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In Fleming and Jamison's book, Banach first proved a lemma which used directional derivative to identity peak point of functions. Then he used the lemma in the proof of Banach-Stone theorem. After several years, Stone extended the result by loosening the condition compact metric space to compact Hausdorff space. But I can't seem to find any proof by Stone on how he proved the theorem with just only Hausdorff space. If anyone came across the original proof, can share the link here? or anyone still remember the idea, can share it?

Idonknow
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2 Answers2

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Here is the refrence to the original article, see theorem 83.

The main idea of the proof is the following. Let $Z$ be a compact Hausdorff space. For a given $p\in Z$ denote $M_Z(p)=\{f\in C(Z):|f(p)|=\Vert f\Vert\}$. Also denote $\mathcal{M}_Z=\{M_Z(p):p\in Z\}$. Given $M\in\mathcal{M}_Z$ we can always recover its point $p\in Z$.

Once we have a surjective isometry $V:C(X)\to C(Y)$ we can establish a bijective correspondence between sets $\mathcal{M}_Y$ and $\mathcal{M}_X$. This correspondence gives rise to the bijection $\rho:Y\to X:q\mapsto p$ which turns out to be a homemorphism.

Norbert
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  • How do you know theorem 83 given in your link is proved by Stone? Also how is it link to the Banach-Stone Theorem? – Idonknow Dec 16 '13 at 05:13
  • Because this article by M. H. Stone and what is more before Th 83 he said that this proof differs from the similar fact proved by Banach – Norbert Dec 16 '13 at 07:05
  • In the article, theorem $83$, what does bicompact H-spaces mean? Also, the author defined (I guess) separable bicompact H-spaces as compact metric spaces. Then what about compact Hausdorff spaces? – Idonknow Dec 20 '13 at 14:08
  • Bicompact is deprecated name for compact. In early twentie's compact meant sequentially compact. H-space is a abbreviation of Hausdorff. – Norbert Dec 20 '13 at 14:38
  • When the author claims that $|g|=\sum_{v=1}^{n}{|f_v|}$ by inferring the inequalities in pg $470$, why we have $\sum_{v=1}^{n}{|f_v({\tau})|}=g(\tau)$? – Idonknow Dec 20 '13 at 17:14
  • since $a\operatorname{sign}(a)=|a|$ for all $a$, then $g(\tau)=\sum_{v=1}^n f_v(\tau)\operatorname{sign}(f_v(\tau))=\sum_{v=1}^n|f_v(\tau)|$ – Norbert Dec 20 '13 at 17:32
  • Again on page $470$, first sentence 'We can now show that the closed sets ...'. How do we know the set is closed? – Idonknow Dec 20 '13 at 19:19
  • this set is closed as preimage of one point (hence closed) set ${\Vert f^\Vert}\subset\mathbb{R}$ under continuous map $e: \Re^ \to\mathbb{R}:f^\mapsto |f^(\tau^*)|$ – Norbert Dec 20 '13 at 20:38
  • Can you elaborate further? I don't understand. – Idonknow Dec 21 '13 at 09:23
  • Do you understand that $e$ is continuous? – Norbert Dec 21 '13 at 11:20
  • No, I don't understand why $e$ is continuous – Idonknow Dec 21 '13 at 11:25
  • If $f_n^$ converges to $f^$ in $\Re^$ (i.e. in $\sup$ norm), then, it converges pointwise (in paticular at point $\tau^$). This is equivalent to write that $f_n^\underset{\Re^}{\to} f^\implies e(f_n^)\to e(f^*)$. So $e$ is continuous Is that clear? – Norbert Dec 21 '13 at 11:30
  • Okay. Then what does this continuity tell us about the closeness of the set? – Idonknow Dec 21 '13 at 11:33
  • This very basic fact from toplogy that preimage under continuous map of closed set is again closed. – Norbert Dec 21 '13 at 11:56
  • I think the domain of your map is not correct. Isn't the domain consist of elements $\tau^{}$ such that $|f^{}|=|f^{}(\tau^{})|$? – Idonknow Dec 21 '13 at 18:42
  • @Idonknow, yes you are right., but the argument remains the same. – Norbert Dec 21 '13 at 20:03
  • I still don't understand why $e$ is continuous. Perhaps you can explain in details? – Idonknow Dec 23 '13 at 07:34
  • Can we still use limit to prove continuity? Because the domain now is not metric space, so we cannot use any distance notion (which includes limit). – Idonknow Dec 23 '13 at 07:52
  • you can use nets – Norbert Dec 23 '13 at 07:55
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In here, the following article of Stone (1937) is referred:

Stone, M.H., Applications of the theory of boolean rings to General Topology, Trans. Amer. Math. Soc., 41 (1937), 375 – 481.

user91126
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  • You beat me to this! Anyway I'll post my answer :) – Norbert Dec 15 '13 at 21:33
  • ahahah! Norbert, I can be quicker, but surely you're wiser than me ;) – user91126 Dec 15 '13 at 21:45
  • @Federico: Actually I had read the article before you provided it in your answer but I didn't find any stone proof on the Banach-Stone Theorem. – Idonknow Dec 16 '13 at 05:08
  • @Idonknow Dear Idonknow, I agree with Norbert' comment to his answer. Maybe you found the statement slightly different from actual formulation because that of Stone is a generalization of the Banach' theorem, but Theorem 83 is what you're looking for. Since Stone doesn't provide any reference (and since he wrote this article in 1937, 5 years after Banach), it is natural to suppose that the proof is by him. Moreover, as the article I linked proves, Theorem 83 is surely the reason why we call it Banach-Stone theorem now. :) – user91126 Dec 16 '13 at 11:36