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First let's defined embedding: $Y$ embeds into $X$, where $X$ and $Y$ are normed spaces, if there exists a 1-to-1 linear map from $Y$ into $X$ that is bicontinuous.

Suppose that $\ell_1$ embeds into a separable Banach space $X$. Clearly $X$ cannot be reflexive since every closed subspace of a reflexive space must be reflexive. However, can the dual $X^*$ of $X$ still be separable?

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    $\begingroup$ Have you considered the dual of the embedding operator? $\endgroup$ – user21467 Apr 27 '15 at 2:01
  • $\begingroup$ I guess if $T$ is the embedding you mean the dual of $T(\ell_1)$. That would indeed be isomorphic to $\ell_\infty$ and so not separable but that is as much as I know. $\endgroup$ – Anguepa Apr 27 '15 at 2:49
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    $\begingroup$ No, I mean the dual of $T$, namely $T^\ast\colon X^\ast\to\ell_\infty$. $\endgroup$ – user21467 Apr 27 '15 at 3:55
  • $\begingroup$ Oh the adjoint. No i haven't, wouldn't know how to use it. $\endgroup$ – Anguepa Apr 27 '15 at 12:54
  • $\begingroup$ Since $T$ is an embedding, it is injective and has closed range. What does that tell you about $T^\ast$? $\endgroup$ – Daniel Fischer Apr 27 '15 at 12:58
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Sketch: Show that $T^\ast\colon X^\ast\to\ell_\infty$ is surjective (and continuous). Show also that the continuous image of a separable space is separable. Conclude that $X^\ast$ is not separable.

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