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I have done an exercise that goes like this:

Consider the operator $\Phi: \ell_1\to(\ell_\infty)'$ that associates each $x=(x_j)_j\in\ell_1$ to $\Phi (x)\in (\ell_\infty)'$ given by $\Phi(x)(y)=\sum x_j y_j$, for all $y=(y_j)_j\in\ell_\infty$. Show that $\Phi$ is well defined, is linear and bounded. Construct an element $g\in (\ell_\infty)'\backslash \ \Phi(\ell_1)$. Is $\ell_\infty$ separable?

So... I went through all this exercise, but not the final. How can I conclude that $\ell_\infty$ is not separable from this? Or is this question not related to the exercise (what doesn't make sense)?

Martin Sleziak
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Filburt
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  • There is something not clear, first you say that $\phi(l_1)\subset l_\infty$ and then $\phi(x) \in (l_\infty)'$. Does the operator map elements into $l_\infty$ or into its dual? – Tommaso Seneci Jan 16 '18 at 13:23
  • Your function $\Phi$ should be with codomain as dual of l infinity –  Jan 16 '18 at 13:23
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    @TommasoSeneci Sorry for that error. Corrected. This is the usual dual map for $\ell_p$ spaces... – Filburt Jan 16 '18 at 13:30
  • Some context may help. Identifying $\ell_1' \cong \ell_{\infty},$ the first parts of the exercise shows that $\ell_1$ is non-reflexive ($\Phi$ is the cannonical embedding under appropriate identifications). That alone however does not imply $\ell_{\infty}$ is non-separable (e.g. $c_0$ is non-reflexive with separable dual $\ell_1$), so you need to use some other property. – ktoi Jan 16 '18 at 13:40
  • The exercise tells you that a certain mapping is not surjective. How should this imply something else? The zero mapping is also not surjective... some more context is missing. – daw Jan 16 '18 at 13:50

2 Answers2

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An answer for separability can be found here Why is $l^\infty$ not separable?. To make it clear: if $l_\infty$ is separable so is any of its subset (easily provable). So that consider the following set $$ G = \{ x\in l_\infty: x_n \in \{0,1\},\ \forall n\in \mathbb{N} \}. $$ Since we have a bijection $\iota: G \to 2^\mathbb{N}$ and the latter is uncountable, so is $G$. Consider $x\neq y\in G$, $\iota(x)\neq \iota(y)$, so that there is a coordinate $\tilde{n}\in\mathbb{N}$ such that $x_n = 1$ and $y_n=0$ (or viceversa). But then $\|x-y\|=\sup_n |x_n-y_n| \geq |x_{\tilde{n}}-y_{\tilde{n}}| = 1$. This is enough to prove that it cannot exists a dense countable subset. Indeed if there existed one, namely $R$, then for all $x\in G\subset l_\infty$ we could find $(r_n)\subset R$ such that $r_n\to x$. Choose $n=n_x\in\mathbb{N}$ such that $\|r_{n_x} -x\|<1/4$. This means that for $x\neq y$, $$ \|r_{n_x}-r_{n_y}\| \geq \|x-y\|-\|r_{n_x}-x\| - \|r_{n_y}-y\| \geq 1-1/4-1/4 = 1/2. $$ So that the cardinality of $r_{n_x} \in R$ equals the cardinality of $G$ which is uncountable. This contradiction shows that $l_\infty$ is not separable.

Tommaso Seneci
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  • Although the linked answer https://math.stackexchange.com/a/660422/263364 is very nice, your answer do not answer my question. I know how to prove that $\ell_\infty$ is separable. This is not exactly what we are dealing here... But thanks for the answer anyway. – Filburt Jan 16 '18 at 16:55
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It seems a bit of a stretch, but it's possible to show that $\ell_\infty$ is not separable in a way that's somewhat related to the rest of the exercise.

Recall the Banach-Alaoglu theorem:

If $X$ is a Banach space the closed unit ball of $X'$ is weak-* compact.

And recall this:

If $X$ is a separable Banach space then the closed unit ball of $X'$ is metrizable in the weak-* topology (and hence is weak-* sequentially compact).

Now it's easy to find a sequence $(e_n)$ in the closed unit ball of $\ell_1$ such that the sequence $(\Phi(e_n))$ has no weak-* convergent subsequence; hence $\ell_\infty$ is not separable.

This seems a little silly, since we really don't need $\Phi$ to show that the unit ball of $\ell_\infty'$ is not weak-* sequentially compact. But it does give some connection between the last bit of the exercise and the rest of it. (If this is an exercise for a class: If you've covered that result about weak-* metrizability then it seems likely that this is more or less the "intended" solution...)

David C. Ullrich
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  • For now I will give you my +1 and acceptance for the help, although this was not suppose to use any sort of weak topology yet... At the first moment I'd understood that question as "So... after all the 20 minutes we have passed together solving the firsts problems, is $\ell_infty$ separable?". But actually I think it is more like "Now forget that and show that $\ell_\infty$ is not separable." Go figure... – Filburt Jan 16 '18 at 17:03