I am reading Bollobás' Linear Analysis. Chapter 8. Theorem 6., as the title says:
$B(X^{**})$ is the $w^*$-closure of $B(X)$ in $X^{**}$
The proof starts by saying that i) $B(X^{**}$) is $w^*$-closed, and ii) $B \subset B(X^{**})$ where $B$ is the $w^*$-closure of $B(X)$ in $X^{**}$. Then, suppose that there is $x_0^{**} \in B(X^{**}) \setminus B$. By the separation theorem, there is a bounded linear functional $x_0^{***}$ on $X^{**}$ such that $$\langle x_0^{***}, b \rangle \leq 1 < \langle x_0^{***}, x_0^{**} \rangle$$ for every $b \in B$. By considering $x_0^{*}$, the restriction of $x_0^{***}$ onto the subspace $X$ of $X^{**}$, we see that $\langle x_0^{*}, x \rangle \leq 1$ for every $x \in B(X)$, thus $\| x_0^{*} \| \leq 1$. This far everything is clear. To finish up the proof, the book argues as follows: $$1 \geq \langle x_0^{*}, x_0^{**} \rangle = \langle x_0^{***}, x_0^{**} \rangle > 1$$ claiming contradiction.
I don't really understand why $\langle x_0^{*}, x_0^{**} \rangle = \langle x_0^{***}, x_0^{**} \rangle$. I see that we can view $x_0^*$ as an element of $X^{***}$, but what guarantees, that this is the same element as $x_0^{***}$?
Note that as $B$ is a closed convex subset of $(X^{**}, w^{*})$ and as $x_{0}^{**}\not\in B$, it follows from a separation argument that there exists some $x_{0}^{***}$ in the dual of $(X^{**}, w^{*})$ which strictly separates $B$ and $x_{0}^{**}$. But as continuous linear functionals on $(X^{**}, w^{*})$ are elements of the form $x^{***}\in X^{***}$ where there is some $x^{*}\in X^{*}$ such that $\langle x^{***}, x^{**} \rangle = \langle x^{**}, x^{*} \rangle$ for all $x^{**} \in X^{**}$, it follows that there is some $x_{0}^{*}\in X^{*}$ such that $\langle x_{0}^{***}, x^{**} \rangle = \langle x^{**}, x_{0}^{*} \rangle$ for all $x^{**} \in X^{**}$. Consequently, $x_{0}^{*}$ is the desired element.
It is also worth mentioning that this theorem is known as Goldstine's theorem.