Let $f:[a,b] \rightarrow \mathbb R$ such that $f^2$ is càdlàg and of bounded variation. Is $f$ of bounded variation?
If $f$ were to be continuous, I think that would be correct, following the proof in $|f|$ is of bounded variation$\Rightarrow$ $f$ is of bounded variation But that proof doesn't work for càdlàg.
$f(x)=x\sin\frac1x$ is not BV on $[0,1]$ (see here), but $f^2(x)=x^2\sin^2\frac1x$ is because its derivative $$ (f^2)'(x)=\begin{cases}0&\text{if }x=0\\ x\sin^2\frac1x-\sin\frac2x&\text{if }x>0\end{cases} $$ is Riemann integrable on $[0,1]$ and $\int_0^1\lvert (f^2)'\rvert\,dx<\infty$.
Here $f$ is continuous, of course.
