Let $X$ be a normed linear space. Then there is a linear isometry $J_X : X \longrightarrow X^{**}$ such that $J_X(x) (\varphi) = \varphi (x), x \in X, \varphi \in X^*.$ Then $X$ is said to be reflexive if $J_X (X) = X^{**}$ i.e. if $J_X$ is onto.
Prove that $X$ is reflexive if so is $X^*.$
Attempt $:$ Let $\psi \in X^{***}$ be such that $\psi (J_X(x)) = 0,$ for all $x \in X.$ Since $X^*$ is reflexive there exists $\varphi \in X^*$ such that $J_{X^*} (\varphi) = \psi.$ So $0 = \psi (J_X(x)) = J_{X^*} (\varphi) (J_X(x)) = J_X(x) (\varphi) = \varphi (x),$ for all $x \in X.$ Hence $\varphi = 0$ in $X^*.$ Therefore $\psi = J_{X^*} (\varphi) = J_{X^*} (0) = 0$ in $X^{***}.$ So any bounded linear functional on $X^{**}$ which vanishes on $J_X (X)$ is identically zero. So if $\overline {J_X (X)} \subsetneq X^{**}$ then by Hahn-Banach theorem there would exist a $y \in X^{**} \setminus {J_X(X)}$ (so that $\text {dist}\ (y, J_X(X)) \gt 0)$ and a $T \in X^{***}$ such that $T(y) = 1$ and $T(x) = 0,$ for every $x \in J_X (X),$ which will lead us to a contradiction. What will happen if $\overline {J_X(X)} = X^{**}$ i.e. if $J_X (X)$ is dense in $X^{**}\ $?
Any help in this regard will be appreciated. Thanks for your time.
The result is not true without completeness of $X$. Take $X=\ell_0$ the space of all sequences which are eventually $0$ with the $\ell^{2}$ norm. The dual of this space is the usual $\ell^{2}$ space which is reflexive. But $X$ itself is not reflexive since any reflexive space is necessarily complete.
Of course, if $X$ is complete then you don't run into a problem with your proof.