Goldstine theorem

Remarks

The weak* closure $C$ of $J(B_X)$ is convex and lies in $B_{X^{**}}$, since $B_{X^{**}}$ is weak*-closed. If some $\xi \in B_{X^{**}}$ were not in $C$, the Hahn–Banach separation theorem in $X^{**}$ with the weak* topology would give $f \in X^*$ with $\operatorname{Re} \xi(f) > \sup_{x \in B_X} \operatorname{Re} f(x) = \norm{f}$. This contradicts $\norm{\xi} \le 1$.

For $X = c_0$, with $X^{**} = \ell^\infty$, the theorem says that every sequence $y$ with $\sup_n \abs{y_n} \le 1$ is a weak* limit of elements of the unit ball of $c_0$. Indeed the truncations $(y_1, \ldots, y_n, 0, 0, \ldots)$ converge to $y$ against every element of $\ell^1$.

Going deeper

For a student

Combined with the Banach–Alaoglu theorem, the theorem shows that a Banach space is reflexive exactly when its closed unit ball is weakly compact. It is also the key step in the proof of the Milman–Pettis theorem. In general the approximation needs nets, not sequences; for separable $X^*$ sequences suffice.