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?
When does $X^\star$ separate the points of $X$?
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functional-analysis
banach-spaces
1 Answers
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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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0You are fast. ${}$ – 2012-11-01
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0Though, Hahn-Banach needs a normed space, so you don't address the case of Fréchet spaces. – 2012-11-01
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1Actually I just found it in Functional Analysis / Walter Rudin (1991) as a corollary of theorem 3.4. – 2012-11-01
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0It is enough that the space is locally convex – 2012-11-01
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1Hello Matt, the Hahn-Banach theorem holds for Fréchet spaces – 2012-11-01
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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. – 2012-11-01
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0Thank you for your comment. I only get pinged if you use "@username". – 2012-11-01
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0Dear @commenter, thank you for your comment! – 2012-11-01