Let $f:\mathbb R \to \mathbb R$ be a $BV$ function and $g:\mathbb R \to \mathbb R$ be a diffeomorphism. What is the total variation of $f \circ g$?


My guess is
$$
TV(f\circ g) \le TV(f) \Vert (g^{-1})'\Vert\_{L^\infty}
$$
but I don't have a proof.



Total variation of composition of BV function and diffeomorphism
$\newcommand\R{\mathbb R}$Since $g$ is a diffeomorphism of $\R$, it is a homeomorphism of $\R$. So, $g$ is either increasing or decreasing.


Suppose that $g$ is increasing. Take any real $x\_1,\dots,x\_n$ such that $x\_1<\dots<x\_n$. Let $y:=g(x\_i)$ for all $i$. Then $y\_1<\dots<y\_n$. So,
$$\sum\_{i=1}^{n-1}|(f\circ g)(x\_{i+1})-(f\circ g)(x\_i)|
=\sum\_{i=1}^{n-1}|f(y\_{i+1})-f(y\_i)|\le TV(f).$$
So,
$$TV(f\circ g)\le TV(f).\label{1}\tag{1}$$
The latter inequality similarly holds when $g$ is decreasing.


Since $f=(f\circ g)\circ(g^{-1})$ and $g^{-1}$ is a homeomorphism of $\R$, we have
$$TV(f)=TV\big((f\circ g)\circ(g^{-1})\big)\le TV(f\circ g),$$
by \eqref{1} with $f\circ g$ in place of $f$ and $g^{-1}$ in place $g$.


Thus, by \eqref{1},
$$TV(f\circ g)=TV(f).$$


