Let $f : [0,1] \longrightarrow \mathbb R$ be a function defined by
$f(x) = x^\alpha \sin \frac 1 {x^\beta}$ , whenver $x \in (0,1]$ and $f(0) = 0$ where $\alpha > \beta$. Then for what values $\alpha$ and $\beta$, $f$ is a function of bounded variation on $[0,1]$?
I think the theorem which states that "If $f : [a,b] \longrightarrow \mathbb R$ be a continuous function on $[a,b]$ and if $f'$ exists and is bounded on $(a,b)$ then $f$ is a function of bounded variation on $[a,b]$" can be applied to solve this kind of problem. But we also know that there exist functions which are of BV but do not satisfy the above theorem. So, my question is what is the actual way to proceed?
Please help me. Thank you in advance.