If $f$ is continuous and piecewise $C^1$ on $\mathbb{R}$ (only a finite number of pieces) and $f'$ is bounded a.e., is $f$ globally Lipschitz?
So $f$ is only not differentiable in a finite number of places.
If $f$ is absolutely continuous then it is Lipschitz but I don't see it.
Yes. Suppose $|f'|\leq L$ almost everywhere. Take any two points $a,b\in\mathbb R$ so that $a<b$ and let $x_1<x_2<\dots<x_n$ be the points where $f'$ may fail to be continuous. Then, using the fundamental theorem of calculus on each interval separately, we get $$ \begin{split} |f(b)-f(a)| &\leq |f(b)-f(x_n)|+|f(x_n)-f(x_{n-1})|+\dots+|f(x_1)-f(a)| \\&= \left|\int_{x_n}^bf'(x)dx\right|+\dots+\left|\int_a^{x_n}f'(x)dx\right| \\&\leq \int_{x_n}^b|f'(x)|dx+\dots+\int_a^{x_n}|f'(x)|dx \\&\leq L(b-a). \end{split} $$ Therefore $f$ is Lipschitz with Lipschitz constant $L$.
In fact, $f:\mathbb R\to\mathbb R$ is Lipschitz if and only if it has a weak derivative and the weak derivative is in $L^\infty$. In your case we don't have to rely on such heavy weaponry, but I wanted to let you know that Lipschitz functions can be characterized as functions having a bounded derivative in a suitable sense.
$f$ being piecewise $C^1$ and $f^\prime $ being bounded, $f$ is piecewise Lipschitz. And a piecewise Lipschitz function (finite number of pieces) is Lipschitz.
