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I have a question that has been bugging me a lot lately.

In all $\ell^p$ spaces with $1<p\leq\infty$ we have that the canonical basis $e_i\in\ell^p$ as a sequence is bounded weakly-$\star$ and converges to zero. This happens in particular because of the Banach-Alaoglu theorem which tells us that this sequence has limit points in those spaces.

In $\ell^1$ the situation is different since there is no weak-$\star$ topology and thus the closed unit ball is not compact and by proxy we cannot expect any convergence, and in fact there is none.

The problem here is that $\ell^1$ can be embedded in its double dual, a.k.a. the infamous dual of $\ell^\infty$. There there is the weak-$\star$ topology and the Banach-Alaoglu theorem holds. Now the canonical basis $e_i\in\ell^1$ forms a sequence of bounded elements in $(\ell^\infty)^*$ which has to admit limit points.

The problem arises here. Given any element $a\in\ell^\infty$ we have $$ \lim_n\langle a,e_n\rangle=\lim_n a_n $$ which by no means is convergent, but since it is bounded it would admit a convergent subsequence.

It is true that for any sequence I can choose a convergent subsequence but doesn't doing so for all possible sequences usually end up in having an empty subsequence? Doesn't using the "take the subsequence" infinite times lead to an empty set?

Am I to believe that this process of choosing a subsequence for any non-convergent sequence is the construction of an ultrafilter and therefore defines the aforementioned limit as a Banach limit of sorts?

The reason this process doesn't sit well with me is because we are defining the 'subsequence choosing' without explicitly taking a subsequence of $(e_i)_i$. What would be more pleasant would be to choose the exact subsequence of $(e_i)_i$ that when tested against any bounded sequence would converge, which is absurd.

For example we know that $(e_i)_i$ when tested against convergent sequences produces a convergent sequence, but when we consider the space of all the sequences that have two limit points there is no way to take a subsequence that will produce a convergent sequence for all of these sequences.

Another approach would be to fix a surjective maps $j:\mathbb N\to\mathbb N$ such that and look at the space of sequences $a\in\ell^\infty$ such that $$ a_{j}\in \mathbb c $$ where $\mathbb c$ the space of convergent sequences. Iterating this procedure would produce a sequence of subsequences of $(e_i)$ that corresponds to a sequence of surjective maps $(j^k)_{k=1}^n$ and so converges in the weak-$\star$ topology generated by the space of bounded sequences that admits convergent subsequences on those fixed chosen $j^k$. Even though this is defined on all subsets of $\ell^\infty$ that have a convergent sequence on that particular subset of $\mathbb N$, saying that that sequences of subsequences doesn't have empty limit seems wishful thinking.

So is this my question: Since the sequence $e_i\in\ell^1$ must have limit points in $(\ell^\infty)^*$, what are those limit points?

The problem here is that I kinda know that nothing explicit is known about $(\ell^\infty)^*$ aside for $\ell^1$, but I at least imagined something sensible for this sequence.

Edit: As mentioned in the comments by David C. Ullrich everything can be fixed by considering subnets, since subnets of sequences are not always sequences. It is probably not explicitly doable because of the axiom of choice, but still it can be seen why it works now.

Lolman
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    First, $l^1$ (the space of summable sequences) can be identified with the dual of $c_0$ , the space of sequences converging to zero. In this setting, $e_i \rightharpoonup^* 0$. Second, $l^\infty$ is not separable, so Banach-Alaoglu cannot be applied in $(l^\infty)^*$. – daw Jul 04 '22 at 18:58
  • @Lolman: Double dual of $l^1$ is $l^{\infty}$ ? – Paresseux Nguyen Jul 04 '22 at 19:22
  • @daw: Banach-Alaoglu cannot be applied to non-seperable spaces? – Paresseux Nguyen Jul 04 '22 at 19:23
  • @daw you can chek Brezis for a proof that doesn't require separability. If not wikipedia provides a proof for topological vector spaces. – Lolman Jul 04 '22 at 19:25
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    Where are you getting all those subsequences from? Seems to me in many places you need to replace"subsequence" with "subnet". may or may not make it seem more possible... – David C. Ullrich Jul 04 '22 at 19:25
  • @ParesseuxNguyen I think I wrote dual of $\ell^\infty$. – Lolman Jul 04 '22 at 19:26
  • @DavidC.Ullrich since $\ell^1$ is the set of summable sequences, its basis will be but a countable net, therefore a sequence. Subnets of sequences are just subsequences. Moreover in $\ell^1$ we have that on bounded sets the weak topology and the strong topology coincide so nets would play no role. – Lolman Jul 04 '22 at 19:31
  • @DavidC.Ullrich it depends on where you are reading but most of the time I am taking a subsequence of the tested sequence $a_n$, something that we do not want. In the last approach I take subsequences of the sequence that we want, that is the canonical basis. – Lolman Jul 04 '22 at 19:34
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    It's not literally true that a subnet of a sequence is a subsequence. possibly you meant is equivalent to in some sense? (what sense???) – David C. Ullrich Jul 04 '22 at 19:49
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    (Ah, proof they're not the same: $(e_n)$ has a subnet convergent in the weak* topology of $(\ell_\infty)^*$, but no such subsequence.) – David C. Ullrich Jul 04 '22 at 19:55
  • @DavidC.Ullrich OMG. I am screaming. As simple as it may seem I never knew that I could consider subnets of a sequence! Working mostly with sequence spaces never let me see this simple consideration. THANKS! – Lolman Jul 04 '22 at 20:16

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The problem is a compact set is not necessarily a sequentially compact set. Compact space which is not sequentially compact
Secondly, if there is a such desired subsequence $(e_{n_k})$ of $(e_n)$, you can easily choose an $a \in l^{\infty}$ such that $\langle e_{n_k}, a\rangle$ does not converge. In particular, this is the simplest way to show that the space $(l^{\infty})^*$ with the weak-$*$ topology is not a sequentially Hausdorf space.

Paresseux Nguyen
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  • Your example was the main point of my argument. I see that apparently I failed to convey it. DavidC.Ulrich mentioned in the comments "sequences can have nontrivial subnets" and that made everything clear. This is also the reason I recently believed that the relation in your link should have been the other way around. If you incorporate the clarifying statement about subnets in your answer, as trivial as it may be for you, I will accept it. – Lolman Jul 04 '22 at 20:23