If $B_X$ is weakly compact, then it is $w^{*}$-closed in $X^{**}$

Question

Suppose that $X$ is a Banach space, $X^{*}$ is it's dual space, $X^{**}$ is its double dual space, and $w$-topology means weak topology while $w^{*}$-topology means weak* topology. Why is the given statement true? I found it on page 75 of Fabian's "Functional Analysis and Infinite Dimensional Geometry."

Answer 1

I believe we can prove this using Alaoglu's theorem and Goldstine's theorem. The main point is that, if $B_X$ is $w$-compact, then $X$ and $X^\star$ are both reflexive.

So here is the argument. Let $\widehat{B_X}$ be the natural embedding of $B_X$ in $X^{\star\star}$. By assumption, $B_X$ is $w$-compact in $X$, so $\widehat{B_X}$ is $w^\star$-compact in $X^{\star\star}$, and in particular, $\widehat{B_X}$ is $w^\star$-closed in $X^{\star\star}$. Meanwhile, Goldstine's theorem tells us that $\widehat{B_X}$ is $w^\star$-dense in $B_{X^{\star\star}}$. But since $\widehat{B_X}$ is $w^\star$-closed, it must be the case that $\widehat{B_X} = B_{X^{\star\star}}$. This proves that $X = X^{\star\star}$, i.e. $X$ is reflexive.

By Alaoglu's theorem, $B_{X^\star}$ is $w^\star$-compact in $X^\star$. But since $X = X^{\star\star}$, the $w$- and $w^\star$-topologies in $X^\star$ are the same, so $B_{X^\star}$ is in fact $w$-compact in $X^\star$. Applying the conclusion of the previous paragraph to $X^\star$ instead of $X$, we learn that $X^\star$ is also reflexive.

Having shown that $X^\star$ is reflexive, we learn that the $w$- and $w^\star$-topologies in $X^{\star\star}$ are the same. We already know that $\widehat{B_X}$ is $w^\star$-closed in $X^{\star\star}$, so we can conclude that $\widehat{B_X}$ is $w$-closed in $X^{\star\star}$.