In functional analysis, a branch of mathematics, the Goldstine theorem, named after Herman Goldstine, is stated as follows:
- Goldstine theorem. Let be a Banach space, then the image of the closed unit ball under the canonical embedding into the closed unit ball of the bidual space is a weak*-dense subset.
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 space and its bi-dual space [[Lp space]
For all and there exists an such that for all
Proof of Lemma
By the surjectivity of
it is possible to find with for
Every element of satisfies and so it suffices to show that the intersection is nonempty.
Assume for contradiction that it is empty. Then and by the Hahn–Banach theorem there exists a linear form such that and Then  and therefore
which is a contradiction.
Proof of Theorem
Fix and Examine the set
Let be the embedding defined by where is the evaluation at map. Sets of the form form a base for the weak* topology, so density follows once it is shown for all such The lemma above says that for any there exists a such that and in particular Since we have We can scale to get The goal is to show that for a sufficiently small we have
Directly checking, we have
Note that we can choose sufficiently large so that for  Note as well that If we choose so that then we have that
Hence we get as desired.