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This question already has an answer here:

Let $(X,\|\|)$ be a normed space, and $(X^*,\|\|_*)$ and $(X^{**},\|\|_{**})$ be it's topological dual and topological bidual respectively. Now consider a norm $\rho$ in $X^*$which is equivalent to $\|\|_*.$ Does there exists a norm $\mu$ in $X$ such that $\mu_*= \rho$? (Of course, $\mu_*$ denotes the dual norm of $\mu.$)

I know that if $X$ is reflexive, such a norm indeed exists, and it is just $$\mu(x)=\sup_{\rho(x^*)\leq 1} |x^*(x)|.$$

The question is: can we remove reflexivity? I think that for a proof we could use Goldstine's Lemma. A proof or counterexample would be nice.

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marked as duplicate by gerw, Community Jul 20 '17 at 11:12

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  • $\begingroup$ Thanks to all, already checked! $\endgroup$ – John D Jul 20 '17 at 10:52