In functional analysis, a branch of mathematics, the Goldstine theorem, named after Herman Goldstine, is stated as follows: 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 c_0, and its bi-dual space Lp space

Proof

Lemma

For all and \delta > 0, there exists an such that for all

Proof of lemma

By the surjectivity of it is possible to find x \in X with for Now let 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 x map. Sets of the form U form a base for the weak* topology, so density follows once it is shown for all such U. The lemma above says that for any \delta > 0 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 \delta > 0, we have Directly checking, one has Note that one can choose M sufficiently large so that for Note as well that If one chooses \delta so that then Hence one gets as desired.