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For a $p$-adic number $a in frac{n}{p^k}+mathbb{Z}_psubset mathbb{Q}_p$ let $exp(2i pi a) =exp(2i pi frac{n}{p^k})$ and $psi_a(x) = exp(2i pi ax)$. Then $$Hom(mathbb{Z}_p,mathbb{C}^times) = { psi_a , a in mathbb{Q}_p/mathbb{Z}_p}$$

• For $f,h$ (uniformly) continuous $ mathbb{Z}_p to mathbb{C}$ let $langle f,h rangle =lim_{k to infty} p^{-k} sum_{n=0}^{p^k -1} f(n) overline{h(n)}$




• $f ast h(x) = langle f(x-.),overline{h} rangle $




• We obtain that
  $$lim_{k to infty} sum_{n=0}^{p^k -1} langle f,psi_{n/p^k}rangle psi_{n/p^k}(x) = lim_{k to infty} f ast sum_{n=0}^{p^k -1} psi_{n/p^k}(x)= lim_{k to infty} f ast p^k 1_{x in p^k mathbb{Z}_p} = f(x)$$
  and the convergence is uniform.



• 
  

  This is to be interpreted as a particular order of summation for the Fourier series $$sum_{a in mathbb{Q}_p/mathbb{Z}_p}langle f,psi_arangle psi_a(x) tag{1}$$

  
  
  

  
  

  Question : when does $(1)$ converge absolutely ?

  
  

  
  



• If $f$ is locally constant, that is for some $m$, $f(x) = f(x bmod p^m)$ then its Fourier series is a finite sum $sum_{a in p^{-m}mathbb{Z}_p/mathbb{Z}_p}langle f,psi_arangle psi_a(x)$ as $langle f, psi_{n/p^k} rangle = 0$ whenever $ k >m$ ($pnmid n$).




• With $|f|_2^2 = langle f,f rangle$ we have the Hilbert space $L^2(mathbb{Z}_p)$ of which ${ psi_a , a in mathbb{Q}_p/mathbb{Z}_p}$ is an orthonormal basis so $f in C^0(mathbb{Z}_p) implies f in L^2(mathbb{Z}_p) implies sum_{a in mathbb{Q}_p/mathbb{Z}_p}|langle f,psi_arangle|^2 = |f|_2^2$. Unlike the derivative $L^2(mathbb{R}/mathbb{Z})to L^2(mathbb{R}/mathbb{Z})$ there is no obvious unbounded operator such that $T(f) = sum_{a in mathbb{Q}_p/mathbb{Z}_p}c(a) langle f,psi_arangle psi_a$ and $sum_{a in mathbb{Q}_p/mathbb{Z}_p}frac{1}{|c(a)|^2} < infty$. If that operator could be easily defined we'd have $T(f) in L^2(mathbb{Z}_p) implies $ $$sum_{a in mathbb{Q}_p/mathbb{Z}_p}|langle f,psi_arangle| le (sum_{a in mathbb{Q}_p/mathbb{Z}_p}|langle T(f),psi_arangle|^2)^{1/2}(sum_{a in mathbb{Q}_p/mathbb{Z}_p}frac{1}{|c(a)|^2})^{1/2} = C |T(f)|_2$$ so $(1)$ would converge absolutely.

$defZZ{mathbb{Z}}defQQ{mathbb{Q}}$The Fourier series need not converge absolutely. First, some notation. Let $A_n = p^n ZZ_p subseteq ZZ_p$ and let $B_n = p^{-n} ZZ_p/ZZ_p subset QQ_p/ZZ_p$. Let $chi_S$ denote the characteristic function of a set $S$, meaning that $chi_S(s)=1$ if $s in S$ and $chi_S(s)=0$ otherwise. Then the Fourier transform of $chi_{A_n}$ is $p^{-n} chi_{B_n}$.

Let $r_n$ be any sequence of real numbers such that $sum_{n=0}^{infty} r_n$ is conditionally convergent. Then $sum_{n=0}^{infty} r_n chi_{A_n}$ is a continuous function on $ZZ_p$. (Because, if $s in A_m setminus A_{m+1}$, then $sum_{n=0}^{infty} r_n chi_{A_n}(s)=sum_{n=0}^m r_n$. If $s to 0$ then $m to infty$, so this approaches $lim_{m to infty} sum_{n=0}^m r_n$, which is the definition of $sum_{n=0}^{infty} r_n$.)

The Fourier transform of $sum_{n=0}^{infty} r_n chi_{A_n}$ is $sum_{n=0}^{infty} r_n p^{-n} chi_{B_n}$. If $t in B_m setminus B_{m-1}$, then this is $sum_{n=m}^{infty} r_n p^{-n}$ (which is absolutely convergent, since $r_n to 0$). If we sum over all points $t in B_m setminus B_{m-1}$, we get
$$c_m:=(1-1/p) sum_{n=m}^{infty} r_n p^{m-n}.$$
If the Fourier series were absolutely convergent, then in particular we could clump together the $t$'s in the same $B_m setminus B_{m-1}$ and then order the $m$'s arbitrarily. In other words, $sum c_m$ would be aboslutely convergent.

But it is easy to come up with $r_n$ such that $sum c_m$ is not absolutely convergent. For example, the standard $tfrac{(-1)^n}{n}$ works.

Of course, what you wanted was a condition under which the sum would be absolutely convergent, but you seemed unclear as to whether this would always happen, so it seemed worth pointing out that the answer is "no". I suspect there should be some sort of condition in terms of the modulus of continuity of the function; I'll think about it a bit.