Preduals and $c_0$

Question

I know that $c_0$ does not have a predual, but if we put an equivalent norm on $c_0$, can this space have a predual?

@Berci I think he the OP meant a vector space $X$ such that $X^*=c_0$. — BigbearZzz, Feb 25 '19 at 10:12
@TheoBendit Wouldn't the unit ball looks different if we use a different norm? — BigbearZzz, Feb 25 '19 at 11:03

GEdgar · Accepted Answer · 2023-12-01T17:06:37.963

7

No. In a separable dual space, every bounded closed convex set is the closed convex hull of its extreme points. (Krein-Milman property). But in $c_0$ with an equivalent norm, the original unit ball is still a bounded closed convex set with no extreme points.

Reference

Diestel, J.; Uhl, J. J. jun., Vector measures, Mathematical Surveys. No. 15. Providence, R.I.: American Mathematical Society (AMS). XIII, 322 p. $ 35.60 (1977). ZBL0369.46039.

edited Dec 01 '23 at 17:06

answered Feb 25 '19 at 11:32

GEdgar

111,679

Could you please explain why separability of $c_0$ is required here? Being a normed (hence Hausdorff) space isn't enough for Krein-Milman to hold? – P. Camilleri Dec 01 '23 at 17:00
The theorem says that a separable dual space has the Krein-Milman property. The example $l^\infty$ of a dual space without the Krein-Milman property shows that separability is needed in the result. – GEdgar Dec 01 '23 at 17:04

score 3 · Answer 2 · edited Feb 25 '19 at 15:51

No. If the norm is equivalent any bounded sequence in the original space would still be bounded. Now use Banach-Alaoglu. Take the sequence $s_N = \chi_{[0,N]}$. That sequence must have a weak-$\ast$ accumulation point. But the predual $E$ is an isometric subset of $c_0^\ast = \ell^1$ and that sequence satisfies that $\langle \psi, s_N \rangle \to \langle \psi, \chi_{\mathbb{N}} \rangle$ for every $\psi \in \ell_1$ and thus for every $\psi \in E$. But $\chi_{\mathbb{N}} \not\in c_0$.

Preduals and $c_0$

2 Answers2

Linked