1
$\begingroup$

Show that a subspace of $c_{0}$ with the norm induced by $c_{0}$, can’t be isomorphic to $l_1$.

Any idea or hint? I think have to show that the given subspace has a certain property that $l_1$ no has (or the reverse) but I can’t understand which property... thank you!

$\endgroup$
1
  • $\begingroup$ @mechanodroid had a great answer. I just feel like adding that since $c_0$ is subprojective, if $c_0$ contained a copy of $\ell_1$ then $\ell_1$ would contain a copy of $c_0$. However $\ell_1$ has the Schur property whereas $c_0$ does not. $\endgroup$
    – Ben W
    Jan 2, 2019 at 19:22

1 Answer 1

13
$\begingroup$

The dual space of $\ell^1$ is isometrically isomorphic to $\ell^\infty,$ which is nonseparable.

However, the dual of $c_0$ is isometrically isomorphic to $\ell^1,$ which is separable. Hence the dual of any subspace of $c_0$ is separable as well.

This argument actually shows that there is not a linear homeomorphism between $\ell^1$ and a subspace of $c_0$, let alone an isometric isomorphism.

$\endgroup$

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