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.
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.