2
$\begingroup$

I want to prove that if $f:D\rightarrow\mathbb{C}$ is $n$ times differentiable at $x_0$ then there is also has a neighbourhood of $x_0$ where all the functions are $n-1$ times differentiable. And I want to show it by induction. I have set up a proof but as the title suggests I am having trouble finishing it.


The limit of a function, i.e. ,$\lim_{x\rightarrow x_0}f(x),$ where $x_0$ might or might not be in the domain is motivated by the definition of accumulation points.

An accumulation point of the domain of a function $f$ is a point where every neighbourhood contains infinitely many points in the domain.

If $x_0$ is not in the domain and not an accumulation point, then a function $F$ which assigns a value to this point and is in every other Point defined as the previous function $f$ is continuous in $x_0.$

$(*)$We say $\lim_{x\rightarrow y_0} f(x)$ exists if and only if $x_0$ is in an accumulation point and not in the Domain of the function or in the Domain of the function and if it is possible to assign a value to $x_0$ for $F$ which is defined in the same manner as before such that $x_0$ is continuous in $F$.


$(1)$For the base case of my induction I assume there exists a Point in $x_0$ which is $2$ times differentiable in $f$. I think I have to choose $2$ because otherwise it would not make sense to call it a higher-order Derivative.

$(2)$I set up a proof by contradiction by assuming at the same time that there exists no neighbourhood around $x_0$ which is $1-$time differentiable or also just differentiable.

The strategy is to prove that given these circumstances $x_0$ is not an accumulation point in the Domain of $f'$ which would be a contradiction to $(1)$ because for the existence of such a Limit a necessary condition is that $x_0$ is a accumulation point $(*)$.

By $(1)$:

$$\lim_{x\rightarrow x_0}\frac{f^{'}(x_0)-f^{'}(x)}{x_0-x}\text{ exists}$$

By $(2)$

$$\text{Define: } \delta_0\text{-neighbourhood around }x_0:=N(\delta_0)=\{x\in D:|x-x_0|<\delta_0\}$$

$$\forall_{\delta_{0}\in\mathbb{R}}\exists_{x'_0\in D}\exists_{\epsilon>0}\forall_{\delta>0}\exists_{x\in D}:x'_0\in N(\delta_0)\wedge |x'_0-x|<\delta\wedge \frac{f^{'}(x'_0)-f^{'}(x)}{x'_0-x}\geq\epsilon$$

$$\text{Define: } N'(\delta_0)=\{x\in D:x \text{ is differentiable }\wedge x\in N(\delta_0)\}$$

If I could show there exists a $\delta_0>0$ such that $N'(\delta_0)=\{x_0\}.$ I would be done, but how can I prove this? Please help me.

$\endgroup$
  • $\begingroup$ Except for wanting to do it by induction (no reason for that), duplicate of math.stackexchange.com/questions/3106249/… $\endgroup$ – David C. Ullrich Feb 10 at 13:19
  • $\begingroup$ You have said that the Limit can not exists unless $f^{n-1}$ exists is some neighbourhood of $x_0$. Can you please explain why this is true? $\endgroup$ – New2Math Feb 10 at 13:26
  • $\begingroup$ Do you know the definition of a limit? $\endgroup$ – David C. Ullrich Feb 10 at 13:28
  • $\begingroup$ $\lim_{x\rightarrow x_0}\frac{f^{'}(x_0)-f^{'}(x)}{x_0-x}\text{ exists}\iff \exists_{a\in\mathbb{C}}\forall_{\epsilon>0}\exists_{\delta>0}\forall_{x\in D}|x-x_0|<\delta\Rightarrow |\frac{f^{'}(x_0)-f^{'}(x)}{x_0-x}-a|<\epsilon$ $\endgroup$ – New2Math Feb 10 at 13:33
  • $\begingroup$ No I think I have not understood it please tell me why the Definition verrifies the Statement : the Limit can not exists unless$ f^{n−1} $ exists is some neighbourhood of $ x_0$ $\endgroup$ – New2Math Feb 10 at 13:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.