MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:[0,1]\to[0,1]$ be a devil's staircase in the usual sense. (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.)

share|cite|improve this question
up vote 8 down vote accepted

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?

share|cite|improve this answer
Thanks, Anton. This will do nicely. – Nikita Sidorov Nov 6 '10 at 3:28

If a map $f$ is Lipschitz, then it is a standard result in dimension theory that $\dim_H f(Z) \leq \dim_H Z$ for all $Z$. More generally, if a map $f$ is Hölder with exponent $\alpha \in (0,1]$, then one has the inequality $$ \dim_H f(Z) \leq \frac 1\alpha \dim_H Z. \qquad \qquad (*) $$ The proof of this uses the same sorts of calculations as in Anton's answer. Now the function $f$ that you describe has the property that there is a Cantor set $C$ such that $f$ is locally constant on the complement of $C$. The complement of $C$ is open, and hence is a countable union of open intervals, so its image under $f$ is a countable set.

In particular, this implies that $f(C)$ is the entire interval $[0,1]$ with an at most countable set removed. Thus $\dim_H f(C) = 1$, and if $f$ were Hölder continuous with exponent $\alpha>0$, the inequality in (*) would give $$ 1 = \dim_H f(C) \leq \frac 1\alpha \dim_H Z = 0, $$ a contradiction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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