Let $X$ be a separable, infinite dimensional Banach space. Does $X^\star$ (the set of bounded complex linear functionals) separate the points of $X$? (meaning, for every two vectors $x,y\in X$ there is some $\phi \in X^\star$ such that $\phi(x)\neq\phi(y)$). What if $X$ is not a Banach space and is just a Fréchet space?
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Yes, the dual $X^*$ of every banach space $X$ separates the points of $X$. This is an immediate consequence of the Hahn-Banch theorem. A proof can be found in every introductory course on fuctional analysis. Moreover, the Hahn-Banach theorem holds in locally convex spaces. Thus statement is also true for Fréchet spaces.
Mat
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1You are fast. ${}$ – Rudy the Reindeer Nov 01 '12 at 10:25
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Though, Hahn-Banach needs a normed space, so you don't address the case of Fréchet spaces. – Rudy the Reindeer Nov 01 '12 at 10:27
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2Actually I just found it in Functional Analysis / Walter Rudin (1991) as a corollary of theorem 3.4. – user25640 Nov 01 '12 at 10:29
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It is enough that the space is locally convex – user25640 Nov 01 '12 at 10:30
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1Hello Matt, the Hahn-Banach theorem holds for Fréchet spaces – Mat Nov 01 '12 at 10:33
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3The Hahn-Banach theorem holds for Fréchet spaces ... provided that user25640's definition of Fréchet spaces includes local convexity. There are people who only require complete metrizability for Fréchet spaces and then Hahn-Banach fails in general. – commenter Nov 01 '12 at 10:45
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Thank you for your comment. I only get pinged if you use "@username". – Rudy the Reindeer Nov 01 '12 at 10:45
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Dear @commenter, thank you for your comment! – Rudy the Reindeer Nov 01 '12 at 11:08