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I want to prove that a reflexive normed space $X$ is a Banach space. By the definition of the reflexive space, the evaluation map $J:X\to X''$ is a bijection. All I need is to prove that the evaluation map is bicontinous (isomorphism) to transfer the Cauchy sequence properties from the Banach space $X''$ to $X$. But I don't know how to prove that $J$ is an isomorphism. To apply the "Banach isomorphism theorem", $X$ has to be already a Banach Space.

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  • $\begingroup$ Show that it is an isometry, that is, $\lVert x\rVert = \lVert Jx\rVert$. $\endgroup$ – Daniel Fischer May 13 '14 at 12:49
  • $\begingroup$ The map $J$ is an isometric isomorphism by definition of reflexive spaces. $\endgroup$ – Romeo May 13 '14 at 12:50
  • $\begingroup$ Thank you, this answers the question, $J$ having norm $=1$ is important then. $\endgroup$ – user165633 May 13 '14 at 13:13
  • $\begingroup$ @DanielFischer I don't think it's needed. It's enough that it's a normed vector space isomorphism, it doesn't need to be an isometry. Or am I missing something? $\endgroup$ – Rudy the Reindeer May 13 '14 at 15:41
  • $\begingroup$ @MattN. The OP says $X$ is reflexive if the canonical map $J\colon X\to X''$ is bijective. (That is usually called semi-reflexive, and reflexive means it's an isomorphism of locally convex spaces, but let's disregard the finer points of nomenclature here.) So the missing part is to show that the canonical map is a topological isomorphism. If you don't know that normed spaces are bornological, hence quasi-barrelled, and hence semi-reflexivity coincides with reflexivity, you need to show that $J$ is continuous and open. The easiest way to show that, IMO is to show that it is an isometry. $\endgroup$ – Daniel Fischer May 13 '14 at 15:52
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As pointed out in comments, the main step is to prove $\|Jx\|_{X^{**}}=\|x\|_X$ for all $x\in X$. This, together with surjectivity of $J$, imply that $X$ is isometric to $X^{**}$ and therefore inherits completeness from the latter.

For every unit norm functional $\phi$ we have $|\phi(x)|\le \|x\|$. By taking supremum over such $\phi$, we get $\|Jx\| \le \|x\|$. On the other hand, by the Hahn-Banach theorem (which does not require completeness) there is a unit norm functional $\phi\in X^*$ such that $\phi(x)=\|x\|$. Therefore, $\|Jx\|\ge |\phi(x)| = \|x\|$.

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