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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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  • $\begingroup$ You need a stronger definition of smooth. Iff $f: \mathbb A/\mathbb Q \rightarrow \mathbb C$ is $H$ invariant with $H= \prod_p' p^{e_p}\Bbb{Z}_p, e_p \ge 0, \sum_p e_p < \infty$ a finite index subgroup of $\prod_p' \Bbb{Z}_p$ then let the finite group $G = \prod_p' \Bbb{Z_p}/p^{e_p} \Bbb{Z_p}$ and $i$ the diagonal embedding so that $f$ is a function on $\Bbb{A/(i(Q) \times} H) = (\prod_p' \Bbb{Z_p} \times \Bbb{R})/(i(\Bbb{Z}) \times H)= (G \times \Bbb{R})/i(\Bbb{Z}) = (G \times (\Bbb{R} / |G|\Bbb{Z}))/i(\Bbb{Z}/|G|\Bbb{Z})$. So you need uniform local-constantness not only local-constantness. $\endgroup$ – reuns Feb 24 '19 at 2:49
  • $\begingroup$ Uniform local-constantness is in the definition of smooth I gave in my question. This implies $f$ is $H$-invariant for some $H$ as I indicated. $\endgroup$ – D_S Feb 24 '19 at 4:37
  • $\begingroup$ Uniform means $W =(x_0,y_0)+H$ with $H$ not depending on $(x_0,y_0)$. You didn't specify that, this is the main point as it implies $f$ is a function on $G \times \mathbb{R}/|G| \mathbb{Z}$ ie. finitely many copies of $\cong \mathbb{R}/ \mathbb{Z}$ where absolute convergence of the Fourier is implied by having an $L^2$ derivative. $\endgroup$ – reuns Feb 24 '19 at 5:21
  • $\begingroup$ Does the compactness of $\mathbb A/\mathbb Q$ not give us uniform local constantness, given pointwise local constantness? $\endgroup$ – D_S Apr 11 '19 at 5:45
  • $\begingroup$ Sure but you need to be make clear for what metric/topology it is compact and locally constant. It won't be the obvious one. Same for (uniform) continuity and Hölder continuity and absolute convergence of the Fourier series. $\endgroup$ – reuns Apr 11 '19 at 11:37

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