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The following question is an exercise and so I'm just looking for advices and not for answers if it's possible.

I have the following sets in $l^\infty$ $$c_0 := \{x_n \in l^\infty: \lim x_n = 0\} \subseteq c := \{x_n \in l^\infty: \exists \lim x_n\}.$$ And I intend to prove that they are not isometrically isomorphic. I suppose that the problem is that for every $(x_n) \in c_0$ there exists a finite and non-empty set $\{x_{n_i}\}$ such that $|x_{n_i}| = \|x_n\|$ but this is false in $c$, but I don't know how to continue with this idea. Can you help me?

An interesting fact about this two spaces is that although they aren't linearly isometric they are linearly homeomorphic given by $T:c \to c_0, T(x_n) = (\lim x_n, x_n-\lim x_n)$. This is too strange to me.

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  • $\begingroup$ In principle, I don't think that your idea will work, as you don't know what an isomorphism might do to your sequences. What you might want to look at is the convexity of the unit ball ( more concretely, the existence or not of extreme points). $\endgroup$ Commented Apr 26, 2013 at 5:04
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    $\begingroup$ This isn't an exact duplicate - the OP is wanting help with a specific approach to this problem, not for a full answer! $\endgroup$ Commented Apr 26, 2013 at 10:03
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    $\begingroup$ Can somebody link again the post above-mentioned? $\endgroup$ Commented Apr 27, 2013 at 1:48
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    $\begingroup$ The question can be found here. The edit history contains all old versions of a question, which is where I found the link. $\endgroup$ Commented Apr 27, 2013 at 15:56
  • $\begingroup$ @user1729: I don't quite get your point. This is a problem having a unique standard answer hinted at by Martin Argerami in a comment and by Norbert in the answer here and David Mitra in the linked answer. It is the way to make the idea in the post precise and usable. $\endgroup$ Commented Apr 27, 2013 at 16:25

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Hints:

  1. Prove that unit ball of $c$ have a lot of extreme points. In fact there are $\mathfrak{c}$ extreme points but this is not important for the solution.

  2. Prove that unit ball of $c_0$ have no extreme points.

  3. Prove that if $x$ is an extereme point of unit ball of some normed space $X$, and $i:X\to Y$ is a surjective isometry, then $i(x)$ is an extreme point of unit ball of $Y$.

  4. The rest is clear.

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  • $\begingroup$ Thanks for your help, Norbert. $\endgroup$ Commented Apr 27, 2013 at 17:51
  • $\begingroup$ @Norbert Sir, for proving 3) (in the Hints), don't we require onto isometry or is it true that a general linear isometry maps extreme points to extreme points? $\endgroup$ Commented Nov 14, 2021 at 5:00
  • $\begingroup$ @akansha, I think I forgot to mention surjectivity. Now fixed $\endgroup$ Commented Nov 14, 2021 at 6:17
  • $\begingroup$ @Norbert Thank you, sir. $\endgroup$ Commented Nov 15, 2021 at 8:59

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