Goldstine theorem


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|, and its bi-dual space Lp space|.

Proof

Lemma

For all, and, there exists an such that for all.

Proof of Lemma

By the surjectivity of
we can find 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 map. Sets of the form form a base for the weak* topology, so density follows if we can show for all such. The lemma above says that for all there exists an such that. 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.