Function composition: $\Phi$ locally Lipschitz map and $f \in AC([0,1])$

Question

Let $\Phi:\mathbb{R}\to \mathbb{R}$ be a locally Lipschitz map, i.e. a map for which for every compact set $K\subset\mathbb{R}$ there exists a constant $c$ such that $|\Phi(x)-\Phi(y)|\leq c|x-y|$ for every $x, y \in K$.

Prove that if $f \in AC([0, 1])$ then $\Phi ◦f \in AC([0, 1])$.

My idea:

$\Phi$ is locally Lipschitz, so $\Phi\in Lip([0,1])$ hence: \begin{equation} \exists L>0 \quad s.t. \quad |\Phi(x)-\Phi(y)|\leq L|x-y| \end{equation} Furthermore, $f \in AC([0, 1])$, so \begin{equation} \forall \epsilon>0 \qquad \exists\delta=\delta(\epsilon) \quad s.t. \quad \sum(b_i-a_i)<\delta \Rightarrow \sum|f(b_i)-f(a_i)|<\epsilon \end{equation} Fix $\epsilon>0$. I have: \begin{equation} \sum |\Phi(f(b_i))-\Phi(f(a_i))|\leq \sum L|f(b_i)-f(a_i)| \end{equation} \begin{equation} \leq nL\sum |f(b_i)-f(a_i)| \end{equation} \begin{equation} \leq nL\epsilon \end{equation}

Answer 1

Since $f$ is absolutely continuous it is therefore also continuous. Then, since $[0,1]$ is a compact set we have that $f([0,1])$ is a compact set as well, i.e. $\exists m,M\in \mathbb{R}$ such that $m\leq f(x)\leq M$ for all $x\in[0,1]$. Since $\Phi$ is locally Lipschitz, it is Lipschitz on the set $[m,M]$ with constant $L$. Now you can use the rest of your proof, with the justification for,

$$|\Phi(f(b_i))-\Phi(f(a_i))|\leq L|f(b_i)-f(a_i)|$$

being the above.