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Let $\ell_{1}$ and $\ell_{\infty}$ be the usual spaces of, respectively, convergent and bounded sequences, with their usual norms. As we know, the topological dual of $\ell_{1}$ is $\ell_{\infty}$, that is $(\ell_{1})^{*}=\ell_{\infty}$. But, the dual of $\ell_{\infty}$ is not $\ell_{1}$ (because $\ell_{1}$ is not reflexive), that is, $(\ell_{1})^{**}\subset \ell_{\infty}^{*}$ and the inclusion is strict.

Suppose that $X$ is a separable and reflexive subspace of $\ell_{\infty}$. Can we identify $X^{*}$ with a subspace of $\ell_{1}$?

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  • $\begingroup$ As a matter of fact, every reflexive subspace of $\ell_\infty$ is automatically separable. $\endgroup$ – Tomek Kania Jul 13 '16 at 16:38
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No.

In fact, $\ell_1$ contains no infinite dimensional reflexive subspace.

This fact follows since $\ell_1$, hence any closed subspace of $\ell_1$, has the Schur property (weakly convergent sequences are norm convergent), while reflexive infinite dimensional Banach spaces do not have the Schur property (in such spaces, the closed unit ball is sequentially weakly compact but not norm compact).

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  • $\begingroup$ Thanks Mitra!! I don't take into account the Schur property of $\ell_{1}$. $\endgroup$ – user123043 May 6 '14 at 5:55

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