# Derivative of higher order

Suppose that we have a function $$f:[0,1]\to\mathbb{R}$$ that is differentiable only in $$x=1,1/2, 1/3, 1/4, ..., 1/n$$ and in $$x=0$$. So a new function $$f':\mathcal{A} \to\mathbb{R}$$ arises, where $$\mathcal{A}=\{0\}\cup \left\{\frac{1}{n} \right\}_{n\in\mathbb{N}}$$.

Can I say that:

$$\lim_{\mathcal{A}\ni x \to 0}\frac{f'(x)-f'(0)}{x-0}$$

is the second derivative of the function $$f$$ in $$x=0$$? Or does the second derivative, to be defined as such, requires the existence of the first derivative in a neighborhood of $$0$$ of the type $$[0, a]$$, with $$0?

• The second derivative, just as the first derivative and any order one, is a limit: you need the function to be divided in a complete neighborhood of the point to which $\;x\;$ tends...and you don't have it here, not even at zero. – DonAntonio Jan 13 '19 at 23:09
• In order for the limit to make sense, it is sufficient that $x = 0$ is a limit point for the domain of $f '(x)$, and $x=0$ is a limit point for $\mathcal{A}$. – Nameless Jan 13 '19 at 23:12
• That, I think, is false. Within the real numbers we need a continuum to do limits. You don't have it here. – DonAntonio Jan 13 '19 at 23:14
• @Nameless, considering *@DonAntonio comment, are you defining the derivative as the right limit only of the increment ratio ? – G Cab Jan 13 '19 at 23:15
• I think you can say that. But I'd include the set $A$ somehow. Maybe like this: $...$ is the second derivative of $f$ at $x=0$ on the set $A$. – Botond Jan 14 '19 at 18:13

The definition of the derivative is $$\lim_{x\rightarrow0}\frac{f(x)-f(0)}{x-0}$$ and by definition of the limit this is $$\forall\epsilon>0\,\exists\delta>0\;|\;\left|x\right|<\delta\;\Rightarrow\left|\frac{f(x)-f(0)}{x-0}-L\right|<\epsilon$$ The definition most definitely requires all $$\left|x\right|<\delta$$ to work, that is to say, at the very least $$f$$ needs to exist in a neighbourhood $$\left(-\delta,+\delta\right)$$.
The right one-sided derivative still needs a neighbourhood of the form $$\left[0,\delta\right)$$ to work.
• Sorry if I return to this point, but in the book V. Zorich Mathematical Analysis 1, the definition of the derivative of a function $f:E\subset \mathbb{R}\to \mathbb{R}$ in a point $a\in E$ that is a limit point for $E$, is: $$\lim_{E \ni x\to a}\frac{f(x)-f(a)}{x-a}.$$ So, nowhere Zorich says that $E$ is an interval, it can be a generic set with $a$ as limit point and such that $a\in E$. – Nameless Jan 14 '19 at 10:35
• @obscurans Exactly my point. In the link given by the OP it is aksed exactly something like this: to define the limit notion for element in $\;\Bbb Q\;$ and not in $\;\Bbb R\;$ ...and this already would produce some rather interesting results, as having a limit equal to an element not in $\;\Bbb Q\;$, for example. Why that author Zorich defines what he does the way he does is beyond my understanding. Perhaps later I'll take a peek to his book. – DonAntonio Jan 15 '19 at 8:25