Hahn-Banach separation theorem: show that $f(a)<\inf f(B)$ for all $a\in \text{int}(A)$

Question

Hahn-Banach separation Theorem. Let $A,B$ be convex non-empty disjoint subsets of a real normed space $X$. Suppose also that $\text{int}(A)\neq \emptyset$ (non-empty interior). Then there exists a continuous linear functional $f$ on $X$ such that $\sup f(A)\leq \inf f(B)$ and $f(a)<\inf f(B)$ for all $a\in \text{int}(A)$.

The proofs I have seen do not show the last claim that $f(a)<\inf f(B)$ for all $a\in \text{int}(A)$.

How can one show this?

Answer 1

Assume that $f(a)=\inf(f(B))$ and $a \in \rm{int}(A)$.

Then $\sup(f(A)) \le \inf(f(B)) = f(a) \le \sup(f(A))$, hence $f(a)=\sup(f(A))$. Choose $r> 0$ such that the ball $B_r(a):=\{x \in X: \|x-a\|<r\}$ is a subset of $A$. Since $f$ is not the zero functional there is some $y \in X$ (w.l.o.g. $\|y\|=1$) such that $f(y) > 0$. Now $a+(r/2)y \in B_r(a) \subseteq A$ and $f(a+(r/2)y) > f(a)=\sup(f(A))$, a contradiction.