Is $g$ bounded variation in [-1,1]?

Question

Let $g$ be defined by $g(0)=0$ and $g(x)=x^2\sin(1/x^2)$ for $x\neq 0$. Is $g$ of bounded variation in $[-1,1]$?

Answer 1

For $n\in N$ let $a_n=(n\pi)^{-1/2}$ and $b_n=((n+1/2)\pi)^{-1/2}$.Then $1>a_n>b_n>a_{n+1}>0.$ So the variation of $g$ on $[-1,1]$ is at least $\sup_{m\in N} \sum_{n=1}^m |f(a_n)-f(b_n)|=\sup_{m\in N}\sum_{n=1}^m (b_n)^2=\infty.$