2

Just like in these two questions: 1, 2, I was also struggling to understand bases in $\ell^\infty$ in a constructive way, which I understood was not possible. However, in order to get a better understanding of infinite dimensional vector spaces and Hamel bases in general, I have a couple of perhaps simple questions:

Q1: Since any vector space has a Hamel basis, we still have a basis $B$ for $\ell^\infty$, even if we cannot write it down explicitly. This however means that for any vector $v\in\ell^\infty$, $v$ can be written uniquely as a linear combination of (finitely) many vectors $B \supset B_0=\{b_1,\dots,b_n\}$, which we would thus be able to construct given a specific $v_0\in \ell^\infty$ (right?) (EDIT:) not be able to construct, since we do not know $B$. So, knowing such a finite linear combination exist, regardless whether $B_0\subset B$ or not, are we at all able to construct a finite linear combination in general when not all but finitely many terms are $0$, and all of them are distinct, e.g. could we construct a finite $B_0$ be for say $\ell^\infty\ni v_0=(1/n)_{n=1}^\infty$?

Basically, what felt weird was:

Q1.2: Could we have a sequence in $\ell^\infty$ that thus can be written as a (finite) linear combination, but we may never be able to construct even one?

EDIT: But technically just the vector $v$ itself is a finite linear combination of vectors in $\ell^\infty$, so this question became rather trivial. I guess I was somehow confused by the counter intuitive notion of not being able to construct the basis. I think the answer given (together with this edit) suffice.

Q2: Is (more generally) a reason that $\{e_i\}$ does not form a (Hamel) basis for $\ell^\infty$, as mentioned in 1, essentially (but rather informally) that any assumed countable basis $\{b_j\}$ for any Banach space $X$ would have the property that any $B_n:=\text{span}\{b_1,\dots,b_n\}$ would of course be a proper subspace and thus a closed nowhere dense subset of $X$, and thus $X=\cup_n B_n$ would contradict Baire's (since $\ell^\infty$ is Banach)?

  • Q1: We can't construct $B_0$, because we can't construct $B$. Can you express $(1,0,0) \in \mathbb{R}^3$ in a basis $B$ if I don't tell you $B$ itself? Of course, you can come up with your basis and say that it's $1(1,0,0)$, but nothing will guarentee that $(1,0,0) \in B$. – Botond Feb 18 '20 at 12:02
  • Q2: A simple reason would be the $(1,1,...)$ vector, because you can't express it with finite linear combinations of $e_i$. – Botond Feb 18 '20 at 12:05
  • @Botond No, that's clear now, I guess what I thought about was rather coming up with such a linear combination at all, regardless if $B_0\subset B$ or not. – Christopher.L Feb 18 '20 at 12:38
  • @Botond :Well, this became embarrassingly trivial, apart from Q2 which I understood was a somewhat correct argument. I'll let it be as it is, or do you suggest I rather delete it? – Christopher.L Feb 18 '20 at 13:17
  • I would not delete it, because It can be useful for people in the future, and I think it's rude to delete questions which has already been answered. – Botond Feb 18 '20 at 13:19
  • @Botond The second part is certainly true in general (and I'm not a rude person), I just thought this perhaps turned out to be a little bit too trivial, but I hope the first part of your statement is just as true, so that it was not in vain after all. Since my edits also hopefully make things more clear, I'll leave it as it is. (Thank you!) – Christopher.L Feb 18 '20 at 13:24

1 Answers1

1

Q1: Of course we can't say what $B_0$ is without knowing what $B$ is! Given just that $B$ is some basis, $B_0$ could be any independent set that has $v_0$ in its span.

Q2: It's simply obvious that $(e_j)$ is not a Hamel basis for $\ell_\infty$; for example if $x=(1,1,1,\dots)$ then it's clear that $x$ is not a finite linear combination of the $(e_j)$.

In fact it's trivial to show directly that $(e_j)$ is not even a Schauder basis. Say $x = (1,1,1,\dots)$. For any $n$ and scalars $a_1,\dots,a_n$ we have $$||x-\sum_{j=1}^n a_je_j||\ge1.$$So $x\ne\sum_{j=1}^\infty a_j e_j$.

  • No, the argument for Q2 is correct, and $(e_j)$ is not a Hamel basis for $\ell^2$. – MaoWao Feb 18 '20 at 12:09
  • @MaoWao Thanks - it's clear what I was confused about.... – David C. Ullrich Feb 18 '20 at 12:10
  • On Q1: I guess my intuition (which was clearly wrong) made me believe that even though we could not explicitly construct $B$, we could construct a set $B_0$ for each $v$, but I guess then $\cup_{v\in X} B_v$ would be a basis for the whole space, which would contradict my first statement. It just felt strange enough we couldn't construct $B$, but that's fine, but it feels so much more strange that we cannot even find a finite linear comb $v=\sum a_ib_i$, knowing that one exists. – Christopher.L Feb 18 '20 at 12:26
  • On Q1: And of course that seems trivial now that I think about it (when $B_0\subset B$). I felt that knowing $v$ could be written as a finite linear combination, we could somehow construct one, actually regardless of whether $B_0\subset B$ or not. But it felt so counter intuitive to me that such a finite linear combination always existed at all. – Christopher.L Feb 18 '20 at 12:35
  • On Q2: Yes, that is trivial of course, I guess I was really after the more general argument (contrary to my actual question), which I thus assume is more or less correct? – Christopher.L Feb 18 '20 at 12:53
  • @Christopher.L I don't know what "more general argument" you're talking about – David C. Ullrich Feb 18 '20 at 12:55
  • The one given by me in Q2: Any countable basis ${b_j}$ for any Banach space $X$ would have the property that any $B_n:=\text{span}{b_1,…,b_n}$ would of course be a proper subspace and thus a closed nowhere dense subset of $X$, and thus $X=\cup_n B_n$ would contradict Baire's. I have also edited my question somewhat to better reflect my confusions. – Christopher.L Feb 18 '20 at 12:59