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I'm wondering if a dense subset of the dual space is always norming. Precisely, let $E$ be a Banach space. Then for all $e\in E$, $$\|e\|=\sup\limits_{\substack{e'\in E'\\\|e'\|=1}}|e'(e)|.$$

Now assume that $D\subset E'$ is dense. Do we have

$$\|e\|=\sup\limits_{\substack{e'\in D\\\|e'\|=1}}|e'(e)|?$$

If this is not true in this generality, is it at least valid if $E$ is reflexive?

Alex Ortiz
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Noobtron
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    The dense subset $D \subset E'$ may not contain any element of norm $1$. You have $\lVert e\rVert = \sup {\lvert e'(e)\rvert : e' \in D, \lVert e'\rVert \leqslant 1}$. – Daniel Fischer Aug 27 '13 at 11:12

1 Answers1

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Slightly extending the comment by Daniel Fischer: if a subset $D$ of $E'$ is contained in the closed unit ball and $\overline{D}$ (norm closure) contains the unit sphere, then $D$ is norming for $E$. Indeed, given any $e\in E$ we can pick a norming functional $e'$ and then find $d\in D$ arbitrarily close to $e'$.

user98130
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