On Defining the Fourier Transform and Performing Changes of Variable on Quotient Subgroups of $\mathbb{Q}/$

Much to my dismay, in my work the more number-theoretic side of harmonic analysis (ex: the fourier transform on the adeles, on the profinite integers, etc.), I have found myself struggling with technicalities that emerge from frustratingly simple issues—so simple (and yet, so technical) that I haven't been able to find anything that might shine any light on the matter.

Let $$L_{\textrm{loc}}^{2}\left(\mathbb{Q}\right)$$ be the space of functions $$f:\mathbb{Q}\rightarrow\mathbb{C}$$ which satisfy $$\sum_{t\in T}\left|f\left(t\right)\right|^{2}<\infty$$ for all bounded subsets $$T\subseteq\mathbb{Q}$$. Letting $$\mu$$ be any positive integer, it is easy to show that any functions which is both $$\mu$$-periodic (an $$f:\mathbb{Q}\rightarrow\mathbb{C}$$ such that $$f\left(t+\mu\right)=f\left(t\right)$$ for all $$t\in\mathbb{Q})$$ and in $$L_{\textrm{loc}}^{2}\left(\mathbb{Q}\right)$$ is necessarily an element of $$L^{2}\left(\mathbb{Q}/\mu\mathbb{Z}\right)$$, the complex hilbert space of functions $$f:\mathbb{Q}/\mu\mathbb{Z}\rightarrow\mathbb{C}$$ so that: $$\sum_{t\in\mathbb{Q}/\mu\mathbb{Z}}\left|f\left(t\right)\right|^{2}<\infty$$Equipping $$\mathbb{Q}/\mu\mathbb{Z}$$ with the discrete topology, we can utilize Pontryagin duality to obtain a Fourier transform: $$\mathscr{F}_{\mathbb{Q}/\mu\mathbb{Z}}$$. The ideal case is when $$\mu=1$$. There, $$\mathscr{F}_{\mathbb{Q}/\mathbb{Z}}$$ is an isometric hilbert space isomorphism from $$L^{2}\left(\mathbb{Q}/\mathbb{Z}\right)$$ to $$L^{2}\left(\overline{\mathbb{Z}}\right)$$, where: $$\overline{\mathbb{Z}}\overset{\textrm{def}}{=}\prod_{p\in\mathbb{P}}\mathbb{Z}_{p}$$ is the ring of profinite integers, where $$\mathbb{P}$$ is the set of prime numbers, and where $$L^{2}\left(\overline{\mathbb{Z}}\right)$$ is the space of functions $$\check{f}:\overline{\mathbb{Z}}\rightarrow\mathbb{C}$$ which are square-integrable with respect to the haar probability measure $$d\mathfrak{z}=\prod_{p\in\mathbb{P}}d\mathfrak{z}_{p}$$ on $$\overline{\mathbb{Z}}$$.

The first sign of trouble was when I learned that, for any integers $$\mu,\nu$$, the (additive) quotient groups $$\mathbb{Q}/\mu\mathbb{Z}$$ and $$\mathbb{Q}/\nu\mathbb{Z}$$ are group-isomorphic to one another, and thus, that both have $$\overline{\mathbb{Z}}$$ as their Pontryagin dual. From my point of view, however, this isomorphism seems to only cause trouble. In my work, I am identifying $$L^{2}\left(\mathbb{Q}/\mu\mathbb{Z}\right)$$ with the set of $$\mu$$-periodic functions $$f:\mathbb{Q}\rightarrow\mathbb{C}$$ which are square integrable with respect to the counting measure on $$\mathbb{Q}\cap\left[0,\mu\right)$$. As such, $$\mathbb{Q}/\mu\mathbb{Z}$$ and $$\mathbb{Q}/\nu\mathbb{Z}$$cannot be “the same” from my point of view, because $$f\left(t\right)\in L^{2}\left(\mathbb{Q}/\mu\mathbb{Z}\right)$$ need not imply that $$f\left(t\right)\in L^{2}\left(\mathbb{Q}/\nu\mathbb{Z}\right)$$.

In my current work, I am dealing with a functional equation of the form:$$\sum_{n=0}^{N-1}g_{n}\left(t\right)f\left(\frac{a_{n}t+b_{n}}{d_{n}}\right)=0$$where $$N$$ is an integer $$\geq2$$, where the $$g_{n}$$s are known periodic functions, where $$f$$ is an unknown function in $$L^{2}\left(\mathbb{Q}/\mathbb{Z}\right)$$ (i.e., $$f\left(t\right)\in L_{\textrm{loc}}^{2}\left(\mathbb{Q}\right)$$ and $$f\left(t+1\right)=f\left(t\right)$$ for all $$t\in\mathbb{Q})$$, and where $$a_{n},b_{n},d_{n}$$ are integers with $$\gcd\left(a_{n},d_{n}\right)=1$$ for all $$n$$. For brevity, I'll write: $$\varphi_{n}\left(t\right)\overset{\textrm{def}}{=}\frac{a_{n}t+b_{n}}{d_{n}}$$ Because $$\varphi_{n}\left(t+1\right)$$ need not equal $$\varphi_{n}\left(t\right)+1$$, the individual functions $$f\circ\varphi_{n}$$, though periodic and in $$L_{\textrm{loc}}^{2}\left(\mathbb{Q}\right)$$, are not necessarily going to be of period $$1$$. Letting $$p$$ denote the least common multiple of the periods of $$g_{n}$$ and the $$f\circ\varphi_{n}$$s, I can view the functional equation as existing in $$L^{2}\left(\mathbb{Q}/p\mathbb{Z}\right)$$, and as such, I hope to be able to simplify it by applying the fourier transform.

Letting $$e^{2\pi i\left\langle t,\mathfrak{z}\right\rangle }$$ denote the duality pairing between elements $$t\in\mathbb{Q}$$ (or $$\mathbb{Q}/\mu\mathbb{Z}$$) and $$\mathfrak{z}\in\overline{\mathbb{Z}}$$, the idea is to multiply the functional equation by $$e^{2\pi i\left\langle t,\mathfrak{z}\right\rangle }$$, sum over an appropriate domain of $$t$$ (ideally, $$\mathbb{Q}/p\mathbb{Z}$$), make a change of variables in $$t$$ to move one of the $$\varphi_{n}\left(t\right)$$s out of $$f$$ and into $$e^{2\pi i\left\langle t,\mathfrak{z}\right\rangle }$$, pull out terms from this character, and then invert the fourier transform to return to $$L^{2}\left(\mathbb{Q}/p\mathbb{Z}\right)$$ with a vastly simpler equation. My main difficulty can be broken into three parts:

(1) Is taking the least common multiple of the periods to reformulate the functional equation as one over $$L^{2}\left(\mathbb{Q}/p\mathbb{Z}\right)$$ legal?

(2) I know that the fourier transform $$\mathscr{F}_{\mathbb{Q}/\mathbb{Z}}:L^{2}\left(\mathbb{Q}/\mathbb{Z}\right)\rightarrow L^{2}\left(\overline{\mathbb{Z}}\right)$$ is given by:$$\mathscr{F}_{\mathbb{Q}/\mathbb{Z}}\left\{ f\right\} \left(\mathfrak{z}\right)\overset{\textrm{def}}{=}\sum_{t\in\mathbb{Q}/\mathbb{Z}}f\left(t\right)e^{2\pi i\left\langle t,\mathfrak{z}\right\rangle }$$ and that the inverse transform is: $$\mathscr{F}_{\mathbb{Q}/\mathbb{Z}}^{-1}\left\{ \check{f}\right\} \left(t\right)\overset{\textrm{def}}{=}\int_{\overline{\mathbb{Z}}}\check{f}\left(\mathfrak{z}\right)e^{-2\pi i\left\langle t,\mathfrak{z}\right\rangle }d\mathfrak{z}$$ However, I am at a loss as to what formula to use for $$\mathscr{F}_{\mathbb{Q}/p\mathbb{Z}}$$ and its inverse, and for two reasons. On the one hand, because $$\mathbb{Q}/\mathbb{Z}$$ and $$\mathbb{Q}/p\mathbb{Z}$$ are group-isomorphic, what is to stop me from using the same formula for their fourier transforms? On the other hand, if I use a modified formula—say:$$\mathscr{F}_{\mathbb{Q}/p\mathbb{Z}}\left\{ f\right\} \left(\mathfrak{z}\right)=\sum_{t\in\mathbb{Q}/p\mathbb{Z}}f\left(t\right)e^{2\pi i\left\langle \frac{t}{p},\mathfrak{z}\right\rangle }$$ does the fact that $$\mathscr{F}_{\mathbb{Q}/p\mathbb{Z}}\left\{ f\right\} \left(\mathfrak{z}\right)\in L^{2}\left(\overline{\mathbb{Z}}\right)$$ then mean that I can recover $$f$$ by applying $$\mathscr{F}_{\mathbb{Q}/\mathbb{Z}}^{-1}$$, or do I have to also modify it in order to make everything consistent? Knowing the correct formula for the fourier transform on $$L^{2}\left(\mathbb{Q}/p\mathbb{Z}\right)$$ and its inverse is essential.

(3) I would like to think that performing a change-of-variables for a sum of the form:$$\sum_{\mathbb{Q}/r\mathbb{Z}}f\left(\alpha t+\beta\right)$$ (where $$\alpha,\beta,r\in\mathbb{Q}$$, with $$\alpha\neq0$$ and $$r=\frac{p}{q}>0$$) would be a relatively simple matter, but, that doesn't appear to be the case. For example, if $$t\in\mathbb{Q}\rightarrow f\left(\alpha t+\beta\right)\in\mathbb{C}$$ is not a $$r$$-periodic function, then this sum is not well-defined over the quotient group $$\mathbb{Q}/r\mathbb{Z}$$. Worse yet—supposing $$f$$ is $$r$$-periodic take a look at this: write elements of $$\mathbb{Q}/r\mathbb{Z}$$ in co-set form: $$t+r\mathbb{Z}$$, where $$t\in\mathbb{Q}$$. Then, make the change-of-variable $$\tau=\alpha t+\beta$$. Consequently, the set of all $$\tau$$ is:$$\alpha\left(\mathbb{Q}/r\mathbb{Z}\right)+\beta=\left\{ \alpha\left(t+r\mathbb{Z}\right)+\beta:t\in\mathbb{Q}\right\} =\left\{ \tau+\alpha r\mathbb{Z}:t\in\mathbb{Q}\right\}$$. Here is where things get loopy.

(1) Since $$\alpha,\beta\in\mathbb{Q}$$ with $$\alpha\neq0$$, the map $$\varphi\left(t\right)\overset{\textrm{def}}{=}\alpha t+\beta$$ is a bijection of $$\mathbb{Q}$$. As such, I would think that:$$\left\{ \tau+\alpha r\mathbb{Z}:t\in\mathbb{Q}\right\} =\left\{ \tau+\alpha r\mathbb{Z}:\varphi^{-1}\left(\tau\right)\in\mathbb{Q}\right\} =\left\{ \tau+\alpha r\mathbb{Z}:\tau\in\varphi\left(\mathbb{Q}\right)\right\}$$ and hence:$$\alpha\left(\mathbb{Q}/r\mathbb{Z}\right)+\beta=\left\{ \tau+\alpha r\mathbb{Z}:\tau\in\mathbb{Q}\right\} =\mathbb{Q}/\alpha r\mathbb{Z}$$ Using this approach, I obtain:$$\sum_{t\in\mathbb{Q}/r\mathbb{Z}}f\left(\alpha t+\beta\right)=\sum_{\tau\in\mathbb{Q}/\alpha r\mathbb{Z}}f\left(\tau\right)$$

(2) Since $$r=\frac{p}{q}$$, decompose $$\mathbb{Z}$$ into its equivalence classes mod $$q$$:$$\left\{ \tau+\alpha r\mathbb{Z}:\tau\in\mathbb{Q}\right\} =\bigcup_{k=0}^{q-1}\left\{ \tau+\alpha r\left(q\mathbb{Z}+k\right):\tau\in\mathbb{Q}\right\}$$ and so:$$\sum_{t\in\mathbb{Q}/r\mathbb{Z}}f\left(\alpha t+\beta\right)=\sum_{k=0}^{q-1}\sum_{\tau\in\mathbb{Q}/\alpha rq\mathbb{Z}}f\left(\tau+\alpha rk\right)$$ On the other hand, for each $$k$$:$$\left\{ \tau+\alpha r\left(q\mathbb{Z}+k\right):\tau\in\mathbb{Q}\right\} =\left\{ \tau+\alpha rk+\alpha rq\mathbb{Z}:\tau\in\mathbb{Q}\right\}$$ and, since $$\tau\mapsto\tau+\alpha rk$$ is a bijection of $$\mathbb{Q}$$, the logic of (1) would suggest that:$$\left\{ \tau+\alpha r\left(q\mathbb{Z}+k\right):\tau\in\mathbb{Q}\right\} =\left\{ \tau+\alpha rq\mathbb{Z}:\tau\in\mathbb{Q}\right\}$$ for all $$k$$. But then, that gives:$$\sum_{k=0}^{q-1}\sum_{\tau\in\mathbb{Q}/\alpha rq\mathbb{Z}}f\left(\tau+\alpha rk\right)=\sum_{t\in\mathbb{Q}/r\mathbb{Z}}f\left(\alpha t+\beta\right)=q\sum_{\tau\in\mathbb{Q}/\alpha rq\mathbb{Z}}f\left(\tau\right)$$ which hardly seems right.

• You had to make your $f,g_n,a_n,b_n,d_n$ concrete to get help. The Fourier transform (over $\Bbb{Q/Z}$ for any topology, not only the discrete one..) will not simplify in general, of course for some particular $g_n,a_n/d_n$ it will – reuns Jun 27 '19 at 23:12
• I don't follow. What do you mean "will not simplify in general"? – MCS Jun 27 '19 at 23:16