TITLE: Do results from any $L^p$ space for functions hold in the equivalent $\ell^p$ spaces for infinite sequences?
QUESTION [6 upvotes]: 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 $\ell^2$?

REPLY [6 votes]: 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.