Assuming $X$ is reflexive, that is the linear map $J_X:X \mapsto X^{**}$, $J_X(x)(f) = f(x)$ for all $f \in X^*$ is bijective. Is it true that $J_X^{-1}T^{**}J_X = T$?
Here's my attempt: Let $x \in X$, $y := (J_X^{-1}T^{**}J_X)(x)$. We have $(J_X^{-1}T^{**}J_X)(x) = J_X^{-1}(J_X(x)T^*) = y$. Hence $J_X(y) = J_X(x)T^*$, for any $f \in X^*$ it follows that $f(y) = f(T(x))$. Hahn-Banach tells us that functionals separate points, so this implies that $y = T(x)$, from which the statement follows. Is the proof correct?