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Let $f \in BV(\mathbb R)$ and $g: \mathbb R \to \mathbb R$ be Lipschitz. How can I estimate the total variation of $f\circ g$, that is $$ \int_{\mathbb R} \left|\frac{d}{dx}f(g(x))\right| dx \ ? $$ For example is it true that $$ \int_{\mathbb R} \left|\frac{d}{dx}f(g(x))\right| dx \le TV(f) \Vert g' \Vert_{L^\infty} $$ holds?


Does the estimate improve if we also assume $g$ to be invertible?


As a bonus question: if we additionally assume that $g$ is a homeomorphism, is it true that $$ \int_{\mathbb R} \left|\frac{d}{dx}f(g(x))\right| dx = TV(f) $$ hold?

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    $\begingroup$ You possibly mean "homeomorphism", not homomorphism. Yes, this is simply a change of variables formula. In particular it explains why the above inequality does not hold (compare $g(x)=cx$ for small and large $c$). $\endgroup$ Jan 29, 2021 at 22:18
  • $\begingroup$ @FedorPetrov Thanks, I fixed the typo. Why doesn't the inequality hold? Which one holds instead? $\endgroup$
    – Jun
    Jan 29, 2021 at 22:35
  • $\begingroup$ choose $g(x)=cx$ with small $c$ $\endgroup$ Jan 29, 2021 at 23:06
  • $\begingroup$ @FedorPetrov This only suggests that the inequality might hold if you replace $\Vert g' \Vert$ with $\Vert (g^{-1})'\Vert$ $\endgroup$
    – Jun
    Jan 29, 2021 at 23:32
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    $\begingroup$ If $g(x)$ is periodic, this is usually even infinite $\endgroup$ Jan 30, 2021 at 0:14

1 Answer 1

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If $f$ has a jump at 0, and $g:[0,1]\to\mathbb R$ crosses zero infinitely often, then var$(f\circ g)=\infty$.

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  • $\begingroup$ In this case $g$ is not invertible. Does the estimate change if $g$ is also assumed to be invertible? $\endgroup$
    – Jun
    Jan 30, 2021 at 9:58
  • $\begingroup$ If $g$ is a homeo, $TV(f\circ g)=TV(f)$. $\endgroup$ Jan 30, 2021 at 14:17

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