# 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?

• What do you have in mind as "Fourier transform"? Apr 9, 2011 at 19:53
• Whoa, are there multiple definitions? The one I'm referring to is for a function $f(x)$, its Fourier transform is $\hat{f} (y)=\int_{-\infty}^{\infty} f(x) e^{-\imath xy} dx$. For a sequence $f[n]$, its Fourier transform is $\hat{f}[k]=\sum_{m=-\infty}^{\infty} x[n]e^{-\imath k m}$.
– user7815
Apr 9, 2011 at 22:15
• I have never seen such a Fourier transform. Anyway, the proof of Plancherel's theorem would fail if you use the counting measure instead of the Lebesgue measure because it is not $\sigma$-finite. Apr 9, 2011 at 23:54
• I think it fails. $a_n = \frac1n$ is in $\ell^2$ but the "Fourier transform" is not in $\ell^2$. Apr 10, 2011 at 0:01
• @Jonas, but $a_n=1/n$ is not in $\ell^1$. Maybe I wasn't clearer (I'm not a mathematician, so bear with me), but I meant $\ell^1\cap\ell^2$ as in my post.
– user7815
Apr 10, 2011 at 3:35

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 ${\hat {S^1}}=\mathbb{Z}$.