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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.

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    $\begingroup$ A simpler and related question is to interpret the "limit" of the Pontryagin duality relationships $\mathbb{Z}_n^{\vee} \cong \mathbb{Z}_n$ as the Pontryagin duality relationship $\mathbb{Z}^{\vee} \cong S^1$. The closest I know how to get is to take categorical limits / colimits, which gets you the duality $\widehat{\mathbb{Z}}^{\vee} \cong \mathbb{Q}/\mathbb{Z}$. $\endgroup$ – Qiaochu Yuan Jan 14 '16 at 6:23
  • $\begingroup$ What do you mean by $\mathbb{Z}^ˆ∨$? $\endgroup$ – Lao-tzu Jan 14 '16 at 6:42
  • $\begingroup$ Interpreting Fourier inversion as a limit is not only interpreting but is the right way to see Fourier inversion in $L^2$. One defines the Fourier transform by extending it from Schwartz space (or $L^1\cap L^2$) to $L^2$ and similar for the inverse. To be concrete one can use the limit $\lim_{T\to\infty} \int_{-T}^T \hat f(\xi) \exp(i x\xi) d\xi$… $\endgroup$ – Dirk Jan 14 '16 at 11:31
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    $\begingroup$ The Bohr compactification of ${\mathbb R}$ looks nothing like a torus... $\endgroup$ – Yemon Choi Jan 14 '16 at 15:23
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    $\begingroup$ @Dirk But I think the question is hoping to see the continuous FT as a limit of the discrete one, and I think that perspective while obviously attractive can be misleading if applied without caution $\endgroup$ – Yemon Choi Jan 14 '16 at 15:27
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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 $L^2(\mathbb{R})$.)

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  • $\begingroup$ In the 2nd paragraph, do you mean that we extend a function $f \in L^2(r\mathbb{T})$ with its periodic extension to $\mathbb{R}$ and view the extended one as a tempered distribution on $\mathbb{R}$? But if so, the extended one doesn't lie in $L^2(\mathbb{R})$, so how to get an isometric embeddings of $L^2(r\mathbb{T})$ into $L^2(\mathbb{R})$ in line -5? Could you explain what is '$P_n$'? I think you may mean '$P_r$' (depends on the parameter $r$). $\endgroup$ – Lao-tzu Jan 18 '16 at 7:45
  • $\begingroup$ @Lao-tzu: multiplying by sinc takes the periodically extended $L^2(r\mathbb{R})$ into $L^2(\mathbb{R})$. Okay, $P_r$ if you like --- I meant the orthogonal projection onto the embedded $l^2(\frac{1}{2\pi r}\mathbb{Z})$ (but was thinking of taking a sequence of them). $\endgroup$ – Nik Weaver Jan 18 '16 at 15:46
  • $\begingroup$ I think to prove that $P_{r_n} \chi_E \to \chi_E$ (a.e. pointwise or in $L^2$ sense), where $r_n\to\infty$ as $n\to \infty$, and $E$ is any Lebesgue measurable set with finite measure $|E|$, we need the following: $P_r \chi_E =\sum_{\alpha\in\lambda_r^*} 2\pi r |E\cap (\alpha+(-\frac{1}{4\pi r}, \frac{1}{4\pi r}))|\chi_{\alpha+(-\frac{1}{4\pi r}, \frac{1}{4\pi r})} \to \chi_E.$ I think it's easy when $E$ is an interval, but I can't prove it for general $E$. Could you please give some help? $\endgroup$ – Lao-tzu Feb 21 '16 at 2:18
  • $\begingroup$ I had make this into an MO question here: mathoverflow.net/questions/231716/… $\endgroup$ – Lao-tzu Feb 21 '16 at 2:53
  • $\begingroup$ What do you mean by ''converge strongly''? Do you mean the operator norm ||P_n-Id|| go to 0? But this only shows P_n f go to f in L^2. $\endgroup$ – Lao-tzu Feb 24 '16 at 3:15
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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.

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    $\begingroup$ How can elements of an infinite dimensional space (that of functions whose Fourier transforms have supports contained in $[-R,R]$) be "completely determined" by finitely many coefficients ?? $\endgroup$ – Jean Duchon Jan 14 '16 at 14:31
  • $\begingroup$ Moreover, what does your last paragraph actually mean? It seems unfortunately close to hand-waving $\endgroup$ – Yemon Choi Jan 14 '16 at 15:23
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    $\begingroup$ If the Fourier transform of f has compact support, then f is not compactly supported (follows from Heisenberg uncertainty principle, or, more nicely from the uncertainty principle by Donoho and Stark). Hence, there are countable many nonzero samples that determine f completely. If bandlimited functions could be compactly supported, signal processing would be considerably much easier. $\endgroup$ – Dirk Jan 14 '16 at 16:03
  • $\begingroup$ A bandlimited function is completely determined by its samples if the sampling rate is twice the maximum frequency. That's the Nyquist sampling theorem. The result of a Fourier transform has infinitely many frequency components where the frequencies go down to zero: it's time limited samples where the number of frequency components is finite. This is all standard material in signal processing textbooks, which I cited by the name of the theorem. $\endgroup$ – Watson Ladd Jan 15 '16 at 1:55
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    $\begingroup$ @Watson Ladd ... but your sentence "it is completely determined by a finite number of Fourier coefficients" is wrong, whence the downvotes to your answer, by people (not me) who may well know the basics of Fourier analysis. The Nyquist sampling theorem, or some form thereof, is certainly part of that "standard material". $\endgroup$ – Jean Duchon Jan 15 '16 at 12:26

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