Which means $E = E ''$ when $E$ is a reflexive space?

Question

We know that canonical immersion $J_E: E \rightarrow E'' $ is an immersion of a normed vector space $E$ in its bidual. In case the space $ E $ is reflective, this gives an isometric isomorphism. But does that really mean that $ E = E ''$? For example, in an exercise, knowing that $E$ is reflective, I proved that $ M = (M^{\perp})^ {\perp} $ for all closed subspace $M$ of $E$, where $M^{\perp} = \{\phi \in E'\ ,: \phi (x) = 0 \,\, \forall x \in M \} $, but for that I actually proved that $ J_{E}(M) = (M^{\perp})^ {\perp}$. How should I argue to conclude from this that $ M = (M^{\perp})^ {\perp} $?

Answer 1

You can't conclude that $M = (M^\perp)^\perp$ in the sense of equality as sets since elements of the right hand side are functions on $E'$ and not elements of $E$. In this setting, that equality is always to be interpreted as equality up to the canonical identification of $E$ as a subset of $E''$ given by $J_E$. That is, the goal is to prove that $J_E(M) =(M^\perp)^\perp$.