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I'm trying to prove that $X$ is closed in $X''$, where $X$ is a Banach space. I know that $X$ is embeddable in $X''$.

If the isomorphism was bijective, I could show that $X$ is closed since $X''$ closed and closeness is preserved. But in general, it isn't bijective.

Which approach should I take? Hints only please, if possible.

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up vote 2 down vote accepted

Sketch: $X$ is complete. The embedding $(Jx)(\varphi) = \varphi(x)$ is an isometry, so $J(X)$ is complete. Complete subspaces are always closed.

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In fact, a complete subspace of any (metric) space is closed. – Chris Eagle Dec 5 '12 at 22:48
@ChrisEagle thanks. Corrected. – kahen Dec 5 '12 at 22:49

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