Hahn-Banach Separation Theorem and Bishop's Theorem

Question

I am looking at the proof of Bishop's Theorem on pages 122 and 123 of Rudin's Functional Analysis. The following quote is from the the last two sentences of the proof on pg. 123.

"Every continuous linear functional on $C(S)$ that annihilates $A$ also annihilates g. Hence $g\in A$ by the Hahn-Banach Separation Theorem."

I understand the first sentence but I don't see why that and the HBST implies $g\in A$.

Thanks!

Answer 1

By the HBST, since $\left\{ f\right\}$ is compact, $A$ is closed, and $C(X)$ is locally convex, there exists a $\Lambda\in {C(X)}^*$ such that $\Lambda(f)\not\in \Lambda(A)$. Since $A$ is a subspace of $C(X)$, the image of $A$ is $\left\{ 0\right\}$ or $\mathbb{C}$. However, any $\Lambda$ that annihilates $A$ must annihilate $f$ as well implying $\Lambda(f)\in \Lambda(A)$ in both cases. Therefore, $\Lambda$ cannot exist implying $f\in A$.

Answer 2

I assume that $A$ is a subspace in your context. If $g\notin \overline{A}$, then by HBST, there exist a bounded linear functional $F$ such that $F(g)$ and $F(\overline{A})$ are disjoint. Since $\overline{A}$ is a subspace, $F(\overline{A})$ is a subspace of the scalar field. As well, $F(\overline{A})$ is different from the scalar field, since $F(g)$ is disjoint from $F(\overline{A})$. Hence $F(\overline{A})=\{0\}$ and $F(g)\neq 0$, contradicting your hypotheses, that any functional cancelling $A$, also cancels $g$. Therefore $g\in\overline{A}$. On second thought, $A$ must be a closed subspace in your context, to get the claimed conclusion.