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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.)

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2 Answers

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?

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Thanks, Anton. This will do nicely. –  Nikita Sidorov Nov 6 '10 at 3:28
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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.

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