3
$\begingroup$

The (continuous) dual of a normed vector space is always a Banach space, but the converse is not true. That is, not all Banach spaces are isomorphic to the dual space of some normed vector space. For instance $L^1$ is not isomorphic to any dual space.

My question is, are the sequence spaces $c_0$ and $c$ isomorphic to the duals of any spaces?

$\endgroup$
  • $\begingroup$ Without choice it's consistent that $\ell^1$ is the dual of $\ell^\infty$. $\endgroup$ – Asaf Karagila Mar 19 at 8:52
  • $\begingroup$ @AsafKaragila Yeah, I’ve heard that before. But what is the situation with choice? $\endgroup$ – Keshav Srinivasan Mar 19 at 8:53
  • 1
    $\begingroup$ Possible duplicate of Preduals and $c_0$ $\endgroup$ – GEdgar Mar 19 at 12:10
1
$\begingroup$

$c_0$ is not the dual of some normed vector space.

sketch of proof:

This can be proven using the Krein-Milman theorem.

If $c_0$ was the dual of a normed vector space, than its unit ball would be weakly-* compact.

However, it can be shown that the unit ball of $c_0$ has no extremal points, therefore by Krein-Milman it is not weakly-* compact.

edit: for $c$ the situation is more complicated than I initially thought.

As uniquesolution pointed out in the comments, its unit ball has many extreme points, and it is not clear to me if Krein-Milman can be used to show that the unit ball of $c$ cannot be weakly-* compact.

$\endgroup$
  • $\begingroup$ The case of $c$ needs a little more attention, because there are many more extreme points than those you mentioned: in fact, every sequence whose entries have absolute value $1$ which is eventually constant (this means there exists $N$ such that if $n>N$ then $x_n=1$, for example) is an extreme point. $\endgroup$ – uniquesolution Mar 19 at 10:06
  • $\begingroup$ @uniquesolution thank you for that remark. I edited to retract my claim for $c$ until I (or someone else) fixes the argument or provides a new argument. $\endgroup$ – supinf Mar 19 at 10:25
  • $\begingroup$ See the proposed duplicate question. Note $c_0$ is a closed subspace of $c$. If $c$ were a dual, then every closed convex subset of $c$ would have extreme points, so in particular the unit ball of $c_0$ would have extreme points. $\endgroup$ – GEdgar Mar 19 at 12:15
  • 1
    $\begingroup$ @GEdgar Wouldn't one need that the unit ball of $c_0$ is weak$^*$-closed in $c$ to conclude that it is weak$^*$-compact and thus has extreme points? I don't see why norm-closedness should be enough. $\endgroup$ – Jochen Mar 19 at 13:32
  • 1
    $\begingroup$ Krein-Milman property, holds in all separable dual spaces: a bounded norm-closed convex set is the norm-closed convex hull of its extreme points. Reference: Diestel-Uhl, Vector Measures ... math.stackexchange.com/a/3125931/442 ... $\endgroup$ – GEdgar Mar 19 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.