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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 $g\in A$.


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2 Answers 2

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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 $f\in A$.

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I don't think your proof is correct. The last three sentences don't make sense. If $\Lambda$ annihilates $A$, then automatically you get a contradiction, since $\Lambda(f)\not\in \Lambda(A)$, no need for the last three sentences. But what if it does not? The point is, as I explained, that any $\Lambda$ that separates a closed subspace from a convex compact set, must annihilate the subspace, since $\Lambda(subspace)$ must be a proper subspace of $\mathbb{R}$ or $\mathbb{C}$. –  Theo Apr 8 '13 at 5:46
Well I would not say that my proof was incorrect just poorly stated. I make an assumption that $f\not\in A$ and HBST gives me a $\Lambda$ s.t. $\Lambda(f)\not\in\Lambda(A)$. Then either $\Lambda$ does or does not annihilate f. If it does then we know that it also sends f to $0$, a contradiction. Thus, we move on to assume it does not then I show essentially the image of $A$ must be $\mathbb{C}$ by showing there is member of $A$ that has the same value as f, contradiction again. Thus, $\Lambda$ must not have existed showing that $f\not\in A$ and our original assumption was false –  Leo Spencer Apr 8 '13 at 8:43
Yes, now is more clear, and is essentially the same (standard) proof I gave. –  Theo Apr 8 '13 at 13:55

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.

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