For e.g., is $\ell^2$ self-dual like $L^2$? If some $x[n]\in\ell^1\cap\ell^2$, then does it have a Fourier transform in $\ell^2$?
Do results from any $L^p$ space for functions hold in the equivalent $\ell^p$ spaces for infinite sequences?
2 Answers
Both $\ell^p$ and $L^p$ spaces are special cases of the Lebesgue function spaces $\mathcal{L}^p$.
Given a measure space $(X,\mu)$, we can consider the collection of all measurable functions $f$ from $X$ to $\mathbb{R}$ (or $\mathbb{C}$, or $\mathbb{R}^n$, or $\mathbb{C}^n$, or any Banach space) such that $$||f||_p = \left(\int_X |f|^p d\mu\right)^{1/p}\lt\infty.$$ These functions form a pseudo-normed vector space; we mod out by functions with $||f||_p = 0$ to get a normed vector space.
The usual real-valued $L^p$ spaces are just the case where $(X,\mu) = (\mathbb{R},\lambda)$, the Lebesgue measure. The sequence spaces $\ell^p$ are the case with $(X,\mu) = (\mathbb{N},\mu)$, where $\mu$ is the counting measure.
Many of the abstract properties of $L^p$ spaces (like the fact that for $p\gt 1$, $L^q\cong (L^p)^*$, where $\frac{1}{p}+\frac{1}{q}=1$) are proven in the abstract setting of Lebesgue function spaces. As such, they hold for $\ell^p$ spaces as special cases of the general construction.
In particular, yes, $\ell^2$ is self-dual; and the dual of $\ell^3$ is $\ell^{3/2}$, etc.
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2As a complement, the wikipedia page on [$L^p$-spaces](http://en.wikipedia.org/wiki/Lp_space) is quite good. – 2011-04-09
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0thanks a lot. I guess my main concern was how can I interpret theorems and results that specifically are stated/proven for $L^p$ spaces into $\ell^p$ spaces, which is where I need to apply them. I'm not a mathematician, so I'm trying to understand if there is an intuitive way to do it. Specifically, the theorem I'm trying to go through now is Plancharel's. I know that sequences in $\ell^2$ have F.T. in $\ell^2$ (e.g. discretely sampled sinc$\leftrightarrow$rect). But how do I say this follows from Plancharel's when it is originally defined for $L^2$ space? – 2011-04-09
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1@Axiom: Basically, there are *some* results that *do* transfer, because they are "really" results about the Lebesgue function spaces, and not specifically about $L^p$ or $\ell^p$ spaces. On the other hand, as dissonance rightly points out, *other* results don't transfer because they are *specifically* about $L^p$ (or $\ell^p$ spaces) using specific properties about those measure spaces rather than general properties. In the case of Plancherel, it seems like it might be (at least originally) specifically about $L^p$ spaces rather than Lebesgue spaces, but I don't know enough to help you. Sorry. – 2011-04-09
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0thanks arturo. I'll post a new question about Plancherel. – 2011-04-09
To complement what Arturo said, I would like to point out that there are some properties of $\ell^p$ spaces that have no correspondence in $L^p(\Omega)$ spaces. The easiest of such regards inclusion: we have
$$\ell^1 \subset \ell^2 \subset \ldots \subset \ell^\infty$$
but, if $\Omega$ is an open subset of some $\mathbb{R}^n$, it's certainly not true that
$$L^1(\Omega) \subset L^2(\Omega) \subset \ldots L^\infty(\Omega).$$
In fact, if $\Omega$ is bounded (or, more generally, if it is a finite measure space) then
$$L^\infty(\Omega) \subset \ldots \subset L^2(\Omega) \subset L^1(\Omega),$$
that is, inclusions are reversed.
Another specific property of $\ell^p$ spaces regards duality. Riesz theorem asserts that, for $1 < p < \infty$ and $\frac{1}{p}+\frac{1}{p'}=1$, the mapping
$$f\in L^{p'}(\Omega) \mapsto T_f \in [L^p(\Omega)]',\quad \langle T_f, g \rangle= \int_{\Omega}f(x)g(x)\, dx;$$
is an isometric isomorphism. This holds true for every measure space and so for $\ell^p$ also. However, this theorem gives no information about extreme cases $p=+\infty, p'=1$, which have to be studied separately, yielding various results. One of those is the following.
Proposition Let $c_0$ be the subspace of $\ell^{\infty}$ consisting of all sequences $x=(x_n)_{n \in \mathbb{N}}$ s.t.
$$\lim_{n \to \infty}x_n=0.$$
Then the mapping
$$y=(y_n) \in \ell^1 \mapsto T_y \in [c_0]',\quad \langle T_y, x \rangle=\sum_{n \in \mathbb{N}}y_nx_n;$$
is an isometric isomorphism and we can write
$$\ell^1 \simeq [c_0]'.$$
As far as I know, we have no direct generalization of this to $L^1(\Omega)$ spaces. One may conjecture, for example, that the following is an isomorphism:
$$f \in L^1(\mathbb{R}) \mapsto T_f \in [C_0(\mathbb{R})]'$$
(here $C_0(\mathbb{R})$ stands for: "continuous functions on the line vanishing at infinity"). But this is not true, because that mapping is not surjective: $[C_0(\mathbb{R})]'$ contains $\delta$, the linear functional defined by the equation
$$\langle \delta, g \rangle=g(0),\quad g \in C_0(\mathbb{R});$$
and we have no representation for $\delta$ as $\delta=T_f$ for some $f \in L^1(\mathbb{R})$. In fact, suppose a $f$ as such exists. Then, for all $g\in C_0(\mathbb{R})$ whose support does not contain $\{0\}$, we would have
$$\int_\mathbb{R}f(x)g(x)\, dx=0,$$
so that, for every open subset $A$ of $\mathbb{R}-\{0\}$, $f=0$ a.e. on $A$. But this forces $f=0$ a.e. on $\mathbb{R}$ and so $T_f=0$, which is a contradiction since $\delta$ certainly is not null.
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0In fact, if $\mu$ is a non-atomic measure then it is easy to see that $L^1{(\Omega,\mu)}$ is *not* a dual space at all. One way to prove this (certainly not the easiest one) is to show that the unit ball has *no* extremal points (this *is* easy). If it were the case $L^1(\Omega,\mu) = X^{*}$ then the unit ball would be compact in the weak$^{\ast}$-topology by Alaoglu's theorem and Krein-Milman's theorem would yield the existence of extremal points, a contradiction. – 2011-04-09
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0@Theo: Sounds like a nice exercise for me. Unfortunately you use functional analytic tools that I do not know, but I think I can salvage the main idea. I'll let you know. – 2011-04-09
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0Well, this is the standard argument to show that a Banach space is not a dual space. It also applies to $c_{0}$. As I said, you only need to prove that there are no extremal points at all. Explicitly, you need to exhibit each $f$ with $\|f\| \leq 1$ as $f = (1-\alpha)g + \alpha h$ with $\|g\|, \|h\| \leq 1$ and $g \neq f \neq h$. This is easy from the hypothesis that the space have no atoms. [Krein-Milman](http://en.wikipedia.org/wiki/Krein-Milman) and [Alaoglu](http://en.wikipedia.org/wiki/Alaoglu%27s_theorem) are beautiful and extremely useful theorems, and it is good to know about them. – 2011-04-09
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0Ok, I'll gather some information then. I was trying to follow another avenue (for the special case $L^1(\mathbb{R})$), namely: should $L^1(\mathbb{R})$ be isomorphic to $X^\star$ for some separable Banach space $X$, from every bounded sequence in $L^1(\mathbb{R})$ we should be able to extract a weakly convergent subsequence ($\star$) and this is false, as $\chi_{[n, n+1]}$ shows. I don't know if this may work, though: I'm dubious about ($\star$). Oh well. – 2011-04-09
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0$(\star)$ is correct and it follows from Alaoglu (which is easy to prove in this situation using a diagonal argument - which should be an easy adaptation of my proof of the Lemma in [this answer](http://math.stackexchange.com/questions/30717/compactness-of-a-bounded-operator-t-colon-c-0-to-ell1/30780#30780)). What I don't see at the moment is how you can guarantee that $\chi_{[n,n+1]}$ has no weak$^{\ast}$-convergent subsequence if $X$ is arbitrary. – 2011-04-09
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0Thank you for suggestions on ($\star$)! In fact I was thinking exactly at the diagonal trick, the one you use. As for $\chi_{[n, n+1]}$, I thought: for every interval $I$, the mapping $f \mapsto \int_I f(x)\, dx$ is a continuous linear functional on $L^1(\mathbb{R})$. So if $(f_n)$ is a subsequence of $(\chi_{[n, n+1]})$ weakly convergent to a function $f$, then $\int_a^b f(x)dx=0$ for all $a < b$. This means that $f$ needs be null a.e. and this is a contradiction, because on the other hand $\int_{-\infty}^\infty f(x)\, dx=1$. – 2011-04-09
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0What you say is correct but it *doesn't* prove that $L^{1} = X^{\ast}$ can't be true. There is an isometric embedding $X \to L^{\infty} = X^{\ast\ast}$ and you're using the pairing $L^1 \times L^\infty \to \mathbb{R}$ given by $(f,\phi) \mapsto \int f \phi$, so you prove two things, in fact: 1. The constant function $\phi = 1$ can't be in the image of $X$, 2. The unit ball of $L^1$ is not weakly compact. *However*, you want to prove that the unit ball of $L^1$ can't be weak$^{\ast}$-compact (to be continued)... – 2011-04-09
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0If $L^1 = X^\ast$ we have a tautological pairing $L^1 \times X \to \mathbb{R}$ given by identifying $f \in L^1$ with a linear functional $\phi_{f}: X \to \mathbb{R}$. Given a sequence $f_{n}$ this sequence converges to $f$ in the weak$^{\ast}$-topology iff for each $x \in X$ we have $\phi_{f_{n}}(x) \to \phi_{f}(x)$. Alaoglu's theorem asserts that for every bounded sequence $f_n$ there is a subsequence $f_{n_k}$ and $f$ such that for all $x \in X$ we have $\phi_{f_n}(x) \to \phi_{f}(x)$. As you have almost no control about $X$ it seems difficult to prove that *no* subsequence can converge. – 2011-04-09
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0@Theo: Err yes, I see the subtlety now. In fact this is what I vaguely smelled when I said that ($\star$) left me dubious. Ok, so if you believe this approach leads nowhere I'll cease thinking about it and read something about Krein-Millman theorem instead. BTW... Can you give me a hint on how to prove that the unit ball of $L^1$ hasn't any extremal points? Not even on the boundary? This must be a consequence of infinite dimension, I think, because in $\mathbb{R}^3$ the ball has the boundary as extremal, if I see well. Also, thank you very much for this conversation. I'm finding it very useful – 2011-04-09
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0Glad to hear that you find it useful! First let me address Krein-Milman. I've recently given a sketch of its proof in [this answer](http://math.stackexchange.com/questions/31187/compact-convex-problem/31221#31221). You only need one more ingredient, and this is the [separation version of Hahn-Banach](http://en.wikipedia.org/wiki/Hahn-Banach_theorem#Hahn-Banach_separation_theorem). Concerning no extremal points, infinite dimensionality is *not* enough, as the example of $\ell^1$ shows. You need to use that the measure is *non-atomic*. (to be continued)... – 2011-04-09
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0If $f \neq 0$, for $\varepsilon \gt 0$ small enough at least one of the sets $X_{\gt \varepsilon} = \{f \gt \varepsilon\}$ or $X_{\lt - \varepsilon} = \{f \lt -\varepsilon\}$ has positive measure. Let's assume it's $X_{\gt \varepsilon}$. We can thus write $X_{\gt \varepsilon} = A \cup B$ with $A$ and $B$ disjoint and of positive measure. Perturb $f$ on $A$ and $B$ without changing its norm by adding and subtracting appropriately scaled versions of the characteristic functions $\chi_{A}$ and $\chi_{B}$. In other words, write $f = \frac{1}{2}(f+h) + \frac{1}{2}(f-h)$ with carefully chosen $h$. – 2011-04-09
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0@Theo: It is easier than I thought! Using your previous notations, I think that $h=(\mu(A))^{-1}\chi_A-(\mu(B))^{-1}\chi_B$ will do. And this does *not* work in $L^p$ with $1
, right? It shouldn't, since those are reflexive spaces and so their unit ball is weakly-compact. By Krein-Milman, then, this unit ball is the convex hull of its extreme points and, in particular, some extreme point need exist. Correct? I'm just mimicking what you did before, in fact.
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0Well, almost. Multiply your $h$ by $\frac{\varepsilon}{2}$ just to be sure. You're absolutely right about the rest. In fact, by [Clarkson's inequalities](http://en.wikipedia.org/wiki/Clarkson's_inequalities) *all* functions with $\|f\|_{p} = 1$ are extremal points for $1 \lt p \lt \infty$. – 2011-04-09
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0Clarkson, yes. Because if $f=\frac{1}{2}f_1+\frac{1}{2}f_2$ and $\lVert f \rVert_p=1$ then by Clarkson we get $\lVert \frac{f_1-f_2}{2} \rVert_p^p \le 0$ and so $f_1=f_2$ a.e. In conclusion we may say: not all balls are round! $L^p$ ones are but $L^1$ ones have a flat-flat surface. – 2011-04-09
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0So, it seems like everything's settled, then, right? You might want to have a look at [uniform convexity](http://en.wikipedia.org/wiki/Uniformly_convex_space), as well. See you around, buona serata! – 2011-04-09