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I learned that $f$ is a function of bounded variation, when function $f$ is differentiable on $[a,b]$ and has bounded derivative $f'$.

What I want to know is converse part. If $f$ is differentiable on $[a,b]$ and $f$ is a function of bounded variation, Is derivative of $f$ bounded? I guess it's false, but i cannot find a counterexample. If it's true, please show me proof.

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    @PavelM Oh, thanks. I'm going to learn it.2012-12-29

2 Answers 2

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Take $a=0$, $b=1$,

$f(x) := \begin{cases} x^2 \cdot \sin x^{-\frac{3}{2}} & x \in (0,1] \\ 0 & x=0 \end{cases}$

Then $f$ is differentiable and of bounded variation, but $f'$ is unbounded.

Hint To show that $f$ is of bounded variation you can use the following theorem: Let $f: [0,1] \to \mathbb{R}$ differentiable and $f' \in L^1([0,1])$. Then $f$ is of bounded variation and $\text{Var} \, f = \int_0^1 |f'(t)| \, dt$

Remark As Pavel M suggested one can also prove that $f$ is of bounded variation by splitting up the interval $[0,1]$ in intervals $[a_n,b_n]$ such that $f$ is monotone on $[a_n,b_n]$. Then one can easily compute the variation of $f$ on the interval $[a_n,b_n]$ and use the fact that the variation on $[0,1]$ is equal to the sum of the variations on $[a_n,b_n]$.

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    saz, I haven't learned about lebesgue integration yet.(cause I'm learning the previous chapter) But I find that improper Riemann-integrability infers Lebesgue-integrability, so after learning lebesgue integration I can proof the theorem you show me and solve the problem. Thank you for your help!2012-12-30
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$f(x)=\sqrt {x}$ is function of bounded variation but its derivative is unbounded.

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    Is $f(x)$ differentiable at $x=0$?2018-12-01