Interpret Fourier transform as limit of Fourier series

Question

Let $V=\mathbb{R}^n$, $\Lambda_r=2\pi r \mathbb{Z}^n \subset V (r>0)$ a lattice; $V^*\cong\mathbb{R}^n$ the dual vector space of $V$, and $\Lambda_r^*=\frac{1}{2\pi r} \mathbb{Z}^n =\text{Hom}(\Lambda_r, \mathbb{Z})$ the dual lattice in $V^*$.

$\Lambda_r^*$ can be thought of as the Pontryagin dual of the torus $T^n_r=V/\Lambda_r$; also, $V^*$ can be thought of as the Pontryagin dual of $V$ and can be identified with $V$ via the pairing $\left< x,\xi \right>=e^{2\pi i x\cdot\xi}$. Chapter 4 of Gerald B. Folland's book A course in abstract harmonic analysis is a nice introduction to these materials in the context of locally compact abelian groups; see also this blog of Terence Tao.

It's well known that the Fourier transform gives an isometry of Hilbert spaces $$L^2(V)\cong L^2(V^*).$$

Also, Fourier series give an isometry of Hilbert spaces $$L^2(T^n_r)\cong l^2(\Lambda_r^*).$$

We have the following obvious intuition: as $r>0$ becomes larger and larger, the scale of $T^n_r$ also becomes larger and larger, and finally becomes like $V=\mathbb{R}^n$; on the other hand, the dual lattice $\Lambda_r^*$ becomes more and more 'dense' in $V^*=\mathbb{R}^n$ as the distance of adjacent points is $\frac{1}{2\pi r}$, which goes to 0 as $r$ goes to $\infty$.

My question is the following:

Can we make it mathematically rigorous, both on the level of functions and on the level of spaces (e.g. $T^n_r \to V$), that the 'limit' of the isomorphisms $$L^2(T^n_r)\cong l^2(\Lambda_r^*)$$ is the isomorphism $$L^2(V)\cong L^2(V^*)$$ as $r$ goes to $\infty$?

The bad thing is that $V=\mathbb{R}^n$ is noncompact, while we have the notion of Bohr compactification, I hope this can be helpful.

Is there any relation between the tori $T^n_r$ and the Bohr compactification of $\mathbb{R}^n$?

Hopefully, if we can do this, then we can do similar things such as interpreting Fourier inversion as a limit. Some aspect (on the level of functions) is discussed in Exercise 40 (Fourier transform on large tori) of Tao's blog.

Answer 1

Great question. I've often used this heuristic but never thought about whether it had a rigorous meaning.

Let me do this in one dimension; the generalization to higher dimensions is straightforward. My first comment is that the Fourier transform between $l^2(\frac{1}{2\pi r}\mathbb{Z})$ and $L^2(r\mathbb{T})$ genuinely sits inside of the Fourier transform of distributions on $\mathbb{R}$: identify an element $(a_n)$ of $l^2(\frac{1}{2\pi r}\mathbb{Z})$ with the sum of delta functions $\sum a_n\delta_{n/2\pi r}$, and a function $f \in L^2(r\mathbb{T})$ with its periodic extension to $\mathbb{R}$. The integral of $\sum a_n\delta_{n/2\pi r}$ against $e^{-2\pi i xt}$ is $\sum a_n e^{-int/r}$. So the ordinary Fourier transform between the integers and the circle matches up with the distributional Fourier transform after making this identification.

But you want to approximate the $L^2$ Fourier transform on $\mathbb{R}$. I guess the obvious thing to do here is to convolve the embedded $l^2(\frac{1}{2\pi r}\mathbb{Z})$ with the "rectangular function" which takes the value $\sqrt{2\pi r}$ on $[-\frac{1}{4\pi r}, \frac{1}{4\pi r}]$ and is $0$ elsewhere. This isometrically embeds $l^2(\frac{1}{2\pi r}\mathbb{Z})$ into $L^2(\mathbb{R})$. In the transformed picture it corresponds to multiplying a function in the embedded $L^2(r\mathbb{T})$ by Fourier transform of the rectangular function, which is $\frac{1}{\sqrt{2\pi r}}{\rm sinc}(\frac{t}{2r})$. So now we have isometric embeddings of $l^2(\frac{1}{2\pi r}\mathbb{Z})$ and $L^2(r\mathbb{T})$ into $L^2(\mathbb{R})$ which are compatible with taking the Fourier transform before or after embedding. They converge to $L^2(\mathbb{R})$ in the sense that the orthogonal projections onto the embedded spaces converge strongly to the identity operator; this is easy to check in the untransformed picture. (It's clear that $P_nf \to f$ when $f$ is piecewise constant, and such functions are dense in $L^2(\mathbb{R})$.)

Answer 2

Yes. Let $f(x)$ be a function on $\mathbb{R}^n$ such that its Fourier transform has compact support inside of $[-R, R]$. Then $f(x)$ is completely determined by its values on $\frac{2\pi}{R}\mathbb{Z}$ by the Nyquist sampling theorem. (I may be off by a factor of two here). Thus it is completely determined by a countable number of Fourier coefficients. As the spacing between the samples decreases functions with higher frequencies can be exactly represented.

As a limit of Hilbert spaces I think this only works for functions with rapidly decreasing Fourier transforms, but I haven't sat down to work through all the details.