Let $\ell^\infty$ be the Banach space of bounded sequences with the usual norm and let $c,c_0$ be the subspaces of sequences that are convergent, resp. convergent to zero. Show that:

  • The linear functional $\ell_0\colon c\rightarrow \mathbb{C}$ defined for $x = (x_n) \in c$ by $$ \ell_0(x) = \lim_{n\rightarrow \infty} x_n$$ extends to a continuous functional on $\ell^\infty$
  • if $L$ denotes the set of all continuous extensions of the functional $\ell_0$ from (1), then a sequence $x = (x_n) \in \ell^\infty$ belongs to $c_0$ iff $$\ell(x) = 0 \;\; \forall \ell \in L$$
  • Describe $c$ in a similar way My try:

(1): This follows by Banach limits.

(2): $(\Rightarrow)$ follows by extension

$(\Leftarrow)$ Here Im a bit unsure, assume $x \not \in c_0$ if $x \in c$ we get an contradiction. But if $x\not \in c$ what happens then, can we use a subsequence? since we have bounded functionals? can we use $\ell x = \lim_{k \rightarrow \infty} x_{n_k}$ or something like that, would $\ell \in L$?

(3): same as two I suppose, can we use subseqeunces?

Please correct what I'm missed

  • $\begingroup$ (2) is a direct application of the Hahn-Banach theorem: As $c_0$ is a closed subspace of $l^\infty$, for $x\in l^\infty \setminus c_0$ exists an $l\in L$ with $l(x)\neq 0$. $\endgroup$ – Vobo Dec 29 '12 at 12:08
  • $\begingroup$ thanks, but I don't see the importance of $c_0$ being closed. can you please expand? $\endgroup$ – Johan Dec 29 '12 at 13:53
  • $\begingroup$ Hahn-Banach can only seperate a point and a closed set. $\endgroup$ – Vobo Dec 29 '12 at 18:01
  • $\begingroup$ Yes you are correct! Would it work for $c$ as well? is it also closed? $\endgroup$ – Johan Dec 29 '12 at 18:11
  • $\begingroup$ It is indeed. But you should think why, it is the same reason as for $c_0$. $\endgroup$ – Vobo Dec 29 '12 at 19:08

In the general case you have the equivalence: $x\in c$ if and only if there exist a constant $a\in\mathbb{R}$ such that $l(x)=a$ for all $l\in L$.

One is easy, so suppose that $x\in \ell^\infty\setminus c$. Consider the space $$c+x\mathbb{R}=\{z\in\ell^\infty:\ z=y+\lambda x,\ y\in c,\ \lambda\in\mathbb{R} \}$$

Let $b\neq a$ and define a bounded linear functional $\tilde{l}:c+x\mathbb{R}\rightarrow\mathbb{R}$ by $$\tilde{l}(z)=l_0(y)+\lambda b$$

Note that $\tilde{l}$ is a bounded linear functional defined in a subspace of $\ell^\infty$ and such that $\tilde{l}(x)=b\neq a$. From here you can conclude.

  • $\begingroup$ thanks a lot! But this $\tilde{l}$ is it defined for $\ell^\infty$? Or is $c + x\mathbb{R} = \ell^\infty$? $\endgroup$ – Johan Dec 29 '12 at 14:53
  • $\begingroup$ No, but now you can extend it by using Hahn-Banach and then you conclude. I think that there is a more straightforward approach. $\endgroup$ – Tomás Dec 29 '12 at 15:02
  • $\begingroup$ So we can extend two times? $\endgroup$ – Johan Dec 29 '12 at 15:04
  • $\begingroup$ There is no problem in extend it again. But as I say maybe there is a more straightforward argument. $\endgroup$ – Tomás Dec 29 '12 at 15:14

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