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?

  • $\begingroup$ 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
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    $\begingroup$ Possible duplicate of Preduals and $c_0$ $\endgroup$ – GEdgar Mar 19 at 12:10

$c_0$ 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.

  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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

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