Let $C$ be a convex subset of a l.s.c space $E$. If $f$ is a continuous linear form on $E$, then $\overline C\subseteq f^{-1}(\overline{f(C)})$, indeed $\overline C$ is the intersection of all closed half spaces that contains $C$.
This remains true if $f$ is continuous on $\operatorname{span} C$ so my question is if, for a linear form that is discontinuous on $\operatorname{span} C$, we have $\overline C\subseteq f^{-1}(\overline{f(C)})$.
If I recall correctly, being continuous is equivalent to being bounded, therefore we get that $\overline{f(C)}$ is one of $]-\infty,+\infty[$, $]-\infty, a]$ or $[a,+\infty[$ assume, toward contradiction and WLOG the last case as well as $x\in\overline C$ is such that $f(x) < a$.
I am not sure on how to finish the proof, or if it is even true, any ideas would be most welcome.