Let $p$ and $q$ be prime numbers. When $p=q$, Mahler's Theorem gives a complete description of $C\left(\mathbb{Z}_{p};\mathbb{Z}_{p}\right)$, the space of continuous functions from $\mathbb{Z}_{p}$ to $\mathbb{Z}_{p}$. I'm wondering (possibly in vain) if there might be a comparable classification of $C\left(\mathbb{Z}_{p};\mathbb{Z}_{q}\right)$ when $p$ and $q$ are distinct.

I ask only because I've been doing $p$-adic harmonic analysis, but have found myself having to brave the wilds of $L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)$, the space of all $f:\mathbb{Z}_{p}\rightarrow\mathbb{C}_{q}$ so that:$$\sup_{\mathfrak{z}\in\mathbb{Z}_{p}}\left|f\left(\mathfrak{z}\right)\right|_{q}<\infty$$

Pontryagin duality lets me do Fourier analysis on $L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}\right)$; for $p=q$, on the other hand, I can use things like the volkenborn integral, or the amice transform / mazur-mellin transform—$p$-adic distributions, in general. The problem is, without a structure theorem like Mahler's for the $p\neq q$ case, though I can define “integration” on $L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)$ by elements of its dual space (continuous functionals $\varphi:L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)\rightarrow\mathbb{C}_{q})$, I don't see a way to do useful computations for the specific, non-abstract functions that I'm trying to fourier analyze.

So, I guess what I'm really asking is: how do you take the "integral" or "fourier transform" of such a function?

Any thoughts? Reference recommendations? Etc.?


1 Answer 1


${\mathbb Z}_p$ and ${\mathbb Z}_q$ are homeomorphic; hence so are $C({\mathbb Z}_p,{\mathbb Z}_p)$ and $C({\mathbb Z}_p,{\mathbb Z}_q)$.

  • 1
    $\begingroup$ Okay then... what is the formula for the homeomorphism? Because it sure ain't the identity map. xD $\endgroup$
    – MCS
    Jan 31, 2020 at 3:31
  • 2
    $\begingroup$ @MCS This answer may be useful. $\endgroup$
    – Mark
    Jan 31, 2020 at 4:39
  • 2
    $\begingroup$ Every compact, totally disconnected metric space without isolated points is homeomorphic to the Cantor set. $\endgroup$ Jan 31, 2020 at 16:04

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