A question about a necessary condition for higher order derivates in a point $x_0$ of a function $I\subseteq \mathbb{R}\rightarrow \mathbb{C}$.

Question

Why is this implication below true?

$x_0$ is $n$-times differentiable in $f:I\subseteq \mathbb{R}\rightarrow\mathbb{C}\Rightarrow$ $\exists N\subseteq I$ neighbourhood of $x_0$ where $f$ is $n-1$ times differentiable.

I want to prove it or at least understand why. I could not find any explaination in the book, Maybe it is something obvious. But when I have tried to prove it I did not see the crucial Observation,

I have set up a proof by contradiction, by writing out the definitions. Please help me to finish the proof or tell me how to proceed:

Suppose there existsts no such neighbourhood of $x_0$ where the Points within this neighbourhood $N$ are at least $n-1$ times differentiable. This would be equivalent to:

$$N(\delta'):=\{x\in I:|x-x_0|<\delta'\}$$

$$\forall_{\delta'\in\mathbb{R}}\forall_{x'_0\in N(\delta')}\exists_{\epsilon>0}\forall_{\delta>0}\exists_{x\in I}|x-x_0'|<\delta \wedge|f^{n-k}(x)-f^{n-k}(x'_0)|\geq \epsilon$$

Where $2 < k\leq n$ by assumption

Simulataneously we have by assumption that $f$ is $n$-times differentiable in $x_0$,i.e.:

$$\forall_{\epsilon>0}\exists_{\delta>0}\forall_{x\in I}|x-x_0|<\delta\Rightarrow |f^{n-1}(x)-f^{n-1}(x_0)|<\epsilon$$

I want to invoke a contradiction now somehow

I am familiar with the equivalent definitions of differntiability that $f$ is differentiable in $x_0$ if and only if there exist a function $\varphi$ which is continious in $x_0$ and which has the property:

$$f(x)-f(x_0)=(x-x_0)\cdot\varphi(x)$$

And also $f$ is differentiable in $x_0$ if and only if there exists a linear map $L:\mathbb{R}\rightarrow\mathbb{C}$ which has the property that

$$\lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)-L(h)}{h}=0$$

If there is a way how to utilize those definitions then please tell me how to.

I also request you to avoid anything that is related to Taylor because in our sylabus higher order functions were a prerequisite for Taylor-series.

Answer 1

The question is slightly garbled - you meant this:

If $f$ is $n$ times differentiable at $x_0$ then $f^{(n-1)}$ exists in some neighborhood of $x_0$.

This is clear. By definition, $$f^{(n)}(x_0)=\lim_{x\to x_0}\frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0},$$and that limit can't exist unless $f^{(n-1)}$ exists in some neighborhood of $x_0$.