Let $f:[0,1]\to[0,1]$ be a devil's staircase in the [usual sense](http://en.wikipedia.org/wiki/Singular_function). (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.)



Non-Hölder continuous devil's staircases
Let $K$ be the bad set. 
Assume $f$ is Hölder continuous with exponent $\alpha$.


Since Hausdorff dimension of $K$ is zero,
given $\epsilon>0$ we can cover $K$ by open intervals $\left]a\_i,b\_i\right[$ 
with length $\ell\_i=b\_i-a\_i$ has such that 
$$\sum\_n\ell\_n^\alpha<\epsilon\ \ \ \ \ (\*)$$ 
and $\ell\_n<\epsilon$ for any $n$.
Set $v\_i=f(b\_i)-f(a\_i)$.
Since $f$ is Hölder continuous, 
$$v\_i < C{\cdot}\ell\_i^\alpha.\ \ \ \ \ (\*\*)$$
But clearly
$$\sum v\_i=1$$
which contradicts $( \* )$ and $( \* \* )$. 


Did I miss something?


