$f$ is Lipschitz if and only if there exists $L\geq0$ such that $|f'(x)|\leq L$

Question

Let $f:[a,b]\to\mathbb{R}$ be absolutely continuous.
Prove that $f$ is Lipschitz if and only if there exists $L\geq0$ and a set $E\subset[a,b]$ such that, $m(E)=0$ and $f$ is differentiable at each $x\in[a,b]\setminus E\quad$ with $|f'(x)|\leq L$.

My attempt:
For $\Rightarrow$ direction:
Since $f$ is absolutely continuous on $[a,b]$, it is differentiable a.e on $[a,b]$
And notice:
$\begin{align} |f'(x)|&=\left|\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\\ &\leq\lim\limits_{h\to0}\left|\frac{f(x+h)-f(x)}{h}\right|\\ &\leq \lim\limits_{h\to0}\left| \frac{L|(x+h)-x|}{h}\right|\\ &=L \quad\text{ Where, $L$ is the Lipschitz constant} \end{align}$
Is that correct?
And for the other side of the implication I would really appreciate your help

Answer 1

Important: this proof only works when $f$ is differentiable in $\mathbb{R}$.

The reverse implication can easily be seen with the Mean value theorem. Assuming $|f'(x)|\leq L$ (being $L\geq 0$) for any $x\in\mathbb{R}$, using the mean value theorem we know that, given $x,y\in\mathbb{R}$ (we will assume $x<y$), there exists $c\in(x,y)$ such that $$\frac{f(x)-f(y)}{x-y}=f'(c).$$ Using absolute values, and using that $|f'|\leq L$ we get that $$\frac{|f(x)-f(y)|}{|x-y|}\leq |f'(c)|\leq L,$$ so we end up with $$\boxed{|f(x)-f(y)|\leq L\cdot|x-y|}$$ and we conclude $f$ is Lipschitz

Answer 2

Since $f$ is absolutely continuous, we have that $$ f(x) - f(y) = \int_x^y f'(t) \, dt .$$ Then \begin{align} \text{$f$ is Lipschitz with constant $L$} &\Leftrightarrow -L (y-x) \le \int_x^y f'(t) \, dt \le L (y-x) \text{ for all $x < y$} \\ &\Leftrightarrow \int_x^y (L - f'(t)) \, dt \ge 0 \text{ and } \int_x^y (L + f'(t)) \, dt \ge 0\text{ for all $x < y$} \end{align} Also $$ \int_x^y g(t) \, dt \ge 0 \text{ for all $x < y$} \Leftrightarrow g \ge 0 \text{ a.e.} $$ Hence $f$ is Lipschitz with constant $L$ if and only if $L - f'(t)$ and $L + f'(t)$ are greater than or equal to zero for almost every $t$.