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I can't figure out whether $f(x)=x\sin(1/x)$ with $f(0)=0$ is of bounded variation on $[0,1]$ or not.

But I think it is not. Can someone suggest a partition to prove it is not of bounded variation is so? Thanks

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    Do you define it as $0$ at $x=0$? Look for values of (1/x) for which the sin takes values $1$ and $-1$ respectively. – DBFdalwayse Dec 13 '13 at 07:45
  • @DBFdalwayse yes I missed that I edited my posted –  Dec 13 '13 at 07:47
  • :No problem. It helps to do a plot , e.g., in Wolfram:http://www.wolframalpha.com/input/?i=plot+xsin%281%2Fx%29%3D . Maybe you could try for $x^2sin(1/x)$ now. Of course, looking at a plot is not a substitute for the actual calculations. – DBFdalwayse Dec 13 '13 at 07:48

3 Answers3

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Consider the partition defined by $\{x_n\}=\{\frac{1}{n\pi+\pi/2}\}$,

$$ \sin (1/x_n)=\left\{\begin{array} \{1 \qquad n \text{ is even}\\-1 \qquad n \text{ is odd}\>\end{array}\right.\qquad\text{for }n\geq 0$$ Hence $$ f(x_n)=x_n\sin(1/x_n)=\left\{\begin{array} \{x_n \qquad n \text{ is even}\\-x_n \qquad n \text{ is odd}\>\end{array}\right.\qquad\text{for }n\geq 0$$ Therefore $$\sum_{n=1}^m|f(x_n)-f(x_{n-1})|=\sum_{n=1}^m|(-1)^n(x_n+x_{n-1})|= \sum_{n=1}^m(x_n+x_{n-1})\\=x_m+x_0+2\sum_{n=1}^{m-1}x_n\\\geq \sum_{n=1}^{m-1} x_n=\sum_{n=1}^{m-1} \frac{1}{n\pi+\pi/2}.$$ We see that $$\sum_{n=1}^{m-1} \frac{1}{n\pi+\pi/2}\to \infty \qquad \text{ as }\qquad m\to \infty.$$ Since $V_0^1(f)$ is not bounded in this partition, $V_0^1(f)$ is not bounded variation function.

In general, the function $$ f(x)=x^a\sin(1/x^b), \qquad x\in[0,1] $$ is not bounded variation for $0<a\leq b$.

Mohamed
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  • $$\sum_{n=1}^{m-1} \frac{1}{n\pi+\pi/2}\to \infty \qquad \text{ as }\qquad m\to \infty.$$ Why is this true? – Muno May 02 '18 at 01:01
  • Use the Limit Comparison Test with the series $\sum_{n = 1}^{m - 1} \frac{1}{n}$ – BMac Aug 15 '18 at 17:37
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    for the general rule of above can we say that the partition could be also in this form? $$x_n = {\frac{1}{(n\pi + \frac{\pi}{2})^{\frac{1}{b}}} }$$ – domath Oct 10 '19 at 01:19
  • The thing I don't get is that, as far as I'm concerned, the partition of [0,1] should be a finite and also, the first and last points os the partition should be equal to the maximum and minimum of the compact. So how can you take this partition? – H44S Dec 02 '19 at 14:21
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Sure. Try $x_n=\frac1{n\pi+\pi/2}$ for every $n\geqslant0$.

Did
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  • The point being, since it is the $\sin(1/x)$ factor that is the wiggly one, to pick the extreme values of that part, i.e., where it is $\pm1$. – Harald Hanche-Olsen Dec 13 '13 at 07:55
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    But it is interesting to see how/why a higher power of $x$--$x^2$ itself is enough, slows the wiggling enough to make the function BVD. – DBFdalwayse Dec 13 '13 at 08:03
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Take the uniform partition with length of the subinterval $2/(n\pi)$ then find the variation where your sum will be in the form of harmonic sequence, which is divergent; that's why it is not of bounded variation.

Michael Hardy
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Uniqraj
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