Let $f: \mathbb R \to \mathbb R$ be a Lipschitz strictly monotone (so, in particular, invertible) function. Let $u: \mathbb R \to \mathbb R$. If $f \circ u \in BV$ can we conclude that $u \in BV$?



If $f \circ u \in BV$ and $f$ is strictly monotone, then is $u \in BV$?
The answer is no.


E.g., let $f(x):=\min(1,|x|)x$ for real $x$. Then $f$ is Lipschitz and strictly monotone.


Let then $g$ be any function in $BV$ such that $g(1/n)=(-1)^n/n^2$ for all natural $n$; it is easy to see that such a function exists. (For instance, let $g=0$ on $(-\infty,0]\cup(1,\infty)$ and let $g$ be monotonic on $[\frac1{n+1},\frac1n]$ for each natural $n$. Then the total variation of $g$ is $2\sum\_{n=1}^\infty1/n^2<\infty$.)


Finally, let $u:=f^{-1}\circ g$. Then $f\circ u=g\in BV$, whereas $u(1/n)=(-1)^n/n$ for all natural $n$. So, $u$ is not in $BV$, because the total variation of $u$ is no less than $\sum\_{n=1}^\infty|u(\frac1n)-u(\frac1{n+1})|=\infty$.


