# Goldstine theorem

In functional analysis, a branch of mathematics, the Goldstine theorem, named after Herman Goldstine, asserts that the image of the closed unit ball $B_X$ of a Banach space $X$ under the canonical imbedding into the closed unit ball $B_{X^{**}}$ of the bidual space $X^{**}$ is weakly*-dense.

The conclusion of the theorem is not true for the norm topology, which can be seen by considering the Banach space of real sequences that converge to zero c0, and its bi-dual space .

## Proof

Given an $x^{**} \in B_{X^{**}}$, a tuple $(\phi_1, \dots, \phi_n)$ of linearly independent elements of $X^*$ and a $\delta>0$ we shall find an $x \in (1+\delta) B_{X}$ such that $\phi_i(x)=x^{**}(\phi_i)$ for every $i=1,\dots,n$.

If the requirement $\|x\| \leq 1+\delta$ is dropped, the existence of such an $x$ follows from the surjectivity of

$\Phi : X \to \mathbb{C}^{n}, x \mapsto (\phi_1(x), \dots, \phi_n(x)).$

Let now $Y := \bigcap_i \ker \phi_i = \ker \Phi$. Every element of $(x+Y) \cap (1+\delta) B_{X}$ has the required property, so that it suffices to show that the latter set is not empty.

Assume that it is empty. Then $\mathrm{dist}(x,Y) \geq 1+\delta$ and by the Hahn-Banach theorem there exists a linear form $\phi \in X^*$ such that $\phi|_Y = 0$, $\phi(x) \geq 1+\delta$ and $\|\phi\|_{X^*}=1$. Then $\phi \in \mathrm{span}(\phi_1, \dots, \phi_n)$ and therefore

$1+\delta \leq \phi(x) = x^{**}(\phi) \leq \|\phi\|_{X^*} \|x^{**}\|_{X^{**}} \leq 1,$