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It is clear to me that if $X$ is a Banach space and its dual $X^*$ is reflexive, then $X$ is also reflexive (that is, the natural map between $X$ and its double dual $X^{**}$ is a surjective isometric isomorphism). However, I suspect that the conclusion fails to be true if we remove the completeness hypothesis, but I can't formalize this argument.

Are there famous examples of incomplete (and hence non-reflexive) normed vector spaces whose duals are still reflexive? Or does the reflexivity of $X^*$ imply that $X$ is reflexive (and hence Banach) even if without assuming ab initio that $X$ is a Banach space?

triple_sec
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1 Answers1

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Let $Y$ be any reflexive Banach space (or even a Hilbert space), and let $X$ be a proper dense subspace of $Y$. Then $X$ is not complete and its completion is $Y$. We then have $X^* = Y^*$ which is reflexive.

Nate Eldredge
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    Thank you very much. To establish the desired example based on your hints, I took $$Y\equiv\left{(c_n){n=1}^{\infty}\subseteq\mathbb C,\bigg|,\sum{n=1}^{\infty}|c_n|^2<+\infty\right}$$ with the norm $(c_n){n=1}^{\infty}\mapsto\sqrt{\sum{n=1}^{\infty}|c_n|^2}$ and $$X\equiv\left{(c_n){n=1}^{\infty},\big|,\exists N\in\mathbb Z+:c_n=0,\forall n>N\right}.$$ Then $Y$ is an infinite-dimensional reflexive normed vector space, $X$ is a proper dense infinite-dimensional incomplete subspace. Moreover, $X^$ is isometrically isomorphic to $Y^$ and $Y^$ is reflexive, and thus so is $X^$. – triple_sec Jan 09 '14 at 22:27