If $N\subset X^*$, then $({^\perp N})^\perp = \overline{N}^{w*}$

Question

I was trying to prove the following fact:

Suppose that $X$ is a Banach space and $N\subset X^*$ a linear subspace. Then $(^\perp N)^\perp = \overline{N}^{w*}$, where the closure is in the weak* topology.

I found this statement here

$G$ is dense in $X^*$ in weak* sense if and only if $G$ is total set

where the author of the post references to an exercise written (without solution) in the book of Dunford and Schwartz.

Answer 1

$(^{\perp}N)^{\perp}$ is $w^{\ast}$-closed, so $\overline{N}^{w^{\ast}}\subseteq(^{\perp}N)^{\perp}$.

For the other direction, if $x^{\ast}\in X^{\ast}-\overline{N}^{w^{\ast}}$, then an application of Hahn-Banach Theorem gives an $x\in X$ such that $\left<x,x^{\ast}\right>=1$ and $\overline{N}^{w^{\ast}}\subseteq\ker(Qx)$, where $Qx$ is the canonical induced element in $X^{\ast\ast}$. Since $x\in{^{\perp}N}$, then $x^{\ast}\notin(^{\perp}N)^{\perp}$.

Edit:

If we endow $X^{\ast}$ with $w^{\ast}$ topology $(X^{\ast},w^{\ast})$, this is not the norm topology, and the continuous maps on $(X^{\ast},w^{\ast})$ are those $Q(X)$, see here.