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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 $u \in BV$?

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    $\begingroup$ BV? Would you define? $\endgroup$
    – Wlod AA
    Oct 23, 2021 at 8:41
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    $\begingroup$ @WlodAA BV means functions with bounded variation. $\endgroup$
    – Zerox
    Oct 23, 2021 at 11:17
  • $\begingroup$ You can find a complete answer here. The fact that the domain and codomain are finite intervals does not restrict the generality of the result $\endgroup$ Oct 23, 2021 at 12:27
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    $\begingroup$ @DanieleTampieri My question is a bit different though: here I know that the composition $f \circ u$ is BV and I want to conclude that $u$ is $BV$. Do you mean that $f$ being Lipschitz suffices? $\endgroup$
    – Zac
    Oct 23, 2021 at 19:14

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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 $\sum_{n=1}^\infty|u(\frac1n)-u(\frac1{n+1})|=\infty$.

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  • $\begingroup$ Thank you so much. Then what other assumption is needed on $f$ to reach the desired conclusion? $\endgroup$
    – Zac
    Oct 24, 2021 at 5:49
  • $\begingroup$ @Zac : As follows from the comment by Daniele Tampieri, for any given Lipschitz strictly monotone $f$, the implication $f\circ u\in BV\implies u\in BV$ will hold for all $u$ iff $f^{-1}$ is Lipschitz. $\endgroup$ Oct 24, 2021 at 13:32
  • $\begingroup$ Got it. Then I guess the question is what additional assumption on $f$ imply that $f^{-1}$ is Lipschitz? $\endgroup$
    – Zac
    Oct 24, 2021 at 17:16
  • $\begingroup$ @Zac : A Lipschitz strictly monotone function $f$ is absolutely continuous and hence has a version $f'\ge0$ of its almost-everywhere derivative, which is a bounded Lebesgue-locally-integrable function such that $f(x)=f(0)+\int_0^x f'(u)\,du$ for all real $x$, with $\int_0^x:=-\int_x^0$ if $x<0$. Then the inverse $f^{-1}$ of a Lipschitz strictly monotone function $f$ is Lipschitz iff there is a version of $f'$ bounded away from $0$, that is, iff the essential infimum of any version of $f'$ is $>0$. $\endgroup$ Oct 24, 2021 at 17:34
  • $\begingroup$ Previous comment continued: However, all these questions in your comments are in addition to your posted question, which was fully answered. So, if you have any further questions, please post them separately. $\endgroup$ Oct 24, 2021 at 17:36

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