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Let X be a a separable Banach space. I have to prove that for every $x \in X$ there exists a sequence $\{x_n^*\}_{n \in \mathbb{N}} \subseteq \overline{B}_1^{X^*} = \{x^* \in X^*: \|x^*\|_* \leq 1\}$ such that $$ \|x\| = \sup _{n \in \mathbb{N}} |x_n^*(x)|.$$

The problem is, that I'm not allowed to use the Hahn-Banach theorem nor that the canonical embedding is an isometric isomorphism. I shall prove it by using this Theorem:

If $X$ is a separable normed space, then $(\overline{B}_1^{X^*},w^*)$ is metrizable.

How do I find this sequence if I only know that there is such a metric?

Duncan Idaho
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1 Answers1

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Shelah has shown that, should you ignore Hahn-Banach (as well as the axiom of choice on which it depends), then one cannot rule out the existence of a (non-separable) Banach space whose dual is $\{0\}$.

In fact Shelah proved that it is consistent with the Zermelo Fraenkel axioms that the dual of $\ell^\infty$ is $\ell^1$, from where one may easily deduce that the dual of $\ell^\infty/ c_0$ is $\{0\}$.

Your metrizability hypothesis therefore holds in that space but the conclusion you seek is false. This says that you cannot prove your result using your metrizability assumption by itself and instead you must reconsider the separability hypothesis, in some perhaps non-trivial way.

This said, it is my impression (clearly open to debate) that the only way to do what you want is to sneak in some parts of Hahn-Banach, shamelesly pretending that you are not using anything but your bare hands :-)

Ruy
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  • The theorem I have to use (If $X$ is a separable normed space, then $(\overline{B}_1^{X^},w^)$ is metrizable.) is using the Hahn-Banach theorem. – Duncan Idaho Sep 23 '20 at 19:35
  • Maybe you should double check upon this claim because I think the standard proof of mertizablity of the dual ball does not use HB at all. Instead one builds the metric $d(f,g) = \sum_{n=1}^\infty 2^{-n}|f(x_n)-g(x_n)|$, where ${x_n}_n$ is a dense set in the unit ball of the given space, and proves directly that the induced topology is the weak* topology. – Ruy Sep 23 '20 at 22:30