Let $f: \mathbb A/\mathbb Q \rightarrow \mathbb C$ be a smooth function. Smooth means that $f$ is continuous, smooth in the archimedean argument, for every $(x_0,y_0) \in \mathbb A = \mathbb R \times \mathbb A_f$, there exists a neighborhood $W$ of $(x_0,y_0)$ such that $f(x,y) = f(x,y_0)$ for all $(x,y) \in W$.

Being continuous, $f$ has a Fourier expansion

$$f(x) = \sum\limits_{\alpha \in \mathbb Q} c_{\alpha} \psi_{\alpha}(x) \tag{1} $$

I want to understand why this Fourier series converges absolutely in the smooth case.

This answer to my previous question explained that what I want to know reduces to the same question about $\mathbb R/\mathbb Z$: $f$ identifies with a smooth function on a torus: it can be shown that there exists an open compact subgroup $H$ of $\mathbb A_f$ such that $f$ is trivial on $H+\mathbb Q$. As a subgroup of $\mathbb R$, $H \cap \mathbb Q$ is discrete, and strong approximation gives an isomorphism of topological groups

$$\mathbb R/(H \cap \mathbb Q) \rightarrow \mathbb A/(H+\mathbb Q)$$

where the left hand side is a torus. Now as a smooth function on $\mathbb R/(H \cap \mathbb Q)$, $f$ has an absolutely convergent Fourier expansion: if $a$ generates the cyclic group $H \cap \mathbb Q$, then

$$f(x) = \sum\limits_{n \in \mathbb Z} d_n e^{2\pi i a^{-1}nx} \tag{2}$$

Is it really possible to use the absolute convergence of (2) to justify the absolute convergence of (1)? We have

$$c_{\alpha} = \int\limits_{\mathbb A/k} f(x) \psi(-x\alpha) dx$$

$$d_n = \int\limits_{\mathbb R/H \cap \mathbb Q} f(x_{\infty})e^{-2\pi i a^{-1}nx_{\infty}}dx_{\infty}$$

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