My question is;
$f(x)=\left\{\begin{array}{cc}x^{2} \sin \left(\frac{1}{x^{2}}\right), & 0<x \leq 1 \\ 0, & x=0\end{array}\right.$
"Show that this function is Bounded Variation on [0,1] or not. "
i know , It's not a bounded variation function.
Because i read a theorem from a book;
where a,b>0 ;
$f(x)=\left\{\begin{array}{cc}x^{a} \sin \left(\frac{1}{x^{b}}\right), & 0<x \leq 1 \\ 0, & x=0\end{array}\right.$
if a>b; f is a bounded variation
if a≤b ; f is not a bounded variation. So; its clear that this function is not bounded variation.
But i want to solve in different ways. I need your helps. Thanks for your answers.
Define $a_n = \sqrt{\dfrac{2}{(2n+1)\pi}}$ (which is a monotone sequence in $[0,1]$)
Then the total variation $$V^{\!1}_0(f) \geq \sum_{n=2}^N|f(a_{n-1})-f(a_n)| = \frac{2}{\pi}\sum_{n=2}^N\left(\frac{1}{2n-1} + \frac{1}{2n+1}\right) \xrightarrow{N\to\infty}\infty$$