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I'm approaching the study of metric spaces, functional analysis, Hilbert spaces and so on and I have some questions in order to understand if I got it right.

For example, the space $\ell^1([0, 1])$: this is a Banach space, the space of the sequences which obey to $\sum_n \vert x_n\vert < +\infty$. Does the $[0, 1]$ part mean that the $x_n$ terms can assume values only between $0$ and $1$? Did I understand well?

Also I would like some clarification about this: "a sequence in $\ell^1$ is a sequence of sequences". Can you please provide me some example?

I'm trying to understand it correctly. So for example, let's take the sequences

$$a_n = \dfrac{1}{\log(n+2)} \qquad \qquad \qquad b_n = \dfrac{1}{e^n}$$

Then $$a_n = \left(\dfrac{1}{\ln(2)}, \dfrac{1}{\ln(3)}, \dfrac{1}{\ln(4)}, \ldots \right)$$

$$b_n = \left(1, e^{-1}, e^{-2}, \ldots\right)$$

I can hypothetically say they are in $\ell^1([0, 1])$ for each $x_n$ term assumes values between $0$ and $1$. Or maybe I understood nothing and I'm speaking malarkey...

(I meant hypothetically, because then I have to check if their sums converge too).

Please explain me where I am wrong, I tried to find some resounce online too but I cannot compare notes with some true human explanation.

Thank you!

Heidegger
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  • It would be helpful if you could point to a source for those symbols. The way I've usually seen $\ell^1(X)$ (where $X$ is a set) used is to denote the $L^1$ space with the counting measure, but I suspect there's a typo somewher, either in your sorce or in your post... – Pelota Dec 27 '22 at 20:30
  • @Pelota I'm aware of, for example, $L^1([0, 1])$ which is the integral from $0$ to $1$ of a function (et cetera). I was wondering if the notation could apply to $\ell$ too, and in that case, its meaning and if I'm correct in what I wrote. – Heidegger Dec 27 '22 at 20:34
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    Concerning the "a sequence in $\ell^1$ is a sequence of sequences": a sequence in $\ell^1$ is a sequence of elements of $\ell^1$. Since elements of $\ell^1$ are sequences, a sequence of its elements is a sequence of, well, sequences. An example to clarify: denote by $x_k$ the sequence $(\delta_k(0),\dots, \delta_k(n),\dots)$. Then ${x_k}_{k\in \mathbb{N}}$ is a (nonconverging) sequence in $\ell^1$. – Pelota Dec 27 '22 at 20:34
  • You can use $\ell^1(X)$ to restrict (or enlarge) the domain of the functions, not the codomain. Notice in particular that, with your definition of $\ell^1([0,1])$, that's not a Banach space – Pelota Dec 27 '22 at 20:35
  • @Pelota I'm still have troubles in understanding the sequence of sequence. So the ones I wrote are just sequences, but not sequences of sequences? I got the point of the domain/codomain. So $\ell^1([0, 1])$ would be the sequences with domain in $[0, 1]$? Then my examples do not fit... Also, it's not a Banach space because...? :( – Heidegger Dec 27 '22 at 20:41
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    If for you $\ell^1([0,1])={{x_n}: \sum |x_n|<\infty; \forall n x_n\in [0,1]}$, which seems to me to be what you mean when you say "the $x_n$ terms can assume values only between 0 and 1?", then it's not a Banach space because its not a vector space. – Pelota Dec 27 '22 at 21:58

1 Answers1

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For example, the space $\ell^1([0, 1])$: this is a Banach space, the space of the sequences which obey to $\sum_n \vert x_n\vert < +\infty$. Does the $[0, 1]$ part mean that the $x_n$ terms can assume values only between $0$ and $1$? Did I understand well?

That is a notation I've come to understand, yes. I've also, however, seen use of another notation where

$$\ell^p(I) := \left\{ (x_i)_{i \in I} \in \mathbb{C}^I \, \middle| \, \sum_{i \in I} |x_i|^p < \infty \right\}$$

(or another field of your choice). I suppose it depends on what you wish to leave implied for your definition: the field in question, or the indexing set.

Also I would like some clarification about this: "a sequence in $\ell^1$ is a sequence of sequences". Can you please provide me some example?

A sequence in a space is a countable subset thereof, given some order, loosely speaking. For instance, in $\mathbb{R}$, we have

$$ \left( \frac{1}{2^n} \right)_{n \in \mathbb{N}} = \left( \frac 1 2, \frac 1 4 , \frac 1 8 , \cdots \right)$$

as a sequence: each number is a real number. We may define a sequence of sequences as so: say we have

$$\left( x_n \right)_{n \in \mathbb{N}} \text{ where } x_n = \left( \xi_k^{(n)} \right)_{k \in \mathbb{N}}$$

For each $n$, we have a sequence $x_n$ (indexed by $n$), and each of these is a sequence itself indexed by $k$ (the $(n)$ being used to keep track of what members belong to a given sequence). For instance,

$$\left( x_n \right)_{n \in \mathbb{N}} \text{ where } x_n = \left( \frac n k \right)_{k \in \mathbb{N}}$$

Then we have

$$\begin{align*} x_1 &= \left( \frac 1 1 , \frac 1 2 , \frac 1 3 , \frac 1 4 , \cdots \right) \\ x_2 &= \left( \frac 2 1 , \frac 2 2 , \frac 2 3 , \frac 2 4 , \cdots \right) \\ x_3 &= \left( \frac 3 1 , \frac 3 2 , \frac 3 3 , \frac 3 4 , \cdots \right) \\ x_4 &= \left( \frac 4 1 , \frac 4 2 , \frac 4 3 , \frac 4 4 , \cdots \right) \end{align*}$$

(This is more focused on the "sequence of sequences" notion than $\ell^p$, mind you; each of these sequences are in $\ell^p$ for $p>1$ though. Again, notice that each $x_n$ is itself a separate element of the set of all sequences: each $x_n$ is itself a sequence, and we built a sequence of them.)

$$a_n = \left(\dfrac{1}{\ln(2)}, \dfrac{1}{\ln(3)}, \dfrac{1}{\ln(4)}, \ldots \right)$$

This sequence does not lie in $\ell^1$, owing to the summation condition. This is because

$$\ln(n) < n$$

for $n$ sufficiently large, so eventually

$$\frac{1}{\ln(n)} > \frac 1 n \text{ but } \sum_{n=1}^\infty \frac 1 n \text{ diverges}$$

PrincessEev
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  • Big Thank you!! That was insightful. For what concerns $\ell^1([0, 1])$, is this Banach? The other guy above told me it is not, but how can I see it? – Heidegger Dec 27 '22 at 20:52
  • For example, I've also seen $\ell^1(\mathbb{N})$. Does this means the sequences have values in $\mathbb{N}$ or that they are labelled with natural values? Also, according to your definition above, what would be $\mathbb{C}^{\mathbb{N}}$ in this case? – Heidegger Dec 27 '22 at 20:56
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    If we use that $\ell^1([0,1])$ is about the elements lying in $[0,1]$, then it is Banach for the same reason $\ell^1(\mathbb{N})$ (where $\mathbb{N}$ denotes the indexing set) is Banach; see some discussion here, for the simple reason $[0,1] \subseteq \mathbb{C}$. When $[0,1]$ refers to an indexing set, I don't know. – PrincessEev Dec 27 '22 at 21:23
  • Thank you again! :D – Heidegger Dec 27 '22 at 21:24
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    $\ell^1(\mathbb{N})$ may refer to sequences that have elements in $\mathbb{N}$, but the more I search around, the more common interpretation seems to be that $\ell^1(\mathbb{N})$ is all sequences $(x_n)_{n \in \mathbb{N}}$ whose members lie in an implied field (commonly $\mathbb{C}$). (That is, labeled with natural values, most commonly.) Meanwhile, $\mathbb{C}^\mathbb{N}$ is a notation for the collection of all functions $f : \mathbb{N} \to \mathbb{C}$. (Note that, formally, a sequence is a function with domain $\mathbb{N}$.) – PrincessEev Dec 27 '22 at 21:25
  • @PrincessEev if by $\ell^1([0,1])$ we mean the elements of the standard $\ell_1$ such that for each $x\in \ell^1([0,1])$ each of its coordinates is contained in $[0,1]$, then it's not a vector space, much less so a Banach VS. Perhaps I'm misunderstanding what you both mean – Pelota Dec 27 '22 at 22:02