$p$-adic differentiation

Question

$p$-adic norm is not obtained via inner product. Is it still possible to differentiate $|x|_p$ with respect to $x\in\mathbf{Q}_p$ by thinking of the norm as a function of $x$?

Answer 1

A Rough idea:

Given a a continuous function $f: \mathbb{R} \to \mathbb{R}$, an antiderivative is defined by $$F(x)=\int_0^x f(t)dt.$$ In $p$-adic world, we have the following theorem.

Theorem (Dieudonne): Any continuous function $f: \mathbb{Z}_p \to \mathbb{Q}_p$ has a continuously differentiable antiderivative.

The differentiation of $f$ at $a \in \mathbb{Z}_p$ is defined by $$f'(a)= \frac{f(x)-f(y)}{x-y},$$ as $(x,y) \to (a,a)$ and $(x,y) \in \mathbb{Z}_p \times \mathbb{Z}_p $ with $x \neq y$.

If $f: \mathbb{Z}_p \to \mathbb{Q}_p$ is defined by $f(\sum a_np^n)=\sum a_np^{2n}$, then $f(x)=\sum a_np^{2n}$ and $f(y)=\sum b_n p^{2n}$ where $x=\sum a_np^n$ and $y=\sum a_np^n$.

Then you can get $p$-adic differentiation of $f$ at $a \in \mathbb{Z}_p$ using the above formula taking $f=|x|_p$.

Answer 2

The function $$N:\Bbb{Q}_p^*\to \Bbb{R}, \qquad N(x)= |x|_p$$ is locally constant.

The formula $$N'(x)=\lim_{h\to 0} \frac{N(x+h)-N(x)}{h}$$ doesn't hold/apply because you can't divide real numbers by $p$-adic numbers.

Something weird happens at $x=0$.

As a metric space you can embed $\Bbb{Q}_p$ into $\Bbb{R}$, through $\iota(\sum_n a_n p^n) = \sum_n a_n (p+1)^{-n},a_n\in 0,\ldots p-1$, then $N\circ \iota^{-1}$ is locally constant except at $0$ where it is not differentiable (for $p=3$, $\lim_{n\to \infty} \frac{(p+1)^{-n}}{N(p^{n})}\ne \frac{2(p+1)^{-n}}{N(p^{n})}$)

Answer 3

As reuns points out, the absolute value function $x \mapsto \lvert x \rvert_p$ is naturally seen as a function with domain $\mathbb Q_p$ and codomain $\mathbb R$, and for such a function it is not at all clear how to even define a derivative.

However, for functions whose domain and codomain are both subsets of one common ultrametric field (like $\mathbb Q_p$), it is quite possible to define a derivative via the differential quotient just like in old school calculus. As M.A. SARKAR alludes to, it turns out very early in this ultrametric analysis though that the so-called notion of "strict differentiability" is much better adapted to this setting than that "naive" differential quotient; that is, at a given non-isolated point $a$ in the domain of $f$ one should better investigate the limit

$$f'(a) := \lim_{(x,y) \to (a,a)} \frac{f(x)-f(y)}{x-y}.$$

(To see why this notion works better, I remember a good explanation in W. Schikhof's book Ultrametric Calculus which is a good resource for all this anyway.)

Be that as it may, let's take a second look at the absolute value function $x \mapsto \lvert x \rvert_p$. It turns out that its image is $p^{\mathbb Z} \cup \lbrace 0 \rbrace \subset \mathbb Q$ and since there is a canonical injection $\mathbb Q \subset \mathbb Q_p$, we can view this as function $f: \mathbb Q_p \rightarrow \mathbb Q_p$ and apply the notions of naive or strict differentiability to it.

However, as reuns also points out, at every $x \in \mathbb Q_p \setminus \lbrace 0 \rbrace$, this function is locally constant. To be precise, given any such $x$, if we choose $0 < r < \lvert x \rvert_p$, then $f(y) = f(x)$ for all $y \in B_r(x) := \lbrace z \in \mathbb Q_p: \lvert x-z\rvert_p < r \rbrace$ by the ultrametric principle. This immediately entails that both the naive differential quotient, and the above limit one looks at for strict differentiability, are just $=0$ for all $a \in \mathbb Q_p \setminus \lbrace 0 \rbrace$.

At the point $x=0$ however, the function is very badly behaved; in fact, it is not even continuous there if we use the $p$-adic topology on both domain and codomain. E.g. for the sequence $a_n := p^n$, we have $\lim_{n\to \infty} a_n=0$ in the $p$-adic topology on the domain, but $f(a_n) = p^{-n}$ is not a convergent sequence, especially not converging to $f(0)=0$, in the $p$-adic topology on the codomain. This entails also that neither the naive nor the strict differentiability limit exist at $x=0$.

So one could say, via this interpretation, that the absolute value function has derivative $0$ at every $x \neq 0$, and is not differentiable at $x=0$. However, it is dubious (as is actually shown by this weird result) whether such an interpretation of the absolute value function as a function from $\mathbb Q_p$ to $\mathbb Q_p$ (and both viewed as metric spaces w.r.t. the $p$-adic metric) is very meaningful. I doubt that for any serious math, it is ever helpful to view the absolute value function as a function whose codomain does not inherit its topology from the real numbers with their standard metric.

Answer 4

I just managed to work this out:

The Vladimirov derivative $D^{n}f(x)=\frac{1}{Γ_p(−n)}\int\limits_{Q_p}\frac{f(x)−f(y)}{|x−y|^{1+n}_p}d_py$ is indeed defined for $f(x)=|x|_p^{\mu}$, where $\Gamma_p(a)=\frac{1-p^{a-1}}{1-p^{-a}}$ is the $p$-adic gamma function.

To evaluate the derivative first shift $y$ to $y'=x-y$, then redefine $y''=xy'$. This brings the $x$-dependent factor out of the integral, $$D^n|x|^{\mu}_p=\frac{1}{Γ_p(−n)} |x|^{\mu-n}_p\int\limits_{Q_p}\frac{\big(1-|1-y''|^{\mu}_p\big)}{|y''|^{1+n}_p}d_py''.$$ The integral can be evaluated using the usual rules in terms of $p$-adic gamma functions, yielding $$ D^n |x|^{\mu}_p=\frac{\Gamma_p(1+\mu)}{\Gamma_p(1+\mu-n)}|x|^{\mu-n}_p.$$

Note that this generalises $$\left(\frac{d}{dx}\right)^nx^{\mu}= \frac{\Gamma(1+\mu)}{\Gamma(1+\mu-n)}x^{\mu-n}$$ on the real line.

Thank you for your responses.