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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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    Welcome to Math.SE! You are more likely to get effective answers here if you put in a bit more information about your effort to solve the problem. For example, do you know an example of a differentiable function with unbounded derivative on an interval? If yes, did you check if that function has bounded variation?2012-12-29
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    @PavelM thanks for your attetion. I found some examples. for instance, f:=\sqrt{x} on [0,1] is a function of bounded variation because it's monotonic, but f has unbounded derivative. But actually, f is differentiable only on (a,b), not [a,b]. I'm finding counterexample whose domain of derivative is also closed interval, but it isn't going well.2012-12-29
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    As a part of learning the machinery of this website, I suggest the following exercises: (1) click the word *edit* under your post and insert the missing letter in the title. I don't like how **fuction** looks and sounds. (2) Although your post is quite readable as is, the formulas will look better if you enclose them in dollar signs $\$$. For example, you get $f'$ instead of f'. This will be important when you need to use more complex formulas. [Here's a TeX tutorial](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)2012-12-29
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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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    thanks. But I cannot show why f is a function of bounded variation. please tell me how can you show.2012-12-29
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    @gy6565 I added a hint.2012-12-29
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    A remark for @gy6565: how would one come up with such a function? One has to know the standard example of a function with discontinuous derivative: $x^2\sin (1/x)$. Here the derivative is bounded, but we can make the behavior a little worse by either (a) replacing $x^2$ with a smaller power of $x$ (this will increase the **magnitude** of oscillations near zero), or (b) replacing $1/x$ with a higher negative power of $x$ (this will increase the **frequency** of oscillations near zero). One must be careful not to overdo things: we want the integral $\int |f'|$ to converge, so that $f$ is BV.2012-12-29
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    @gy6565 .. for example, going to $x\sin (1/x)$ or to $x^2\sin (1/x^2)$ would make the derivative so ugly it would not be integrable anymore (in the first instance, it would also fail to exist at the origin). This is why *saz* went to $1/x^{3/2}$ only. Students sometimes forget about non-integer exponents and when thinking "I need a number greater than 1" reach for "2".2012-12-29
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    @saz ,Pavel M thank you for your help. But maybe there is something I haven't learned yet. I cannot understand the condition about which f is intergrable and what is L^1[0,1]. my question came up when I studied function of variation for preparation step of Riemann–Stieltjes integral. So I think I should study more. Anyway, I really appreciate your help. Lastly, I'll happy if you tell me math concepts to understand and solve above problem.2012-12-29
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    @gy6565 $f' \in L^1([0,1]) \Leftrightarrow f'$ is Riemann-integrable on $[0,1]$. (Thus the integral $\int_0^1 |f'(t)| \, dt$ exists.)2012-12-29
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    @gy6565 More Math.SE technicalities: (1) The author of the answer on which you comment is always notified. You did not need to explicitly notify saz in your comment. (2) Only one additional user can be notified of a comment. So, you could have notified me with "at"PavelM (no spaces). As it happened, I did not know of your comment came here again just by chance.2012-12-30
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    @gy6565 With that out of the way, here's the math part. You don't need integrals to prove that $f$ has bounded variation (although this is the easiest way). Instead, you can identify the intervals on which $f$ is monotone, namely the intervals between its local maxima and minima. In each such interval the variation is just the difference (max-min). They you add the contributions of intervals together. For this example, the extrema occur at points $x_n\approx 1/n^{2/3}$ (I omit details like $\pi$) and the max-min difference on $n$th subinterval is about $x_n^2$. So you get a convergent series.2012-12-30
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    Thanks, saz. I proved the theorem you give me. But In the process of proof, I use MVT and the the inequality between Riemann sum and upper Riemann sum. I learned that the function should be bounded for Riemann-integrability. but I think the example you show me is not bounded(I mean, derivative of the function). I know it is improperly integrable on (0,1], but I think it isn't Riemann-integrable on [0,1]. Where am I wrong?2012-12-30
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    @PavelM I can't explain how much I thank you. I appreciate your many help. I understand your explanation, and I get to have another question. To check a function of bounded variation, I learned that I should calculate for any partition. But you check on only one partition. I think your partition makes maximum intuitionally. what I want to know is the partition you calculate is sufficient for checking the function of bounded variation.2012-12-30
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    @gy6565 Two facts about variation: (1) when a function is monotone on some interval $[c,d]$, its variation on that interval is $|f(c)-f(d)|$. In other words, the trivial partition is all we need. (2) if the interval $[a,b]$ is divided into subintervals $[a_n,b_n]$, then the variation on $[a,b]$ is the sum of variations on $[a_n,b_n]$. // Your textbook may have proofs of these results, if not, you could try to prove them yourself. The first one is easier than the second.2012-12-30
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    @gy6565 You are right. When I wrote "$f'$ is Riemann-integrable on $[0,1]$" I meant improperly integrable. Sorry. I don't know in which context you encountered the topic "bounded variation", but probably it would be a good idea to get familiar with lebesgue integration (and then you'll also understand what $L^1([0,1])$ means)...2012-12-30
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    @PavelM I added your idea as a remark to my answer. If you would like to write it down as an additional answer, I'll remove it. (And your explanations are great, by the way.)2012-12-30
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    @PavelM I think I understand your whole explanation. Thank you for your help!2012-12-30
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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