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Across many texts, I have seen that $l^2$ can be defined over $\mathbb{N}$ (denoted $l^2(\mathbb{N}$) or $\mathbb{R}$ (denoted $l^2(\mathbb{R}$) or orther spaces. I am not sure what does this notation with the space in the round brackets stand for.

First option: I usually think of $l^2$ as of a space of sequences. Does that mean that for $l^2(\mathbb{N}$, the sequences are sequences of natural numbers and for $l^2(\mathbb{R}$, sequences have real numbers?

Second option: Or does that mean that the sequences in $l^2$are indexed by numbers from the "underlying" set? E.g. if $E = \mathbb{N}$, then the sequences living in $l^2$ are in the form $x_1, x_2, x_3, x_4,......$ such that the members in each sequences are denoted by numbers from $\mathbb{N}$? In this case, if you index a sequence by numbers from $\mathbb{R}$, then it becomes a function, or am I wrong? This seems weird to me since then it would not be sequence space anymore.

Which option is correct? Thank you in advance as I am rather confused by the notation.

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    Can you point out where you saw the notation $l^2 (\mathbb R)$? – Arctic Char Aug 13 '21 at 17:27
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    Are you sure it's not $L^2(\Bbb R)$? – José Carlos Santos Aug 13 '21 at 17:29
  • I dont remember to be honest, if your doubts imply the second option is correct interpretation, then maybe I saw $l^2(R)$ with R meaning just some arbitrary set (not real numbers) or it was wrongly used as a $l^2$ space over $\mathbb{R}$ in that sense that sequences have real members (not complex as is common too.) – Tereza Tizkova Aug 13 '21 at 17:34
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    I am not sure if your reference means the following, but for arbitrary set $A$ of any cardinality you can define $l^2(A)$ to be the set of all $(x_i){i \in A}$ such that it is $2$-summable, in the sense that there is $L>0$ s.t. $\sum{i \in F} |x_i|^2 \le L$ for any finite set $F\subset A$. I think under a natural inner product it becomes a Hilbert space with cardinality $=|A|$ – Nick Aug 13 '21 at 17:35
  • @Nick Yes, that is exactly how I meant it and how I believe the notation is used. It really doesn´t make sense for $A$to be comething of continuous nature. – Tereza Tizkova Aug 13 '21 at 17:36

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If any terminology I use below is unfamiliar, you can easily find all of them on Wikipedia. The concept you need is that of measure spaces and the integrals with respect to arbitrary measures. A measure space $(X,\mathcal{M},\mu)$ is a triple of information, having a set $X$, a $\sigma$-algebra $\mathcal{M}$ on $X$, and a measure $\mu$ defined on the $\sigma$-algebra $\mathcal{M}$. Now, one considers complex-valued functions $f:X\to \Bbb{C}$ (which are $\mathcal{M}$-$\mathcal{B}(\Bbb{C})$ measurable, i.e for every Borel subset $E\subset\Bbb{C}$, $f^{-1}(E)\in\mathcal{M}$). Given any $0<p<\infty$, we define $\mathcal{L}^p(\mu)$ to be the set of all measurable $f:X\to \Bbb{C}$ such that \begin{align} \int_X|f|^p\,d\mu <\infty. \end{align} Often though, we write $\mathcal{L}^p(X)$ instead, but it should be noted that what is important is not the set $X$, but the measure $\mu$, because we can always consider different measures on the same set, so logically, it is important to clarify the measure (but usually from context one infers the measure, so one instead specifies the set).

Now, let us consider special examples:

  1. Consider $X=\Bbb{R}$, $\mathcal{M}$ to be the Lebesuge $\sigma$-algebra and $\mu_{\text{Lebesgue}}$ the Lebesgue measure. Then, $\mathcal{L}^p(\mu_{\text{Lebesgue}})$, which again is typically written as $\mathcal{L}^p(\Bbb{R})$, is the set of Lebesgue-measurable functions $f:\Bbb{R}\to\Bbb{C}$ such that $\int_{\Bbb{R}}|f|^p\,d\mu_{\text{Lebesgue}}\equiv\int_{\Bbb{R}}|f(x)|^p\,dx<\infty$. This is the Lebesgue integral with respect to Lebesgue measure on $\Bbb{R}$.... so VERY roughly speaking, this is the integral we all know and are familiar with.

  2. As another example, consider ANY non-empty set $X$. We consider $\mathcal{M}=\mathcal{P}(X)$ the $\sigma$-algebra to be the full power set of $X$ (i.e every set is measurable and thus all functions on $X$ will be measurable). We also consider the counting measure which is defined as $\mu_{\text{counting}}(E)= \text{number of elements in $E$}$ if $E$ is a finite set, and $\mu_{\text{counting}}(E):=\infty$ otherwise. In this case, $\mathcal{L}^p(\mu_{\text{counting}})$ is the set of functions $f:X\to\Bbb{C}$ such that \begin{align} \int_X|f|^p\,d\mu_{\text{counting}}<\infty \end{align}

By unwinding the definition of the counting measure and the integral, we see that $\mathcal{L}^p(\mu_{\text{counting}})$ consists of precisely those functions $f:X\to\Bbb{C}$ such that \begin{align} \sup\left\{\sum_{x\in F}|f(x)|^p\,:\, \text{$F$ is a finite subset of $X$}\right\}<\infty.\tag{$*$} \end{align}

It is tradition that whenever we use the counting measure on a set $X$, rather than writing $\mathcal{L}^p(\mu_{\text{counting}})$ or $\mathcal{L}^p(X)$ or $\mathcal{L}^p(X,\mathcal{P}(X),\mu_{\text{counting}})$, we instead use the notation $\ell^p(X)$.

So, to repeat once again, given a non-empty set $X$ and $0<p<\infty$, we define $\ell^p(X):=\mathcal{L}^p(X,\mathcal{P}(X),\mu_{\text{counting}})$, and this is precisely the set of all functions $f:X\to \Bbb{C}$ such that the supremum in $(*)$ above is finite.

So, for example we can take $X=\Bbb{R}^n$, or $X=\Bbb{C}$, or $X=\Bbb{R}$, or $X=\text{set of all functions $\Bbb{R}^4\to\Bbb{R}^5$}$ or literally any other set you can think of. Notice that if $X$ is a finite set, then the above finiteness condition in $(*)$ above is trivially satisfied, so that $\ell^p(X)$ is simply the set of all functions $X\to\Bbb{C}$ (this is certainly in bijection with $\Bbb{C}^{|X|}$, where $|X|$ is the number of elements in $X$). Most commonly though, we take $X=\Bbb{Z}$ or $X=\Bbb{N}$. Recall that a function from $\Bbb{N}$ or $\Bbb{Z}$ is nothing other than a sequence (in fact a sequence is defined as a function having some subset of $\Bbb{Z}$ as its domain).

Emphasizing this once again, the notation $\ell^p(X)$ is NOT because we're thinking of a sequence space. The notation $\ell^p(X)$ means that we're considering the counting measure on $X$, and considering a certain function space (the case of $X=\Bbb{Z}$ or $\Bbb{N}$ is a (useful) hyper special case).


One other fact I should point out: if $f\in\ell^p(X)$, then $\{x\in X\,:\, f(x)\neq 0\}$ is at most countable (another way of saying it is that an uncountable sum of strictly positive numbers is always infinite). So, in this sense, for an element $f\in \ell^p(X)$, there are only (at most) countably many values to deal with, so it's roughly a sequence, with many zeros.

peek-a-boo
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The way I learned it (expanding from my comment), using the lower case $l$ in $l_p(X)$ means $X$ is the index set for your Hilbert basis. This is different than the upper case $L$ definition where the set in parenthesis refers to the measure space in question.

So here, the set in parenthesis is used only to indicate the cardinality of the Hilbert basis. The reason for this is because $l_2(\aleph_x)$ can be viewed as the "prototypical" Hilbert space, as all Hilbert spaces over that field with that cardinality Hilbert basis are isomorphic.

In particular, any separable infinite dimension Hilbert space is isomorphic to $l_2(\mathbb{N})$. In this notation, $l_2(\mathbb{R})$ would refer to the (assuming Continuum Hypothesis) smallest non-separable Hilbert spaces.

Alan
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  • What is the $\aleph_x$ cardinality particularly? Is the $x$ just an arbitrary natural number? And, in that case, does that imply the Hilbert basis would have $\aleph_x$ elements? To the last paragraph: I have read that any Hilbert space is isoMETRIC to $l^2(E)$ - I guess that is somehow connected with the isomorphism to $l^2(\mathbb{N})$? – Tereza Tizkova Aug 19 '21 at 08:43