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.
Best Answer
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 $L^2(\mathbb{R})$.)