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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

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    No, it is not harder. But after you asked if $c$ is closed, I assumed that you are not aware how to prove that $c_0$ is closed. As this is quite basic analysis, you should know how to do it.2012-12-30

1 Answers 1

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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.

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    There is no problem in extend it again. But as I say maybe there is a more straightforward argument.2012-12-29