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Let $(X,||\cdot||)$ be a reflexive Banach space. Prove that $X$ is separable if and only if $X'$ (the dual space of $X$) is separable. Does anyone have a hint for me? I have no idea where to begin

Angelo Lucia
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2 Answers2

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The following theorem should be found in any textbook on Banach spaces / basic functional analysis, so you should be able to work out the proof and/or look it up.

Thm: Let $X$ be a Banach space. If $X^\prime$ is separable, then $X$ is separable.

Once you prove such theorem, your question becomes an easy corollary.

Angelo Lucia
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  • Why do we need X to be complete? – Not Euler Jun 08 '19 at 19:01
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    Good question: we don't! If $X$ is a normed linear space (not necessarily complete), and $X'$ is separable, then $X$ is separable. But notice that the dual of a normed linear space is a Banach space, so to go back to the original question: if $X$ is reflexive then it has to be complete. See https://math.stackexchange.com/questions/398970/show-reflexive-normed-vector-space-is-a-banach-space – Angelo Lucia Jun 12 '19 at 18:43
  • In general $L(X,Y)$ is complete whenever $Y$ is a Banach space right? – Not Euler Jun 12 '19 at 18:46
  • Yes: https://math.stackexchange.com/a/2057758/14794 – Angelo Lucia Jun 12 '19 at 18:48
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Check on this book of functional analysis http://www.amazon.com/Introductory-Functional-Analysis-Applications-Kreyszig/dp/0471504599

Page 245 $\rightarrow$ $\mathbf{Theorem\,of\,Separability\,in\,Normed\,Space}$

Lugo
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