1. ## Non-Archimedian calculus

When doing calculus with respect to a non-Archimedian metric (such as the p-adic metric), is it true that differentiability does NOT imply continuity (example: the p-adic norm seems to be differentiable but not continuous)?

2. ## Re: Non-Archimedian calculus

Originally Posted by SlipEternal
When doing calculus with respect to a non-Archimedian metric (such as the p-adic metric), is it true that differentiability does NOT imply continuity (example: the p-adic norm seems to be differentiable but not continuous)?
Are you thinking in these terms ?.

3. ## Re: Non-Archimedian calculus

Yes, the p-adic metric is an example of an ultrametric.

4. ## Re: Non-Archimedian calculus

Never mind. I see what I was doing wrong. The p-adic norm is neither continuous nor differentiable at 0. So, differentiability still implies continuity.

5. ## Re: Non-Archimedian calculus

I spoke too soon. The p-adic norm was just a poor example. Instead, consider $f:\Bbb{Q}_p \to \Bbb{Q}_p$ defined by $f(x) = x|x|_p$ (where $\Bbb{Q}_p$ is the set of p-adic numbers, and $|x|_p$ is the p-adic norm of $x$).

First, I will consider continuity at 0. Consider the sequence $a_n = p^n$. This sequence converges to 0 with respect to the p-adic metric, but $f(a_n) = 1$ for all $n$ while $f(0) = 0$. Hence, $f(x)$ is not continuous at $x=0$. Yet:

$f'(0) = \lim_{|h|_p \to 0} \dfrac{f(h) - f(0)}{h} = \lim_{|h|_p \to 0} \dfrac{h|h|_p}{h} = \lim_{|h|_p \to 0} |h|_p = 0$

Hence, the derivative exists at $x=0$. So, differentiability does not imply continuity. Did I make a mistake somewhere?

Edit: In general, if $x \neq 0$, then if $|h|_p<|x|_p$, $|x+h|_p = |x|_p$, so
\begin{align*}\lim_{|h|_p \to 0} \dfrac{f(x+h)-f(x)}{h} & = \lim_{|h|_p \to 0} \dfrac{(x+h)|x+h|_p-x|x|_p}{h} \\ & = \lim_{|h|_p \to 0} \dfrac{(x+h)|x|_p - x|x|_p}{h} \\ & = \lim_{|h|_p \to 0} |x|_p = |x|_p\end{align*}

Hence, $f'(x) = |x|_p$ agrees with the derivative found for $x=0$, as well.

6. ## Re: Non-Archimedian calculus

Ohhh!!! I see what I am doing wrong. I am looking at convergence in the reals rather than the p-adics. $\left| |h|_p \right|_p \to \infty$ as $|h|_p \to 0$, so $f'(0)$ does not exist. My mistake.