Let $f:[0,1]\to[0,1]$ be a devil's staircase in the [usual sense][1]. (That is, $f$ is continuous, non-decreasing, $f'=0$ on a set of full Lebesgue measure.) We also require the complement to the set where $f'$ vanishes to have **Hausdorff dimension zero**. *Question*. Is it true that $f$ is **not** Hölder continuous? (This looks plausible, since $f$ has `very little room' where it can grow so it has to grow very fast - at least, at some points.) [1]: http://en.wikipedia.org/wiki/Singular_function