# Is composition of an absolutely continuous function and a $C^1$ function still absolutely continuous?

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?

• What have you tried? – 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?

• 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! – eleguitar Feb 12 at 21:35
• Just a question. Is the restriction of a $C^1$ function on a compact subset of the domain always Lipschitz continuous? – eleguitar Feb 12 at 21:36
• 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. – Jordan Green Feb 12 at 21:43
• Thank you so much Jordan, really appreciated! – eleguitar Feb 12 at 21:45