Understanding the $\sigma(X^{\ast\ast},X^\ast)$ topology

Question

I'm trying to prove that

$X$, a Banach space, is dense in its bidual $X^{\ast \ast}$ with respect to the $\sigma(X^{\ast\ast},X^\ast)$ topology.

I'd like some help. In particular, I know that the $\sigma(X^{\ast\ast},X^\ast)$ topology is the weakest on $X^{\ast \ast}$ such that all elements of $X^\ast$ are continuous. I also know that $X$ can be isometrically embedded in $X^{\ast \ast}$ by the map $J$ defined by $(Jx)(\phi) = \phi(x)$ for $x \in X$ and $\phi \in X^\ast$. How do I connect these pieces?

Answer 1

That statement is known as 'Goldstine theorem': The image of the $X$-unit ball $B$ under the canoncial evaluation map $J \colon X \rightarrow X''$ is weak*-dense in the closed unit ball $B''$ of $X''$.

You have to prove that for $x'' \in X''$ and any weak*-open set $U''$ with $x'' \in U''$ there exists $x \in B$ with $J(x) \in U$. It is enough to consider open sets given by some $x_1',\ldots,x_n' \in X'$, i.e. $$U''= \cap_{k=1}^n \{y'': |y''(x'_k)-x''(x'_k)| < 1\}.$$

• The next step is to show that we can assume that $x_1',\ldots,x_n'$ are linearly independent.
• Use that $x \mapsto (x_1'(x),\ldots,x_n'(x))$ is surjective onto $\mathbb{R}^n$ if $x_1',\ldots,x_n'$ are linearly independent.