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.

  • $\begingroup$ 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 $\endgroup$ – reuns Jun 27 '19 at 23:12
  • $\begingroup$ I don't follow. What do you mean "will not simplify in general"? $\endgroup$ – MCS Jun 27 '19 at 23:16

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