Let $f$ be absolutely continuous and Lipschitz on $[0,1]$. I would like to show that $f'\in L^\infty([0,1])$.
Since $f$ is absolutely continuous, then it is differentiable almost everywhere on $(0,1)$ and $$\int_x^y f'=f(y)-f(x)$$ for every $0\le x<y\le 1$. Furthemore, since $f$ is Lipschitz, then the quantity $$\sup_{x,y\in[0,1]}\frac{|f(x)-f(y)|}{|x-y|}=\sup_{x,y\in[0,1]} \frac{1}{|x-y|} \left|\int_x^y f'\right| =: M$$ is finite. But by Lebesgue's differentiation theorem, $$\lim_{y\to x} \frac{1}{|x-y|} \int_x^y f' \to f'(x).$$ Thus, $$|f'(x)|\le M$$ almost everywhere, implying that $f'\in L^\infty([0,1])$.
Is this proof correct, and are there other proofs of the same result?