Does locally Lipschitz implies absolutely continuous?

Question

Is there a result that implies locally Lipschitz continuity implies absolutely continuity?

Answer 1

No. $f(x) = x^2$ is locally Lipschitz on $\Bbb R$, but not uniformly continuous and therefore also not absolutely continuous.

Answer 2

For non-compact intervals (which is the only framework where local Lipschitz continuity makes sense), you can consider \begin{align}f:[a,\infty)&\to \Bbb R\\ f(x)&=e^x\\ g:\left(0,1\right]&\to\Bbb R\\ g(x)&= x^{-1}\end{align}

which are $C^1$ and thus locally Lipschitz, but not AC. Of course, on compact intervals Lipschitz and locally-Lipschitz are the same thing.