# Absolute convergence of the Fourier series of a smooth adelic function

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

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

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

• 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. – reuns Feb 24 '19 at 2:49
• 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. – D_S Feb 24 '19 at 4:37
• 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. – reuns Feb 24 '19 at 5:21
• Does the compactness of $\mathbb A/\mathbb Q$ not give us uniform local constantness, given pointwise local constantness? – D_S Apr 11 '19 at 5:45
• 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. – reuns Apr 11 '19 at 11:37