0
$\begingroup$

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 $0<a\leq 1$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – DonAntonio Jan 13 '19 at 23:09
  • $\begingroup$ 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 $\mathcal{A}$. $\endgroup$ – Nameless Jan 13 '19 at 23:12
  • $\begingroup$ That, I think, is false. Within the real numbers we need a continuum to do limits. You don't have it here. $\endgroup$ – DonAntonio Jan 13 '19 at 23:14
  • $\begingroup$ @Nameless, considering *@DonAntonio comment, are you defining the derivative as the right limit only of the increment ratio ? $\endgroup$ – G Cab Jan 13 '19 at 23:15
  • 1
    $\begingroup$ 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 $A$. $\endgroup$ – Botond Jan 14 '19 at 18:13
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $a\in E$. $\endgroup$ – Nameless Jan 14 '19 at 10:35
  • $\begingroup$ I also found this: math.stackexchange.com/questions/1505310/… $\endgroup$ – Nameless Jan 14 '19 at 11:46
  • 1
    $\begingroup$ @obscu Good answer. +1 Yet the OP thinks he knows better and one doesn't need continuum to work out this in real analysis. Fine. $\endgroup$ – DonAntonio Jan 14 '19 at 19:35
  • $\begingroup$ @DonAntonio +1. Of course you can change the definition of the derivative if you want, but then you are no longer doing real analysis. You can say whatever functions you want satisfy your version of differentiability, but theorems of real analysis no longer apply. For example, in the other question you linked, you can't use the Fundamental Theorem of Calculus any more. $\endgroup$ – obscurans Jan 15 '19 at 2:23
  • $\begingroup$ @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. $\endgroup$ – DonAntonio Jan 15 '19 at 8:25

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.