1
$\begingroup$

Suppose $f$ be a real-valued function, such that $f'$ exists everywhere in the domain.
I am thinking about the problem in following steps-
1) Can $f'$ have jump discontinuity?-No, since if it has jump discontinuity, then $f'$ will violate IVP(Darboux's theorem)
2) Can $f'$ have countable number of discontinuity?
3) Can $f'$ have uncountable number of discontinuity?
So, first of all notice that (by ($1$)) $f'$ can have only infinite discontinuity.
I am trying to construct a sequence of function- $f_n:[0,1]\to\Bbb{R}$ by
$f_1(x)= \begin{cases} x^2\sin\left(\frac{\pi}{x+1-{1\over3}}\right), & \text{$x\in[0,{1\over3}]$}\\ (x-{1\over3})^2\sin\left(\frac{\pi}{x+1-{2\over3}}\right), & \text{$x\in[{1\over3},{2\over3}]$}\\ (x-{2\over3})^2\sin\left(\frac{\pi}{x+1-1}\right), & \text{$x\in[{2\over3},1]$}\\ \end{cases} $
My basic motivation is to divide $[0,1]$ into $3^n$ subintervals of equal length for each $n\in\Bbb{N}$. And then define each $f_n$ in the above manner i.e. basically if I want to define $f_n$, then I will divide $[0,1]$ into $3^n$ number of subintervals of equal length, then suppose $[a_n,b_n]$ is one of those sub-intervals. Then define $f_n$ on that perticular $[a_n,b_n]$ by-
$f_n(x)=(x-a_n)^2\sin\left(\frac{\pi}{x+1-{b_n}}\right)$.
Now, I take the whole sequence $\{f_n\}$.
My question is- Is it covergent? If yes and coverge to $f$, then I will take the function $f$. Will $f$ work as the required function? I have no idea about these questions. Even I can't go further. May be my method is wrong.
Can anybody solve this problem? Thanks for assistance in advance.

$\endgroup$
  • 1
    $\begingroup$ Related post. $\endgroup$ – José Carlos Santos Mar 28 '19 at 19:41
  • $\begingroup$ Either 2) or 3) need to be edited. $\endgroup$ – user Mar 28 '19 at 20:01
0
$\begingroup$

Hint: As you noticed, the function $x^2\sin(1/x)$ naturally comes to mind here. It may help to do this: Let $C$ denote the Cantor set. Define

$$f(x)= \begin{cases}d(x,C)^2\sin(1/d(x,C)),&x\notin C\\0,&x\in C\end{cases}$$

This $f$ may not quite do it, but it seems close to what you seek.

$\endgroup$
  • $\begingroup$ What do you mean by $d$ here? Distance? $\endgroup$ – Biswarup Saha Mar 28 '19 at 20:19
  • $\begingroup$ Yes, $d(x,C)$ is the distance from $x$ to $C.$ $\endgroup$ – zhw. Mar 28 '19 at 20:22
  • $\begingroup$ You are saying this $f$ may not work, but close to the answer. Basically, I can't generalize(even visualize) it. Can you generalize the function explicitly so that it will work as the required answer? your answer is likely to be work, but I can't prove it $\endgroup$ – Biswarup Saha Mar 28 '19 at 20:33
  • $\begingroup$ It's a bit of a challenging problem. You have to expect to spend some time on it. Can you show $f'(x)=0$ for every $x\in C?$ $\endgroup$ – zhw. Mar 28 '19 at 20:50
  • $\begingroup$ @Biswarup Saha: "Can you generalize the function explicitly ..." --- I think the method used in this answer, in the paragraph labeled Proof Sketch, could possibly be adapted to work for what you want. $\endgroup$ – Dave L. Renfro Mar 28 '19 at 21:22

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.