What is a $k$-continuously derivable function that is not $k+1$-continuously derivable

Question

Do we have $C^{\infty}\subsetneq C^k\subsetneq ...\subsetneq C^1\subsetneq C^0\ \forall k\in\Bbb N$?

Is it possible to find for each $k$ some function $f_k:\Bbb R \mapsto \Bbb R$ such that $f_k\in C^k$ but $f_k\notin C^{k+1}$

$\mid x^k\mid\in C^{k-1}$ and $\notin C^{k}$ works when $k$ is odd. Is there an example for $k$ even?

Answer 1

If $f \in C^{k}$ but $f \notin C^{k+1}$, then setting $F(x) = \int_0^{x}f(t)\, dt$, $F \in C^{k+1}$ but not in $C^{k+2}$. So as long as you can find $f \in C^{0}$ but not $C^{1}$, you can repeatedly integrate.

For a concrete example similar to what you were going for, you can set $f_{k} = x^{k} \cdot |x|$. Then for $k > 0$, $\frac{d}{dx}f_{k}(x) = (k+1)f_{k-1}(x)$, so $f_{k} \in C^{k}$ but not in $C^{k+1}$.

Answer 2

Take some function $f: \mathbb{R} \to \mathbb{R}$ which is differentiable $(n+1)$ times, and find its $n$th Taylor polynomial $P_n(x)$. Then form the function $$ g(x) = \begin{cases} f(x) & \text{when } x \leq 0 \\ P_n(x) & \text{when } x > 0 \end{cases}$$ By construction, we have that the $n$th derivative of these functions at $x = 0$ agree, but the $(n+1)$th derivative of $P_n$ is zero.

For an explicit example, consider $f(x) = e^x$, and $P_n(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}$. Then $g$ is differentiable $n$ times, but the $(n+1)$th derivative of $g$ has value $1$ at $x = 0$, and $0$ just to the right of $0$.