Let $\displaystyle f(x)$ be defined by $\displaystyle f(0)=0$ and $\displaystyle f(x)=x^2 sin(1/x)$ for $\displaystyle x \neq 0$. Prove that $\displaystyle f$ is of bounded variation on [-1,1].
please help me on this one.
$\displaystyle f'(x)=2x\sin 1\slash x-\cos 1\slash x \Longrightarrow |f'(x)|\le 2+1=3$ on $\displaystyle [-1,1]$, so using the mean value theorem for differentiable functions, $\displaystyle |f(x)-f(y)|\le 3|x-y|$ . Now take any partition of the given interval and from the above you get at once that f has bounded variation there.
Tonio