Extension of Plancherel theorem to $\ell^2$ Can Plancherel's theorem, which was originally defined for $L^2$ spaces (rather, functions in $L^1\cap L^2$) be extended to $\ell^2$ spaces? How would one do that, or is it very obvious/intuitive? If so, could you please explain?
 A: The correct setting for Fourier analysis is in terms of Pontryagin duality.  Let $G$ be a locally compact Abelian group and let $\hat G$ be the group of characters $G\to S^1$, given the compact-open topology; $\hat G$ is called the Pontryagin dual of $G$.  Then $\hat G$ is also locally compact Abelian.
Let $\mu$ be a Haar measure on $G$ and $\hat \mu$ a Haar measure on $\hat G$.  Then the Fourier transform in general is a map $L^2(G)\to L^2(\hat G)$ given by $$f\mapsto \left(\hat f:\chi\mapsto \int \chi(x)f(x)d\mu\right).$$  In general, if $f$ is in $L^1(G, \mu)\cap L^2(G, \mu)$ then $\hat f$ will be in $L^2(\hat G, \hat \mu)$ (this is the general version of the Plancherel theorem).
Now $\ell^2(\mathbb{Z})$ is exactly $L^2(\mathbb{Z}, \mu)$ where $\mu$ is the counting measure on $\mathbb{Z}$, which happens to also be the Haar measure.  The Pontryagin dual of $\mathbb{Z}$ is $S^1$, so the Fourier transform sends $\ell^2(\mathbb{Z})\to L^2(S^1, \lambda)$, where $\lambda$ is the Lebesgue measure on the circle (which also happens to be the Haar measure).  Phrased this way, the Plancherel theorem and essentially every other theorem from Fourier analysis carries over to $\ell^2(\mathbb{Z})$.  
It is a good exercise to check that ${\hat{\mathbb{Z}}}\simeq S^1$ and ${\hat {S^1}}=\mathbb{Z}$.
