If $f(x)$ is absolutely continuous on $[0,1]$, prove that $f(x)$ is Lipschitz if and only if $\sup\left|f'(x)\right|<\infty$. Does the conclusion follow if $f(x)$ is of bounded variation, but not absolutely continuous?
This question represents the last of my "not entirely confident" answers I gave on my recent qualifying exam. Generally, I'm concerned about whether I've applied all the concepts correctly in my proof.
ATTEMPT: Given that $f(x)$ is absolutely continuous, we have that $f'(x)$ exists a.e. and that $\int_x^y f'(t)dt=f(y)-f(x)$ for all $y, x\in[0,1]$, $y>x$.
Suppose that $L=\sup\left|f'(x)\right|<\infty$ where the supremum is taken over all $x\in[0,1]$ such that $f'(x)$ exists. Without loss of generality, let $y,x\in [0,1]$ with $y>x.$ Then we have that $\int_x^y f'(t)dt = f(y)-f(x) \leq \int_x^y L dt = L(y-x)$. Hence $f(x)$ is Lipschitz.
Suppose $f(x)$ is Lipschitz; hence there is an $L<\infty$ such that $\left|f(y)-f(x)\right|\leq L\left|y-x\right|$ for $y,x \in [0,1]$. Then if $f$ is differentiable at $x\in[0,1]$, we have that, for $y\not = x$, $$\left|f'(x)\right|= \lim_{y \to x} \left|\frac{f(y)-f(x)}{y-x}\right|\leq L\left|\frac{y-x}{y-x}\right|=L<\infty$$ hence $\sup{\left|f'(x)\right|} \leq L<\infty$. This concludes the proof.
For the follow-up question, the conclusion does not follow, as a function that is not absolutely continuous cannot be Lipschitz (I provided a proof that all Lipschitz functions are absolutely continuous, but will not reproduce it here). To show that this set is nonempty, let $f$ be the Cantor function on $[0,1]$. Then $f$ is a bounded, monotone increasing function, hence of bounded variation (with total variation 1) with $f'(x)=0$ a.e. But $\int_0^1 f'(x)dx=0<f(1)-f(0)=1-0=1$ shows that $f$ is not absolutely continuous on $[0,1]$.
