Let $C\subset [0, 1]$ be the usual Cantor ternary set and define $f(x)=d(x, C)^{\alpha}$, when $\alpha>0$. I need to determine the values $\alpha$ s.t. $f$ is absolutely continuous on $[0, 1]$. It follows easily from the reverse triangle inequality that $f$ is absolutely continuous when $\alpha=1$. But the general case is not clear to me. What I've found is:
1) The function $x\mapsto d(x, C)$ is differentiable at every $x\in [0, 1]$ which is not in $C$ or which is not a midpoint of an interval removed in the construction of Cantor set. So $f$ is differentiable a.e. in $[0, 1]$.
2) Let $[a, b]$ be an interval which is removed in the construction and let $m$ be the midpoint of $[a, b]$. The derivative of $x\mapsto d(x, C)$ is $1$, when $x\in ]a, m[$ and $-1$, when $x\in]m, b[$.
Also I know that function $g$ is absolutely continuous iff $g$ is differentiable a.e., $g^{\prime}$ is integrable and satisfies the fundamental theorem of calculus. Can I use this characterization to deduce when $f$ is absolutely continuous?
Thank you in advance.
i) $g(t)=t^\alpha$ is differentiable on $(0,\infty)$.
ii) $f:[0,1]\rightarrow \mathbb{R},\ f(x)=d(x,C)^\alpha$ is continuous and $C$ is compact so that $f(x)=d(x,x_f)^\alpha$ where $x_f\in C$. We call $x_f$ a foot of $x$.
Then $$A_1=\{ x\in [0,1] -C| x\ {\rm has\ a\ unique\ foot}\}$$
$$ A_2=\{x\in [0,1]-C| x\ {\rm has\ at\ least\ two\ foots} \} $$ so that $[0,1]-C=A_1\cup A_2$ is a disjoint union.
If $x\in A_2$, then $x$ has a foot $x_f$. WLOG, we can assume that $x_f<x$. Note that $(x_f,x)\subset A_1$. Hence $A_2$ is disconnected. In further, $f$ is differentiable on $(x_f,x)$.
So $A_2,\ C$ have measure $0$ so that $f$ is differentiable a.e.