Is $f(x)=x\sin(\frac{1}{x})$ with $f(0)=0$ of bounded variation on $[0,1]$? I can't figure out whether $f(x)=x\sin(1/x)$ with  $f(0)=0$ is of bounded variation on $[0,1]$ or not.
But I think it is not. Can someone suggest a partition to prove it is not of bounded variation is so? Thanks

Sure. Try $x_n=\frac1{n\pi+\pi/2}$ for every $n\geqslant0$.
