While reading through "Linear Operators: General theory" by "Jacob T. Schwartz", reading the corollary to II.10.1 which states that for a compact convex subset $C$ of some Hausdorff l.c.s $E$ (with topological dual $E'$ and algebraic dual $E^\ast$), we have $C=\bigcap_{f\in E'} f^{-1}(f(C))$, I got to think a bit more about this form. Note that $C\subseteq \bigcap_{f\in Y} f^{-1}(f(C))$ is obvious for any subset $Y$ of $E^\ast$, for instance if $Y=E'$.
If $C$ is an open convex subset of $E$, then for any $x\notin C$, $\{ x \}$ and $C$ are disjoint and therefore by the (Geometrical version of the) Hahn-Banach separation theorem, there is $f\in E'$ such that $f(x)\notin f(C)$, therefore \begin{align*} C&\subseteq \bigcap_{f\in E^\ast} f^{-1}(f(C))\\ &\subseteq \bigcap_{f\in E'} f^{-1}(f(C))\\ &\subseteq C \end{align*}
If $C$ is a closed convex subset of $E$, then for any $x\notin C$, $\{ x \}$ is compact convex and disjoint from $C$, so again we can separate the two sets and similarly to before we get $C=\bigcap_{f\in E'} f^{-1}(f(C)) = \bigcap_{f\in E^\ast} f^{-1}(f(C))$.
Lets say that a convex set $C$ is full (or filled) if $C=\bigcap_{f\in E^\ast} f^{-1}(f(C))$ and note that the set of filled convex subsets of $E$ is closed under arbitrary intersection, indeed if $\{ A_i\}_{i\in\mathcal I}$ is a familly of filled convex sets then \begin{align*} \bigcap_{i\in\mathcal I} A_i &= \bigcap_{i\in\mathcal I} \bigcap_{f\in E^\ast} f^{-1}(f(A_i))\\ &=\bigcap_{f\in E^\ast} f^{-1} \left( \bigcap_{i\in\mathcal I} f(A_i) \right)\\ &\subseteq \bigcap_{f\in E^\ast} f^{-1} \left( f\left(\bigcap_{i\in\mathcal I} A_i\right) \right)\\ \end{align*}
We can therefore define the filled convex hull of any set $C$ as the intersection of all filled convex sets that contains $C$, it turns out to be exactly $\bigcap_{f\in E^\ast} f^{-1}(f(C))$.
If $C$ is a fully convex set, then so is $\lambda C$ for any $\lambda \in \mathbb R$, similarly $C+x$ is also fully convex if $C$ is fully convex and $x\in E$.
My main question is if the set of fully convex set is also closed under set addition. If $A$ and $B$ are fully convex sets then is $A+B$ also fully convex ? This is actually completely equivalent to the following separation theorem :
If $A$ and $B$ are fully convex disjoint sets, then there is $f\in E^\ast$ such that $f(A)$ and $f(B)$ are disjoint.
Indeed $A+B$ fully convex for all $A$, $B$ is equivalent to $A-B$ being fully convex for all $A$, $B$, which is true if for any point $x\notin A-B$, there is $f\in E^\ast$ such that $f(x)\notin f(A-B)$ or equivalently $0\notin f(A-x) - f(B)$ which itself is equivalent to $f(A-x)$ being disjoint from $f(B)$.
It may be useful to note that $f^{-1}(f(A))=A+f^{-1}(\{0\})$
