Does Taylor's theorem (Peano's form) generalize to other fields?

Question

Let $\mathbb F$ be a field with a non-trivial absolute value, such as $\mathbb F_p(X)$ or $\mathbb Q$ or $\mathbb Q_p$, and let $f:\mathbb F\to\mathbb F$ be a function. Limits, continuity, and derivatives are defined as usual: $\lim_{x\to a}f(x)=b$ means that for any real $\varepsilon>0$ there exists $\delta>0$ such that, for all $x\in\mathbb F$ where $0<|x-a|<\delta$, $|f(x)-b|<\varepsilon$.

We say that $f$ is $n$th-order Peano differentiable at $a$, if there's a polynomial $\sum_{k=0}^nc_kx^k$ and a function $h$ continuous at $h(0)=0$ such that

$$f(a+x)=\sum_{k=0}^nc_kx^k+x^nh(x).$$

Suppose $f$ has both ordinary derivatives and Peano derivatives up to order $n$ at $a$. Must they be proportional,

$$f^{(k)}(a)=k!\,c_k\quad?$$

Related: In characteristic $2$, can a function have a non-zero second derivative?

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A function may have Peano derivatives but not ordinary derivatives of order $n$. Here's an example with $\mathbb F=\mathbb R$ (I suppose a similar example could be constructed for $\mathbb F=\mathbb Q$ using $2^{\lfloor-1/x^2\rfloor}$ on the dyadic rationals):

$$f(x)=\begin{cases}e^{-1/x^2},\quad x\in\mathbb Q\setminus\{0\}; \\ 0,\quad\text{otherwise};\end{cases}$$

this satisfies $f(0+x)=0+0x+\cdots+0x^n+x^nh(x)$ where $\lim_{x\to0}h(x)=0$. But $f''(0)$ doesn't exist, because $f'(x)$ doesn't exist in a neighbourhood of $0$, because $f(x)$ is not continuous in a neighbourhood of $0$.

Conversely, a function may have ordinary derivatives but not Peano derivatives of order $n$. An example with $\mathbb F=\mathbb Q_p$ is given here:

$$f(x)=\begin{cases}p^{2m},\quad x=p^m+yp^{2m+1},\quad y\in\mathbb Z_p,\quad m\in\mathbb N; \\ 0,\quad\text{otherwise};\end{cases}$$

this satisfies $f'(x)=0$ for all $x$, hence $f^{(n)}(0)=0$ for all $n>1$. But $\lim_{x\to0}f(x)/x^2$ doesn't exist, so $f$ can't have a $2$nd-order Peano derivative; $f(0+x)=0+0x+c_2x^2+x^2h(x)$ would imply that the limit exists as $c_2$.

Answer 1

No, the derivatives after the first can both exist but with different values.

A p-adic number $x \in \mathbb{Q}_p$ with $0 < |x| < 1$ always has a normal form

$$ x = \sum_{i=k}^\infty a_i p^i $$

where $k$ is a positive integer, every $a_i \in \{0,1,\ldots,p-1\}$, and $a_k \neq 0$.

For each such $x$, define

$$ \tilde f \!(x) = x \sum_{i=2k}^\infty a_i p^i $$

Note $|\tilde{f}\!(x)| \leq p^{-3k} = |x|^3$. This implies $\lim_{x \to 0} \frac{\tilde{f}\!(x)}{x} = \lim_{x \to 0} \frac{\tilde{f}\!(x)}{x^2} = 0$.

Extend this to $f: \mathbb{Q}_p \to \mathbb{Q}_p$ by

$$ f(x) = \begin{cases} 0 & x=0 \\ \tilde f \!(x) & 0<|x|<1 \\ 0 & |x| \geq 1 \end{cases} $$

$f$ has first and second Peano derivatives $0$, since $\frac{f(x)}{x^2}$ converges to $0$ as $x \to 0$, so we can extend that quotient to a function $h$ continuous at $h(0)=0$.

For the limit derivatives, first note if $0 < |x| < 1$ and $|d| \leq |x|^2$, we can write $d = \sum_{i=2k}^\infty b_i p^i$ where every $b_i \in \{0,1,\ldots,p-1\}$ (here $b_{2k}$ can be $0$). Then

$$ f(x+d) = (x+d) \sum_{i=2k}^\infty (a_i+b_i) p^i = (x+d)\left(\frac{f(x)}{x} + d\right) $$

$$ f'(x) = \lim_{d \to 0} \frac{f(x+d) - f(x)}{d} = \lim_{d \to 0} \frac{\left(x + \frac{f(x)}{x}\right) d + d^2}{d} = x + \frac{f(x)}{x} $$

$$ f'(0) = \lim_{d \to 0} \frac{f(d)}{d} = 0 $$

$$ f''(0) = \lim_{d \to 0} \frac{f'(d) - f'(0)}{d} = \lim_{d \to 0} \frac{d + \frac{f(d)}{d}}{d} = 1 $$

So at $0$, the second Peano derivative is $0$ but the second limit-derivative is $1$.