\begin{equation} F(x)= \begin{cases} x^a \sin(x^{-b})& 0<x\leq 1;\\ 0& x=0. \end{cases} \end{equation}
I wonder how to prove $F(x) $ is bounded variation on $[0,1]$ if and only if $a>b$.
There is a conclusion that $f(x)\in BV[a,b]$ if $f$ is differentiable at every point, and $f'$ is bounded on$[a,b]$, by which ,I guess, we may prove our given $F(x)\in BV[0,1]$.
I tried by using the conclusion above, but I get the derivative of $F(x)$ at $x=0$ is $x^{a-1}\sin(x^{-b})$, which seems that we need to classify that $b>0$,$b=0$,or $b<0$, and we need to handle the term $x^{a-b-1}$ which has bounded derivative on$[0,1]$ if and only if $a>b+1$. I cannot work it out in this way.
