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
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 [1] 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,[2] 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 [3] Note as well that If we choose so that then we have that
Hence we get as desired.
See also[]
Banach–Alaoglu theorem – The closed unit ball in the dual of a normed vector space is compact in the weak* topology
^Rudin, Walter. Functional Analysis (Second ed.). Equation (3) and the remark after. p. 69.{{cite book}}: CS1 maint: location (link)
^Folland, Gerald. Real Analysis: Modern Techniques and Their Applications (Second ed.). Proposition 5.2. pp. 153–154.{{cite book}}: CS1 maint: location (link)
Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN978-0-07-054236-5. OCLC21163277.