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Does a differentiable function on $[a,b]$ has bounded variation ? I recall that differentiable on $[a,b]$ means differentiable on $(a,b)$ and $\lim_{x\to a^+}f(x)$ and $\lim_{x\to b^-}f(x)$ exist.

I know that if $f'$ is integrable on $[a,b]$, then it works (since we can majorate $V_{[a,b]}(f)$ by $\int_a^b|f'|$). But if $f'$ is not integrable ? Can we find a differentiable function on $[a,b]$ that has no bounded variation ?

Marios Gretsas
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Herman
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  • Not an answer, but this might help you to think about how crazy $f'$ can be. This answer gives as an example that "most" choices for $f'$ are very unfamiliar functions. So for any counterexample search don't try to find $f$; find any of the very, very common (horrible, horrible) choices of $f'$ that give you the $f$ you want. (The particular example there isn't the counterexample you want because $|f'| < 1$ there. However, there are very many such functions that aren't even $L^1$.) – Eric Towers Oct 26 '19 at 23:13

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A typical example is this:

Take the function $f(x)=x^2 \sin{\frac{1}{x^2}}$ if $x \neq 0$ and $f(0)=0$ on $[0,1]$

Marios Gretsas
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  • Yeah, realized a moment after posting. – Thomas Andrews Oct 26 '19 at 22:55
  • Thank you. I'm not so sure how to prove that it has no bounded variation. I tried to find the maxima and the mini of the function, but this lead to the equation $\tan(\frac{1}{x})=\frac{1}{2x^2}$, and I'm really not sure how to solve that. – Herman Oct 27 '19 at 07:15