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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? –  user53153 Dec 29 '12 at 17:54
    
@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. –  gy6565 Dec 29 '12 at 18:32
    
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 –  user53153 Dec 29 '12 at 20:38
    
@PavelM Oh, thanks. I'm going to learn it. –  gy6565 Dec 29 '12 at 21:03
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2 Answers 2

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. –  gy6565 Dec 29 '12 at 18:39
    
@gy6565 I added a hint. –  saz Dec 29 '12 at 18:52
    
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. –  user53153 Dec 29 '12 at 19:04
    
@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". –  user53153 Dec 29 '12 at 19:27
    
@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. –  gy6565 Dec 29 '12 at 20:02
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$f(x)=\sqrt {x}$ is function of bounded variation but its derivative is unbounded.

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