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

Question

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 $\ell^2$?

Answer 1

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.

Answer 2

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.