 Solution

For a normed space $X$, let $X^{\ast }$ be its continuous dual space and $X^{\ast \ast }=(X^{\ast }{)}^{\ast }$ its bidual space. Its canonical embedding into the bidual isThe definition of a reflexive space in this setting isThe Hahn-Banach theorem makes $J_X$ a linear map that is an isometry. For any normed space $Z$, its continuous dual space is a Banach space: a Cauchy sequence $({\ell }_n)$ in the operator norm has scalar limits $\ell (z)={\lim }_n{\ell }_n(z)$, and the uniform Cauchy bound proves both boundedness of $\ell$ and $\Vert {\ell }_n-\ell \Vert \to 0$. Apply this with $Z=X^{\ast }$ to see that $X^{\ast \ast }$ is complete. If $J_X$ is onto, the isometry transfers this completeness to $X$. Therefore