Let $(X, \| \cdot \|)$ be a normed space, and let $A, B ⊂ X$ be disjoint convex sets such that $B$ is closed and $A$ is compact.
Prove that there exists $\varphi ∈ X^*$ such that $$\sup_{a\in A} \operatorname{Re}(\varphi)(a) < \inf_{b\in B} \operatorname{Re}(\varphi)(b).$$
I've been struggling a lot with this question, and would love some guide or help. My attempt was to use the geometric version of Hahn-Banach theorem, but I couldn't quite get this result. The problem is obviously showing the existence of a functional such that we have "$<$" instead of "$\leq$" using the extra properties of $A,B$.
Any help would be blessed. thanks!
Hints:
Let $A - B = \{a-b \;|\; a \in A, b \in B\}$. Show that $A-B$ is closed, convex, and $0 \notin A - B$.
Since $A-B$ is closed, there is a small $\varepsilon > 0$ so that $T := \{x\in X \;|\; \|x\| < \varepsilon\}$ and $A-B$ are disjoint. Separate them with a functional. $T$ is open, so the functional is continuous. And because of the $\varepsilon$ in the definition of $T$, in fact $A-B$ is strictly separated from $\{0\}$.