6

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.

Diego Silvera
  • 1,717
  • 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). – Martin Argerami Apr 26 '13 at 05:04
  • 1
    This isn't an exact duplicate - the OP is wanting help with a specific approach to this problem, not for a full answer! – user1729 Apr 26 '13 at 10:03
  • 1
    Can somebody link again the post above-mentioned? – Diego Silvera Apr 27 '13 at 01:48
  • 1
    The question can be found here. The edit history contains all old versions of a question, which is where I found the link. – user1729 Apr 27 '13 at 15:56
  • @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. – Martin Apr 27 '13 at 16:25
  • @Martin: My point was that an answer to the question should have been an expansion of Martin Argerami's comment: Your idea does not work for this reason, rather this is what you should do. Closing this question will only provide half of what is needed. – user1729 Apr 27 '13 at 18:26
  • This can also be found as Exercise 3.75 in Functional Analysis and Infinite-Dimensional Geometry or Exercises 3.132 in Banach Space Theory: The Basis for Linear and Nonlinear Analysis. – Martin Sleziak May 13 '18 at 12:15

1 Answers1

5

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.

Norbert
  • 56,803