If $f$ is continuous and piecewise $C^1$ is it Lipschitz? If $f:\mathbb{R} \to \mathbb{R}$ is continuous and piecewise $C^1$ is it Lipschitz? It would have to be at least locally I suppose. How to ensure that taking two points in the different pieces do not mess things up?

You could argue that
$$ f(x) - f(y) = \int_y^x g(z) \, dz ,$$
where $g(z) = f'(z)$ when $f'(z)$ is defined, and it doesn't matter how $g(z)$ is defined when $f'(z)$ isn't defined.  Then $g$ is bounded on any bounded set, and hence $f$ is locally Lipschitz.
