Absolute convergence of Fourier series of periodic adelic function I've asked several questions about this topic before, including here and here.  I've gotten many helpful responses, but I still do not completely understand what is going on.
Let $f: \mathbb A_{\mathbb Q}/\mathbb Q \rightarrow \mathbb C$ be a smooth function.  Here smooth means that as a function of $\mathbb A_{\mathbb Q} = \mathbb R \times \mathbb A_f$, $f$ is smooth in the first argument and "uniformly locally constant" in the second argument.  This last condition means that for every point $(x_0,y_0) \in \mathbb R \times \mathbb A_f = \mathbb A_{\mathbb Q}$, there exist neighborhoods $U$ and $V$ of $x_0$ and $y_0$ such that $f(x,y) = f(x,y_0)$ for all $x \in U$ and $y \in V$.
Let $\Psi$ be a fixed nontrivial character of $\mathbb A_{\mathbb Q}/\mathbb Q$.
Being continuous and therefore square integrable, $f$ has a Fourier expansion
$$f(x) = \sum\limits_{\alpha \in \mathbb Q}c_{\alpha} \Psi(\alpha x) \tag{1} $$
for uniquely determined complex numbers $c_{\alpha}$.  The sum (1) on the right converges to $f$ in the $L^2$-norm on $\mathbb A_{\mathbb Q}/\mathbb Q$.  We can recover the coefficients $c_{\alpha}$ by the usual orthogonality relations:
$$c_{\alpha} = \int\limits_{\mathbb A_{\mathbb Q}/\mathbb Q}f(x)\Psi(-\alpha x)dx$$

What I want to understand:

*

*1 . Under the assumption that $f$ is smooth, show that the right hand side of (1) converges absolutely (and therefore (1) is actually a pointwise, even a uniform, limit).


*2 . Identify $f$ as a smooth function $\phi$ on some torus $T$, as explained in the links above (which I still don't totally understand).


*3 . Relate the adelic Fourier coefficients of $f$ to the classical Fourier coefficients of $\phi$.

For #1, I know that from the compactness of $\mathbb A/\mathbb Q$, it follows that there is an open compact subgroup $H$ of $\mathbb A_f$ such that $f(x+h) = f(x)$ for all $x \in \mathbb A = \mathbb A_{\mathbb Q}$ and $h \in H$.  It follows that $f$ is well defined on the quotient $\mathbb A/(i(\mathbb Q) + H)$, where $i$ is the diagonal embedding of $\mathbb Q$.
Special case I have worked out: I have worked out the special case $H = \prod\limits_p \mathbb Z_p$.  We may assume $\Psi$ is the canonical additive character $(\Psi_{\infty}(x) = e^{2\pi i x}$, and $\Psi_p(x) = e^{2 \pi i [x]}$, where $[x]$ is the reduction of $x$ in $\mathbb Q_p/\mathbb Z_p = \mathbb Q/\mathbb Z$).  The inclusion $\mathbb R \rightarrow \mathbb A$ induces an isomorphism of topological groups
$$\mathbb R/\mathbb Z \rightarrow \mathbb A/(i(\mathbb Q)+H)$$
Injectivity comes from the fact that $\mathbb Z = \mathbb Q \cap H$.  Surjectivity comes from the equation $\mathbb A = \mathbb R + H + i(\mathbb Q)$.  Now we can identify $f$ as a function on $\mathbb R/\mathbb Z$, which by smoothness has an absolutely convergent Fourier expansion
$$f(x_{\infty}) = \sum\limits_{n \in \mathbb Z} d_n \Psi_{\infty}(nx_{\infty})$$
$$d_n = \int\limits_{\mathbb R/\mathbb Z} f(x_{\infty})e^{-2\pi i n x_{\infty}}dx_{\infty}$$
Now for $\alpha \in \mathbb Q$, we have
$$c_{\alpha} = \int\limits_{\mathbb A/\mathbb Q} f(x)\Psi(-\alpha x)dx = \int\limits_{(\mathbb A/\mathbb Q)/((i(\mathbb Q)+H)/\mathbb Q)} \space \space \int\limits_{(i(\mathbb Q)+H)/\mathbb Q} f(x+b)\Psi(-\alpha(x+b))\space  db \space dx$$
Using the third isomorphism theorem, the isomorphism $(i(\mathbb Q)+H)/\mathbb Q = H$, and the fact that $f$ is invariant under $i(\mathbb Q)+H$, this is equal to
$$\int\limits_{\mathbb A/(i(\mathbb Q)+H)} f(x)\Psi(-\alpha x) \int\limits_{H}\Psi(-\alpha b) \space db \space dx = \Bigg( \int\limits_{\mathbb R/\mathbb Z} f(x_{\infty}) \Psi_{\infty}(-\alpha x_{\infty}) \space dx_{\infty} \Bigg) \Bigg( \int\limits_H \Psi(-\alpha b) db \Bigg)$$
$$ = \Bigg( \int\limits_{\mathbb R/\mathbb Z} f(x_{\infty}) e^{-2\pi i \alpha x_{\infty}} \space dx_{\infty} \Bigg)\Bigg( \int\limits_H \Psi(-\alpha b) db \Bigg)$$
Now $b_p \mapsto \Psi(-\alpha b_p)$ is the trivial character on $\mathbb Z_p$ if and only if $\alpha \in \mathbb Z_p$.  It follows that
$$ \int\limits_H \Psi(-\alpha b) db  = \prod\limits_p \int\limits_{\mathbb Z_p} \Psi_p(-\alpha b_p)db_p = \begin{cases} 1 & \textrm{ if  $\alpha \in \mathbb Z$} \\ 0 & \textrm{if $\alpha \not\in \mathbb Z$} \end{cases}$$
Therefore, the adelic Fourier coefficients $c_{\alpha}$ are related to the classical Fourier coefficients $d_n$ by
$$c_{\alpha} = \begin{cases} d_{\alpha} & \textrm{ if $\alpha \in \mathbb Z$} \\ 0 & \textrm{ if $\alpha \not\in \mathbb Z$} \end{cases}$$
This gets us #1 and #3 in this special case.
 A: I have worked out the case of a general $H$, which turns out to not be so different from what I did above.  It turns out exactly as user reuns suggested in his comments on my mathoverflow question about the same topic.
So there is an open compact subgroup $H$ of $\mathbb A_f$ such that $f(x+h) = f(x)$ for all $x \in \mathbb A$ and $h \in H$.  We may shrink $H$ and assume it is of the form
$$p_1^{e_1}\mathbb Z_{p_1} \times \cdots \times p_s^{e_s}\mathbb Z_{p_s} \times \prod\limits_{p \neq p_i} \mathbb Z_p$$
If we let $H_0 = \prod\limits_p \mathbb Z_p$, with $m = p_1^{e_1} \cdots p_s^{e_s}$, then $[H_0 : H] = m$, and $(n_1, ... , n_s, 0, ...)$ with $0 \leq n_j < p_j^{e_j}$ is a complete set of representatives for $H$ in $H_0$.  We see that 
$$(x_{\infty},n_1, ... , n_s) \mapsto (x_{\infty}, n_1, ... , n_s, 0, ... )$$
induces an isomorphism of topological groups
$$\mathbb R/m\mathbb Z \times \mathbb Z/p_1^{e_1}\mathbb Z \times \cdots \mathbb Z/p_s^{e_s}\mathbb Z \longrightarrow \mathbb A/(\mathbb Q+H)$$
Injectivity is straightforward, and surjectivity follows by the same reasoning that $\mathbb R/\mathbb Z \cong \mathbb A/(\mathbb Q + H_0)$: the key fact is that $\mathbb A = \mathbb R + \mathbb Q + H_0$.
Our assumption on $f$ is that each map $x_{\infty} \mapsto f(x_{\infty}, n_1, ... , n_s)$
is smooth and $m\mathbb Z$-periodic, and therefore for each $\underline{n} = (n_1, ... , n_s)$, we have an absolutely convergent Fourier expansion
$$f(x_{\infty}, n_1, ... , n_s) = \sum\limits_{k \in m\mathbb Z} d_{\underline{n},k}e^{2\pi i kx_{\infty}/m}$$
$$d_{\underline{n},k} = \int\limits_{\mathbb R/m\mathbb Z} f(x_{\infty}, n_1, ... , n_s)e^{-2\pi i kx_{\infty}/m}$$
Now we can relate these classical Fourier coefficients to the adelic Fourier coefficients.  We choose the character $\Psi$ so that...  By the same reasoning as in the special case in the statement of my question (skipping a few details), we have
$$c_{\alpha} = \int\limits_{\mathbb A/\mathbb Q} f(x) \Psi(-\alpha x)dx = \int\limits_{\mathbb A/(\mathbb Q+H)} f(x)\Psi(-\alpha x) \int\limits_H \Psi(-\alpha h) \space dh \space dx$$
Now
$$\int\limits_H \Psi(-\alpha h) \space dh = \prod\limits_{i=1}^s \int\limits_{p_i^{e_i}\mathbb Z_{p_i}} \Psi_{p_i}(\alpha h_{p_i}) dh_{p_i} \prod\limits_{p \neq p_i} \int\limits_{\mathbb Z_p} \Psi_p(\alpha h_p) dh_p = \begin{cases} 1 & \textrm{ if $\alpha \in m \mathbb Z$} \\ 0 & \textrm{ if $\alpha \not\in m\mathbb Z$} \end{cases}$$
So the Fourier coefficient $c_{\alpha}$ is zero unless $\alpha \in m\mathbb Z$.  If this is the case, then
$$c_{\alpha} = \int\limits_{\mathbb A/(\mathbb Q+ H)} f(x)\Psi(-\alpha x)dx = \sum\limits_{\underline{n}} \int\limits_{\mathbb R/m\mathbb Z} f(x_{\infty},n_1, ... , n_s) e^{-2\pi i \alpha x_{\infty}/m} dx_{\infty}$$
where the sum is over all $\underline{n} = (n_1, ... , n_s)$ with $0 \leq n_i < p_i^{e_i}$.  We have shown that 
$$c_{\alpha} = \begin{cases} \sum\limits_{\underline{n}} d_{\underline{n},\alpha} & \alpha \in m \mathbb Z \\ 0 & \alpha \not\in m\mathbb Z \end{cases}$$
and since each sum $\sum\limits_{k \in m\mathbb Z} d_{\underline{n},k}$ is absolutely convergent, so are the $c_{\alpha}$.  
