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

Consider the partition defined by $\{x_n\}=\{\frac{1}{n\pi+\pi/2}\}$,
$$ \sin (1/x_n)=\left\{\begin{array} \{1 \qquad n \text{ is even}\\-1 \qquad n \text{ is odd}\>\end{array}\right.\qquad\text{for }n\geq 0$$ Hence $$ f(x_n)=x_n\sin(1/x_n)=\left\{\begin{array} \{x_n \qquad n \text{ is even}\\-x_n \qquad n \text{ is odd}\>\end{array}\right.\qquad\text{for }n\geq 0$$
Therefore $$\sum_{n=1}^m|f(x_n)-f(x_{n-1})|=\sum_{n=1}^m|(-1)^n(x_n+x_{n-1})|= \sum_{n=1}^m(x_n+x_{n-1})\\=x_m+x_0+2\sum_{n=1}^{m-1}x_n\\\geq \sum_{n=1}^{m-1} x_n=\sum_{n=1}^{m-1} \frac{1}{n\pi+\pi/2}.$$ We see that $$\sum_{n=1}^{m-1} \frac{1}{n\pi+\pi/2}\to \infty \qquad \text{ as }\qquad m\to \infty.$$ Since $V_0^1(f)$ is not bounded in this partition, $V_0^1(f)$ is not bounded variation function.
In general, the function $$ f(x)=x^a\sin(1/x^b), \qquad x\in[0,1] $$ is not bounded variation for $0<a\leq b$.
