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For an exercise, I need to show that the canonical basis is not a valid basis in $l^{\infty}$. Concretely, the exercise states :

Consider the Banach space $l^{\infty}$ of sequences $x = \{x_n\}_{n = 1}^{\infty} \subset \mathbb{C}$ with the usual norm $\left\| x \right\|_{\infty} = \sup_{n \in \mathbb{N}} |x_n| < \infty$. Then, we define the sequence $\{e_m\}_{m \in \mathbb{N}} \in l^{\infty}$ as

$e_m := \{0, ..., 1, 0, ...\}$ , $m \in \mathbb{N}$

i.e. with 1 at the $m$-th position and zeros everywhere else. Show that the system $\{e_m\}_{m \in \mathbb{N}}$ is not a basis for $l^{\infty}$.

My thoughts so far, please correct me if I am wrong!

  1. We need to show that every $x \in l^{\infty}$ can be written as $x = \sum_{n = 1}^{\infty} c_n(x) e_n$
  2. For the dense subset of all those sequences which have only finitely many non-zero entries, $l_c^{\infty} \subset l^{\infty}$, we have that $c_n(x) = x_n$. Then, the expression $x = \sum_{n = 1}^{\infty} c_n(x) e_n$ is unique.
  3. We now need to show that for above choice of $c_n$ and for every $x \in l^{\infty}$ we have that $\lim_{N \rightarrow \infty} \left\| x - \sum_{n = 1}^{N} c_n(x) e_n \right\|_{l^{\infty}} = 0$
  4. With the chosen $c_n(x) = x_n$ we have that $x - \sum_{n = 1}^{N} c_n(x) e_n = \{0, ..., 0, x_{N+1}, x_{N+2}, ...\}$
  5. With $x \in l^{\infty}$ and $\left\| x \right\|_{\infty} = \sup_{n \in \mathbb{N}} |x_n| < \infty$, we have that $\left\| x - \sum_{n = 1}^{N} c_n(x) e_n \right\|_{l^{\infty}} = \sup_{n \in \mathbb{N}} \sum_{n = N + 1}^{\infty} |x_n|$
  6. This however is not a Cauchy sequence, hence there is no convergence on this side s.t. $\lim_{N \rightarrow \infty} \left\| x - \sum_{n = 1}^{N} c_n(x) e_n \right\|_{l^{\infty}} = 0$ does not hold.

I am not sure whether my conclusion in 6) is right and suffices for this proof. However, I suppose that you can prove this in a much more compact way (and if so, how?).

Help is much appreciated :-)

Edit :

The easiest way to solve the problem is to show that $l^{\infty}$ is non-separable since the existence of a Schauder basis as defined in above problem implies that the underlying (sequence) space is separable. I refer to this post to show the non-separability : Why is $L^{\infty}$ not separable?

Thanks @AlvinL and the others!

aladin
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  • Hamel basis or Schauder basis? – AlvinL Jun 12 '22 at 13:45
  • Schauder-basis :-) – aladin Jun 12 '22 at 13:47
  • Existence of Schauder basis implies separable, whereas $\ell _\infty$ is...? – AlvinL Jun 12 '22 at 13:49
  • Ah, okay I think we have not yet discussed separability in this context. Do you have a reference to a proof for $l_{\infty}$ being non-separable? – aladin Jun 12 '22 at 13:56
  • $c_n(x) = x_n$ stays the same throughout my proof, i leave it to be $x_n$ :-) – aladin Jun 12 '22 at 14:07
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    Your assertion 2. (that the set points with finite support is dense) is wrong. – GEdgar Jun 12 '22 at 14:13
  • It looks like you take any $x\in \ell ^\infty$ and then $c_n(x)$ is the $n$-th coordinate of $x$. Why do you need to mention $\ell _c^\infty$ for that? I don't think that's dense, either. – AlvinL Jun 12 '22 at 14:14
  • I followed the explanation of a similar problem for the $l^1$ space and deemed it not necessary but maybe useful. But you are right, it does not matter in this case. – aladin Jun 12 '22 at 14:16
  • You shouldn't write a $0$ at the end of $e_m := {0, ..., 1, 0, ..., 0}$. It's an infinite sequence, there's not a "last" entry. – jjagmath Jun 12 '22 at 14:18
  • Alright, edited $e_m$. Okay, i understand the concerns. So what do you guys recommend? Is there a much much simpler way? As I said, I followed mostly a solution to a similar problem but was not sure about it either, hence my post! :-) – aladin Jun 12 '22 at 14:27
  • Well, try the constant sequence $x_n = 1$, for example. Do you have norm-convergence? – AlvinL Jun 12 '22 at 14:38
  • In $(5)$ there should be $\sup_{n\ge N+1}|x_n|$ on RHS. – Ryszard Szwarc Jun 12 '22 at 15:34
  • I am confused now. I also see the mistake in my argumentation in 2.

    However, I cannot seem to find a reasonable conclusion here. Is there a better way to prove the statement, i.e. with non-separability? Can someone show or give a reference?

     – aladin Jun 13 '22 at 11:05
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    @aladin I suggested a specific example. Convince yourself the constant sequence can't be represented by the $e_n$ in the required manner. – AlvinL Jun 13 '22 at 11:48
  • @AlvinL Thanks for your comments, I have overlooked them somehow. Thank you very much for the hint to non-separability in your comment above. I think I can prove the statement easily now using non-separability.

    However, for the constant sequence, I do not see why it wouldnt converge in the norm since $\lim_{N \rightarrow \infty} \left| x - \sum_{n = 1}^N x_n \cdot e_n \right| = 0$ as the sum expression perfectly replicates $x = {x_n}$ right? Or where am I wrong? :-)

     – aladin Jun 13 '22 at 13:07

1 Answers1

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The sequence $\sum _{n=1}^N e_n$ is eventually zero for any $N$. Thus, for the constant sequence $x:=(1,1,\ldots)$, $$ \left\| x - \sum _{n=1}^N e_n\right\|_\infty = 1 $$ for every $N\in\mathbb N$. What do we conclude?

The above also disproves the claim of $c_{00}$ being a dense subset of $\ell_\infty$.

Alternatively, existence of Schauder basis implies separable. But $\ell_\infty$ is not separable.

AlvinL
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