1

Let $f : \mathbb{Z}_p\to \mathbb{R}$, where $\mathbb{Z}_p$ are the $p$-adics. I have encountered the class of all $f$ satisfying

$$ \forall x\in\mathbb{Z}_p, f(x) = \lim_{n\to\infty}f(x\bmod p^n) $$ in conversation with someone. Is this some previously-known notion? I know the $p$-adics can be written as an (inverse) limit. This condition feels like it is saying that $f$ is compatible with this limit (at least in some asymptotic sense), e.g. seems fairly natural.

Also, I call this a notion of continuity because clearly $x = \lim_{n\to\infty} x\bmod p^n$, so this property allows one to commute $f(x) = f(\lim_{n\to\infty} x_n) = \lim_{n\to\infty}f(x_n)$, at least for $x_n := x\bmod p^n$.

Mark Schultz-Wu
  • 7,234
  • What does "$f(x\bmod p^n)$" even mean? The domain of $f$ is $\mathbb Z_p$. What well-defined $p$-adic number is denoted by the notation "$x\bmod p^n$", for $x \in \mathbb Z_p$? – Torsten Schoeneberg Jul 11 '24 at 20:51
  • @TorstenSchoeneberg For $x = \sum_{i\geq 0} x_ip^i$, $x\bmod p^n = \sum_{i = 0}^{n-1}x_ip^i$. – Mark Schultz-Wu Jul 11 '24 at 21:31
  • So I read in your part 2, thank you. But see comments under there. This map is still ill-defined until we fix what set the coefficients $x_i$ are to come from, and I guess this is related to the fact the criterion does not "fully" catch continuity. – Torsten Schoeneberg Jul 11 '24 at 21:35
  • I don't think the choice of coefficients matter. The $p$-adics are an inverse limit of $\mathbb{Z}/p^n\mathbb{Z}$. They therefore come with a morphism $\psi_n:\mathbb{Z}_p\to\mathbb{Z}/p^n\mathbb{Z}$ for each $n$. To check if $x_0$ is $0$, check if $\psi_1(x)$ is the $0$ coset. To check if $x_1$ is 0, check if $\psi_2(x) - \psi_1(x)$ is the 0 coset. Of course, $\psi_2(x)-\psi_1(x)$ doesn't literally make sense, but as $\mathbb{Z}_p$ is an inverse limit you should be able to map between elements in $\mathbb{Z}/p^2\mathbb{Z}$ and $\mathbb{Z}/p\mathbb{Z}$in a way that makes this all well-defined. – Mark Schultz-Wu Jul 11 '24 at 21:40
  • My point is that for one and the same function $f$ and point $x\in \mathbb Z_p$, this criterion will be met for some choice of coefficients / truncation, but not for others. – Torsten Schoeneberg Jul 12 '24 at 22:56
  • @TorstenSchoeneberg Can you describe an example? The comment I left above is my attempt at giving a definition that is independent of any choice of representatives. – Mark Schultz-Wu Jul 13 '24 at 00:08
  • I have summarized my thoughts on this in an answer to the follow-up question https://math.stackexchange.com/q/4944341/96384. This contains examples as you asked for; in fact, my whole point in the answer is that coefficients DO matter. – Torsten Schoeneberg Aug 16 '24 at 17:28

1 Answers1

3

Edit #2: My apologies! This argument doesn't actually work at all and your counterexample works over $\mathbb{R}$ as well; if we take $d(x)$ to be the number of nonzero $p$-adic digits then $f(x) = e^{-d(x)}$ satisfies the desired condition but is not continuous.

The issue is exactly lack of uniformity. For example if we consider $x = 1 - 2^n = \dots 111000 \dots 1 \in \mathbb{Z}_2$ then $f(x \bmod 2^k)$ does converge to $f(x) = 0$ but that converge doesn't begin until $k = n + 1$. So the $K(\varepsilon)$ I was looking for below don't exist and the compactness argument doesn't produce them (the bound only applies to the finitely many $x$ in the open subcover, not to arbitrary $x$).

This is equivalent to saying that $f$ is continuous with respect to the $p$-adic topology on $\mathbb{Z}_p$, which is the topology induced by the $p$-adic metric.

Edit: I'm not actually sure what the cleanest way to prove this is but here's one approach. The convergence $f(x \bmod p^k) \to f(x)$ means that for every $x \in \mathbb{Z}_p$ and every $\varepsilon > 0$ we can find a positive integer $K_x(\varepsilon)$ such that if $k \ge K_x(\varepsilon)$ then $|f(x) - f(x \bmod p^k)| < \varepsilon$.

Now we can consider the open cover of $\mathbb{Z}_p$ by balls of radius $p^{-K_x(\varepsilon)}$ centered at $x$; by compactness this has a finite subcover, from which it follows that we can find a positive integer $K(\varepsilon)$ (namely the maximum of $K_x(\varepsilon)$ over the finitely many points $x$ in this finite subcover) such that if $k \ge K(\varepsilon)$ then $|f(x) - f(x \bmod p^k)| < \varepsilon$, and we have removed the dependence on $x$.

It follows that for any $\varepsilon > 0$, if $k \ge K \left( \frac{\varepsilon}{2} \right)$ and $|x - y| \le p^{-k}$ then $x \equiv y \bmod p^k$ so $f(x \bmod p^k) = f(y \bmod p^k)$, and we also have $|f(x) - f(x \bmod p^k)| < \frac{\varepsilon}{2}$ and the same for $y$. This gives $|f(x) - f(y)| < \varepsilon$. So $f$ is uniformly continuous, hence continuous.

(I don't know if this detour into uniform continuity is necessary.)

Qiaochu Yuan
  • 468,795