For a normed vector space $ E $ and an element $ x \in E $, prove that if $ L(x) = 0 $ for every continuous linear functional $ L $, then $ x = 0 $.

Question

Question. Let $ E $ be a normed vector space. Is it true that for a given $ x \in E $, if $ L(x) = 0 $ for every $ L \in E' $, then $ x = 0_{E} $?

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One way to prove this is to find an $ L \in E' $ such that $ \| L \| = 1 $ and $ L(x) = \| x \| $. The existence of such an $ L $ is a corollary of the Hahn-Banach Theorem, but is it really necessary to use this result here?

Answer 1

Remark $ 3 $ on Page $ 8 $ of this document says that it is an utterly doomed enterprise to establish the statement of the OP without using the Hahn-Banach Theorem ($ \mathsf{HB} $).

If $ \mathsf{ZF} $ is consistent, then $ \mathsf{HB} $ is independent of $ \mathsf{ZF} $. Assuming that $ \mathsf{ZF} $ is consistent, there exists a model of $ \mathsf{ZF} + \neg \mathsf{HB} $. The document mentions that in every such model, there exists an infinite-dimensional normed vector space whose dual space is $ 0 $.

Answer 2

Here is a relatively concrete normed space: $l^\infty / c_0 =: E$. There are concrete nonzero elements, say the coset $(1,1,1,\cdots)+c_0$. But using only ZF, one cannot prove the existence of a nonzero bounded linear functional on $E$.