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Does the space $C([0,1])$ with the usual norm embed isomorphicall/isometrically into its dual?

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  • $\begingroup$ @Shalop OP clearly says isomorphism 'into' the dual space. So the non-separability argument doesn't work. $\endgroup$ – Kavi Rama Murthy Aug 7 '18 at 23:33
  • $\begingroup$ @KaviRamaMurthy Yes you are right $\endgroup$ – Shalop Aug 7 '18 at 23:34
  • $\begingroup$ Yes. Every normed space embeds isometrically as a subspace in its second dual. $\endgroup$ – Michael Greinecker Aug 8 '18 at 8:56
  • $\begingroup$ If I recall correctly, every operator from a $C(K)$ space to a $L_1(\mu)$ space is weakly compact. Does that contradict your claim? $\endgroup$ – Adrián González-Pérez Aug 8 '18 at 10:17
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    $\begingroup$ The dual of a $C(K)$ space is an $L_1(\mu)$ space, and any non-reflexive subspace of an $L(\mu)$ space contains $\ell_1$ isomorphically. Were $C[0,1]$ embeddable into $C([0,1])^*$, $c_0$ would so also. But $c_0$ does not contain a copy of $\ell_1$. $\endgroup$ – David Mitra Aug 8 '18 at 14:48
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No, it does not because the dual of $C[0,1]$ is weakly sequentially complete, but $C[0,1]$ is clearly not and this property passes to subspaces. Consequently, $C[0,1]$ does not isomorphically embed into its dual.

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