Consider the following version of the geometric Hahn-Banach theorem: let $X$ be a locally convex topological vector space over the real or the complex numbers, $A$ and $B$ two non-empty, disjoint and convex subsets of $X$ with $A$ compact and $B$ closed, then there is a continuous linear functional $f$ on $X$ such that $$ \sup_{x\in A} \Re f (x) < \inf_{x \in B} \Re f (x).$$
What if I want the reverse the inequality? Is it true that there is a continuous linear functional $g$ on $X$ such that $$ \sup_{x\in B} \Re g (x) < \inf_{x \in A} \Re g (x)?$$
My naive idea is to take $g=-f$, but does this work and why?
Just let $g = -f$. As $f$ is linear continuous, $g$ is also. As we have \begin{align*} \Re g(x) &= \Re (-f)(x) \\ &= - \Re f(x)\\ \sup_{x\in B} \Re g(x) &= \sup_{x\in B} -\Re f(x)\\ &= -\inf_{x\in B} \Re f(x) \\ &< -\sup_{x\in A} \Re f(x) & \text{by the inequality for $f$}\\ &= \inf_{x\in A} -\Re f(x)\\ &= \inf_{x \in A} \Re g(x) \end{align*}