# Proof that if $f$ is $n$ times differentiable at $x_0$, then there is also a neighbourhood of $x_0$ where it is $n-1$ times differentiable.

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.

• Except for wanting to do it by induction (no reason for that), duplicate of math.stackexchange.com/questions/3106249/… – David C. Ullrich Feb 10 at 13:19
• 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? – New2Math Feb 10 at 13:26
• Do you know the definition of a limit? – David C. Ullrich Feb 10 at 13:28
• $\lim_{x\rightarrow x_0}\frac{f^{'}(x_0)-f^{'}(x)}{x_0-x}\text{ exists}\iff \exists_{a\in\mathbb{C}}\forall_{\epsilon>0}\exists_{\delta>0}\forall_{x\in D}|x-x_0|<\delta\Rightarrow |\frac{f^{'}(x_0)-f^{'}(x)}{x_0-x}-a|<\epsilon$ – New2Math Feb 10 at 13:33
• 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 $x_0$ – New2Math Feb 10 at 13:36