Suppose $y:[a,b]\to \mathbb{R}^n$ is absolutely continuous, where $[a,b]\subset \mathbb{R}$ is a compact interval. Let $\phi:\mathbb{R}^n\to \mathbb{R}$ be a $C^1$ function. Is it true that $$\phi \circ y:[a,b]\to \mathbb{R}$$ is also absolutely continuous?

Can I have an hint in order to prove/disprove it?

Thanks a lot in advance.

  • $\begingroup$ What have you tried? $\endgroup$ – Viktor Glombik Feb 12 at 20:27

Hint: since $y([a,b])$ is compact and connected, the restriction of $\varphi$ to $y([a,b])$ is Lipschitz. What can you say about the composition $\psi \circ y$ if $\psi$ is Lipschitz?

  • $\begingroup$ Suppose $L>0$ is the Lipschitz constant of $\psi$ on $y([a,b])$. Fix $\epsilon>0$. Since $y$ is AC then there exist $\delta>0$ s.t. for every finite collection of open disjoint intervals $\{(x_i,z_i)\}, i=1, \dots, n$ each one contained on $[a,b]$ one has $\sum_{i=1}^{n} \left\lVert x_i-z_i\right\rVert<\frac{\epsilon}{L}$. Got it! Thanks a lot! $\endgroup$ – eleguitar Feb 12 at 21:35
  • $\begingroup$ Just a question. Is the restriction of a $C^1$ function on a compact subset of the domain always Lipschitz continuous? $\endgroup$ – eleguitar Feb 12 at 21:36
  • 1
    $\begingroup$ Good question. I realized my argument that the restriction of $\varphi$ to $y([a,b])$ is Lipschitz also uses the fact that $y([a,b])$ is connected, because I was using the Mean Value Theorem. I’ve edited accordingly. In general, restricting a $C^1$ function to a compact, connected set gives a Lipschitz function. $\endgroup$ – Jordan Green Feb 12 at 21:43
  • $\begingroup$ Thank you so much Jordan, really appreciated! $\endgroup$ – eleguitar Feb 12 at 21:45

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