We have a function $f:[0,1] \rightarrow \mathbb{R}$. We know it is continuous on $[0,1]$. Aside from a set $S$ of measure $0$, we can compute its right derivative, and can show that this right derivative is non-decreasing on $[0,1] \setminus S$.

Is this sufficient to show that $f$ is convex on $[0,1]$?

[If possible, it would be helpful to know also if my reasons for thinking $S$ has measure $0$ are correct: we have another function $g:[0,1] \rightarrow [0,1]$ which is monotonically non-decreasing everywhere in $[0,1]$. We can show that $S$ is equal to those locations where $g$ is discontinuous. I believe this is sufficient to show that $S$ has measure $0$ but it would be great if someone could confirm?]

Many thanks for any comments!


Take the Cantor function $c$. This has zero derivative ae. (hence non-decreasing), but is not convex (since it is constant on some intervals).

A real valued monotonic function can only have a countable number of discontinuities, which has measure zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.