$, then there is $y\in\mathrm{span}(A-x)$ such that $y\notin\mathrm{cone}(A-x)=\bigcap_{f\in E^\ast} f^{-1}(\mathrm{cone}(f(A-x)))$ therefore there is $f\in E^\ast$ such that $f(y)\notin \mathrm{cone}(f(A-x))=\mathrm{cone}(f(A)-f(x))$ and since $\mathrm{span}(A-x)=\bigcap_{f\in E^\ast} f^{-1}(\mathrm{span}f(A-x))$, we get that $f(y)\in \mathrm{span}(f(A-x))=\mathrm{span}(f(A)-f(x))$. From this we deduce $\mathrm{span}(f(A)-f(x))\neq \mathrm{cone}(f(A)-f(x))$ and so $f(x)\notin \mathrm{ri}(F(A))$ i.e. $x\notin f^{-1}(\mathrm{ri}(f(A)))$. This concludes the proof for the relative interior. The only remaining one are the Sum and Scalar Augmentation (and the rest depend on them). It is easy to prove the $\subseteq$ direction for both, indeed for all $f\in E^\ast$, $f(A+B)=f(A)+f(B)$, therefore $A+B\subseteq f^{-1}(f(A)+f(B))$ as well as $f(\Gamma A)=\Gamma f(A)$, therefore $\Gamma A\subseteq f^{-1}(\Gamma f(A))$. The other direction of the proof is much more challenging for me.