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)?

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- May 10th 2014, 01:39 PMSlipEternalNon-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)?

- May 10th 2014, 02:10 PMPlatoRe: Non-Archimedian calculus
Are you thinking in these terms ?.

- May 10th 2014, 02:51 PMSlipEternalRe: Non-Archimedian calculus
Yes, the p-adic metric is an example of an ultrametric.

- May 12th 2014, 09:54 AMSlipEternalRe: 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.

- May 12th 2014, 10:38 AMSlipEternalRe: Non-Archimedian calculus
I spoke too soon. The p-adic norm was just a poor example. Instead, consider $\displaystyle f:\Bbb{Q}_p \to \Bbb{Q}_p$ defined by $\displaystyle f(x) = x|x|_p$ (where $\displaystyle \Bbb{Q}_p$ is the set of p-adic numbers, and $\displaystyle |x|_p$ is the p-adic norm of $\displaystyle x$).

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

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

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

Edit: In general, if $\displaystyle x \neq 0$, then if $\displaystyle |h|_p<|x|_p$, $\displaystyle |x+h|_p = |x|_p$, so

$\displaystyle \begin{align*}\lim_{|h|_p \to 0} \dfrac{f(x+h)-f(x)}{h} & = \lim_{|h|_p \to 0} \dfrac{(x+h)|x+h|_p-x|x|_p}{h} \\ & = \lim_{|h|_p \to 0} \dfrac{(x+h)|x|_p - x|x|_p}{h} \\ & = \lim_{|h|_p \to 0} |x|_p = |x|_p\end{align*}$

Hence, $\displaystyle f'(x) = |x|_p$ agrees with the derivative found for $\displaystyle x=0$, as well. - May 12th 2014, 11:57 AMSlipEternalRe: Non-Archimedian calculus
Ohhh!!! I see what I am doing wrong. I am looking at convergence in the reals rather than the p-adics. $\displaystyle \left| |h|_p \right|_p \to \infty$ as $\displaystyle |h|_p \to 0$, so $\displaystyle f'(0)$ does not exist. My mistake.