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$f(x)$ is everywhere differentiable on $[a,b]$ then give examples for each (they are independent)

(1) $f'(x)$ is not Riemann integrable

(2) $f''(x)$ does not exist

(3) $f'(x)$ is not continuous

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  • $\begingroup$ Could you add more details? For example, for (2) do you mean $f''(x)$ does not exist anywhere in the domain or just somewhere in the domain? Similarly for (3). $\endgroup$ – Rory Daulton Sep 21 '14 at 22:53
  • $\begingroup$ somewhere in the domain...anywhere would be too strict $\endgroup$ – gter Sep 21 '14 at 22:55
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(1) is a famous pathological function called Volterra's function, you can read about it on Wiki at: http://en.wikipedia.org/wiki/Volterra%27s_function It's a pretty interesting function IMO

(2) seems like the second integral of something like $\sin(1/x)$ would work http://www.wolframalpha.com/input/?i=double+integral+of+sin%281%2Fx%29+ so that $f''(x) = \sin(1/x)$ and it isn't defined at $x = 0$;

(3) Edit: as per Ian's suggestion, the Volterra function works for this as well and abs(x) doesn't

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  • $\begingroup$ smart! like (2)! $\endgroup$ – gter Sep 21 '14 at 23:30
  • $\begingroup$ Your (3) doesn't work since your $f(x)$ is not everywhere differentiable on $[a,b]$. The question wants $f'(x)$ existing but not continuous. Do you have a better example? $\endgroup$ – Rory Daulton Sep 22 '14 at 0:07
  • $\begingroup$ @RoryDaulton You can use $x^2 \sin(1/x)$ just as in the Smith-Volterra-Cantor construction for the answer to 1. $\endgroup$ – Ian Sep 22 '14 at 0:08
  • $\begingroup$ Perhaps you should change your answer to reflect your comment. Congratulations on good answers! $\endgroup$ – Rory Daulton Sep 22 '14 at 0:09
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For (1) see this.

For (2) try $\frac {x^3}{|x|}$ (the same as $x|x|$).

For (3) see this.

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  • $\begingroup$ @Rory: your (2) underlines that [a,b] does not contain 0, otherwise f is not defined, not to mention f''. So f is either x^2 or -x^2. for what reason does f'' not exist, i don't understand? $\endgroup$ – gter Sep 21 '14 at 23:14
  • $\begingroup$ @MattRigby: No, my exponent in the numerator is $3$, not $2$, so the function is the same as $\text {sgn}(x) \cdot x^2$. This has no second derivative at zero. $\endgroup$ – Rory Daulton Sep 21 '14 at 23:56
  • $\begingroup$ @deligne: my function is $2x^2$ for $x \ge 0$, $-2x^2$ for $x \lt 0$. The first derivative is defined everywhere and is $2|x|$. The second derivative is undefined at zero (since $|x|$ has no derivative at zero), so it works for your question (2). If you want an interval not containing zero, use $(x-c)^3/|x-c|$ which has a derivative everywhere but no second derivative at $x=c$. $\endgroup$ – Rory Daulton Sep 22 '14 at 0:00

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