Is every $\alpha$-Hölder continuous function of bounded variation absolutely continuous?

Question

Let $f:[a,b] \to \mathbb{R}$ be $\alpha$-Hölder continuous function of bounded variation, does it follow that $f$ is absolutely continuous ? Here $\alpha \in (0,1)$ is fixed .

Here are some sources :

1. If $f$ is $\alpha$-Hölder continuous for $\alpha \geqslant 1$, then the absolute continuity follows straight from the definition.
2. If $f$ is only $\alpha$-Hölder continuous for $\alpha \in (0,1)$ but not necessarily of bounded variation, the Weirstrass function (Hölder continuity of Weierstrass Function) is a counterexample.
3. If $f$ is only (uniformly) continuous and of bounded variation, then the devil's staircase is a counterexample. This counterexample is not $\alpha$-Hölder continuous, see https://mathoverflow.net/questions/45020/non-h%C3%B6lder-continuous-devils-staircases .

Answer 1

Nope, consider the Cantor function corresponding to a set of Hausdorff dimension $\alpha$. Its derivative is supported on a Cantor set which has zero measure, so it cannot be absolutely continuous.

Recall that a non-decreasing function $f:[0,1]\to\Bbb R$ is BV iff $f'=\mu$ is a finite Borel measure; $f$ is absolutely continuous iff in addition that $\mu\in L^1$, i.e. absolutely continuous to Lebesgue measure.

Now one can check that $f$ is H"older $\alpha$ iff $\mu$ is upper Ahlfors $\alpha$ regular, i.e. there is a $C>0$ such that $\mu(a,a+h)\le Ch^\alpha$ for all $h>0$.

Note that if we take a Cantor set with dimension strictly equals to $\alpha$, then the restriction of Hausdorff measure becomes a finite Borel measure. It is clearly not an absolutely continuous measure so achieves the job.