If $f^{(n)}(x)$ continous at a certain point ,then $f^{(n)}(x)$ continous over a neighbourhood of the point?

Question

Let $n$ is a non-negative integer,$f$ diferentiable $n$ times in a neighbourhood of $x_{0}\in \mathbf{R},(x_{0}-\delta ,x_{0}+\delta ),$ and $f^{(n)}(x),$ the $n$th derivative of $f$ ,is continuous at $x=x_{0}.$ Is there a real number $\delta_{1},0<\delta_{1}<\delta,$ such that $f^{(n)}(x)$ is continous over $(x_{0}-\delta_{1} ,x_{0}+\delta_{1} )?$

When $n=0,$it is not difficulty to find a counterexample:

Let $$x_{0}=0,\qquad f(x)=\begin{cases} x^{2}& \text{if}\quad x \text { is a irratioanl number},\\ 0& \text{if}\quad x \text { is a rational number}. \end{cases} \quad $$

Go a step further，I need to find some counterexamples for higher orders, but I'm stumped for that .Can anyone give me any hints on how to start it? Any help will be appreciated.

Answer 1

Suggestion: Given $a<b,$ there exists a differentiable $g$ on $[a,b]$ such that $g'$ is bounded on $[a,b],$ $g'(a) = g'(b)=0,$ and $g'$ is discontinuous at $(a+b)/2.$ For example, on $[-1,1]$ we could take

$$g(x) = (x+1)^2(x-1)^2 x^2 \sin(1/x)$$

(with $g(0) =0$ of course). Now let $a_1 > a_2 > \cdots \to 0.$ For each $n$ we can choose a $g_n$ on $[a_{n+1},a_n]$ that does the above on this interval. Paste these $g_n$'s together and perhaps you'll see how to finish up for an example in the case $n=1.$ (This should yield examples for $n> 1$ as well.)