I want to show that $f(x) = x \sin(\frac{1}{\sqrt{x}})$ , with $f(0)=0$ on $[0,1]$ is of bounded variation. To do that I Don't want to use of showing this is continuously differentiable hence, BV. (SO I don't want to use derivatives to show that)

I want to try proving this either with the def of BV , or either showing this is Absolutly continuous or is Lipschitz continuous.

  • $\begingroup$ Split into intervals where $f$ is increasing and decreasing. If $f$ is increasing on $[a,b]$, then the variation is just $f(b)-f(a)$. (This is easily proved from the definition of variation.) Bound the variation of $f$ by a sum over these intervals. $\endgroup$
    – Dzoooks
    Oct 10, 2019 at 17:28

1 Answer 1


Proof to be detailed

The derivative of $f$ is

$$f^\prime(x) = \sin\left(\frac{1}{\sqrt x}\right) - \frac{1}{2 \sqrt x}\cos\left(\frac{1}{\sqrt x}\right)$$

The map $$g(x) = \tan x - \frac{x}{2}=\frac{1}{\cos x}\left(\sin x - \frac{x}{2}\cos x\right) = \frac{1}{\cos x}f^\prime\left(\frac{1}{x^2}\right)$$

is defined on $\mathbb R \setminus \{n\pi + \frac{\pi}{2} \mid n \in \mathbb N\}$. On each interval $I_n=(n\pi - \frac{\pi}{2}, n\pi + \frac{\pi}{2})$, we have $g^\prime(x) = \frac{1}{\cos^2 x} - \frac{1}{2} >0$ which with the consideration of the limits at the bound of the interval allows to conclude that $g$ is first negative, vanishes at one point $x_n \in I_n$ and then is positive.

From that we deduce that $f$ is increasing on $J_{2n}$ and decreasing on $J_{2n+1}$ where $J_n = (\frac{1}{\sqrt{x_{n+1}}}, \frac{1}{\sqrt{ x_n}})$.

Now you can prove that $$\left\vert f\left(\frac{1}{\sqrt{x_{n+1}}}\right) - f\left(\frac{1}{\sqrt{x_{n}}}\right) \right\vert \le \frac{k}{n^2}$$ where $k$ is a constant.

That proves that $f$ is of bounded variation as $\sum \frac{1}{n^2}$ converges.

  • $\begingroup$ Is this claim also correct? that if I choose $$x_n=\{\frac{1}{n\pi+\frac{\pi}{2}}\}_{n\geq0}$$ then I am partitioning the domain at its peak and valleys. as $\sin(\frac{1}{\sqrt{x}})$ would be $1$ or $-1$. So I can write given the above choice: $$\sup \sum |f(x_{n})-f(x_{n-1})| = \sum |f(x_{n})-f(x_{n-1})|$$ and I can show that this RHS series converges. $\endgroup$
    – domath
    Oct 10, 2019 at 23:36
  • $\begingroup$ in above I meant $$x_n=\{(\frac{1}{n\pi+\frac{\pi}{2}})^2\}_{n\geq0}$$ $\endgroup$
    – domath
    Oct 11, 2019 at 0:02
  • $\begingroup$ No because considering the extreme value of $\sin$ doesn’t mean that those are the one of $f$. And indeed they are not if you look at the derivative. $\endgroup$ Oct 11, 2019 at 6:21

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