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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.

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  • $\begingroup$ You are fast. ${}$ $\endgroup$ – Rudy the Reindeer Nov 1 '12 at 10:25
  • $\begingroup$ Though, Hahn-Banach needs a normed space, so you don't address the case of Fréchet spaces. $\endgroup$ – Rudy the Reindeer Nov 1 '12 at 10:27
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    $\begingroup$ Actually I just found it in Functional Analysis / Walter Rudin (1991) as a corollary of theorem 3.4. $\endgroup$ – user25640 Nov 1 '12 at 10:29
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    $\begingroup$ Hello Matt, the Hahn-Banach theorem holds for Fréchet spaces $\endgroup$ – Mat Nov 1 '12 at 10:33
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    $\begingroup$ The 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. $\endgroup$ – commenter Nov 1 '12 at 10:45

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