$f$ is Lipschitz continuous with constant $M$ iff $f$ is absolutely continuous and $|f'| < M$

Question

I have problems proving one of the directions. First, we can use the following arguments:

If $f$ is Lipschitz continuous, then it has bounded variation, and therefore it is differentiable almost everywhere on some interval $[a,b]$. Therefore, there exists a function $f'(x) : f(b) - f(a) = \int_a^b f'(t) dt$

Now, since $f$ is Lipschitz continuous with constant $M$

$$\frac{|f(x) - f(y)|}{|x - y|} \leq M$$

and taking limits $y \to x$ we get

$$|f'(x)| \leq M$$

For the converse, I tried to work this last argument backwards but I run into problems when getting rid of the limit.

Answer 1

Going the other way, if it is absolutely continuous and $|f'| \leq M$ then $|f(x)-f(y)| = \left | \int_x^y f'(z) dz \right | \leq M|x-y|$ which gives Lipschitz.