# Do results from any $L^p$ space for functions hold in the equivalent $\ell^p$ spaces for infinite sequences?

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 $\ell^2$?

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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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As a complement, the wikipedia page on $L^p$-spaces is quite good. –  t.b. Apr 9 '11 at 5:16
thanks 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? –  user7815 Apr 9 '11 at 14:57
@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. –  Arturo Magidin Apr 9 '11 at 19:13
thanks arturo. I'll post a new question about Plancherel. –  user7815 Apr 9 '11 at 19:48

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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In 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. –  t.b. Apr 9 '11 at 11:23
@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. –  Giuseppe Negro Apr 9 '11 at 12:10
Well, 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 and Alaoglu are beautiful and extremely useful theorems, and it is good to know about them. –  t.b. Apr 9 '11 at 12:54
Ok, 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. –  Giuseppe Negro Apr 9 '11 at 14:01
$(\star)$ 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). 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. –  t.b. Apr 9 '11 at 14:12