$f:[a,b] \to \mathbb{R}$ lipschitz continuous $\Leftrightarrow$ $\exists \ g:\|g\|_\infty<+\infty$ so that $f(x)=\int_{[a,x]}g \ d \lambda$

Question

I want to show:

$f:[a,b] \to \mathbb{R}$ lipschitz continuous $\Leftrightarrow$ $\exists \ g:\|g\|_\infty<+\infty$ so that $f(x)=\int_{[a,x]}g \ d \lambda$

Anyone got any hints how to prove this?

Answer 1

That's not so easy - you have to use the fundamental theorem of Lebesgue integral calculus as follows. Since $f$ is Lipschitz-continuoous, it is also absolute continuous. This implies by the fundamental theorem of Lebesgue integral calculus that $f$ is almost everywhere differentiable with derivative $f'$ and $$f(x) = f(a) + \int_a^x f'(y) \, dy.$$ If $L$ is the Lipschitz-constant, then of course $|f'| \le L$.

The "$\Leftarrow$"-direction is "easy": We have $$|f(x)-f(y)| = \Big| \int_x^y g \, d \lambda \Big| \le \|g\|_\infty |x-y|. $$