If $f$ is monotone on $[a,b]$, is $f'$ bounded a.e. on $[a,b]$?

Question

All the counterexamples I can develop for $f'$ being unbounded when $f$ is monotone only fail at one point. So I am wondering if it can only happen at a few points so that $f'$ is still bounded almost everywhere.

Answer 1

If I understand your question, you can take some asymptotic function like $\sqrt{\cdot}$ and scale and shift it. E.g. a scaling/shifting of:$$ f(x) := \begin{cases} -\sqrt{-x} & \text{if } x<0\\ \sqrt{x} & \text{if } x \ge 0\\ \end{cases} $$Within a ball around $x=0$, $f'$ exists and is unbounded. Furthermore, this is indeed a monotonically increasing function.

A flip-side example is a constant function (making $f'$ bounded). Therefore I do not believe monotonicity on $[a,b]$ implies anything about the boundedness of $f'$.

The issue touched on in the comments applies: the extreme value theorem for $f'$ does not directly apply here, but a similar idea holds: if you require $f'$ to be unbounded on a closed interval $[a,b]$, then it must be undefined somewhere (you can consider a convergent sequence $\{x_i\in[a,b]\}_{i=0}^{\infty}\rightarrow x_c$ whose $f'$ values $\{f'(x_i)\}_{i=0}^{\infty}$ increase without bound; then $x_c$ must be contained in $[a,b]$ by definition of closed set).

Answer 2

Of course $f'$ must be in $L^1$ because $\int_a^b f'(t)\; dt \le f(b) - f(a)$. But $f'$ can be unbounded on every interval. You can get this by taking the sum of a suitable series of translated and scaled "ramp" functions.