Non-Hölder continuous devil's staircases - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:04:27Z http://mathoverflow.net/feeds/question/45020 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45020/non-holder-continuous-devils-staircases Non-Hölder continuous devil's staircases Nikita Sidorov 2010-11-06T01:51:31Z 2010-11-06T03:20:24Z <p>Let $f:[0,1]\to[0,1]$ be a devil's staircase in the <a href="http://en.wikipedia.org/wiki/Singular_function" rel="nofollow">usual sense</a>. (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 <strong>Hausdorff dimension zero</strong>.</p> <p><em>Question</em>. Is it true that $f$ is <strong>not</strong> Hölder continuous? </p> <p>(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.)</p> http://mathoverflow.net/questions/45020/non-holder-continuous-devils-staircases/45024#45024 Answer by Anton Petrunin for Non-Hölder continuous devil's staircases Anton Petrunin 2010-11-06T03:06:37Z 2010-11-06T03:06:37Z <p>Let $K$ be the bad set. Assume $f$ is Hölder continuous with exponent $\alpha$.</p> <p>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 $$\sum_n\ell_n^\alpha&lt;\epsilon\ \ \ \ \ (*)$$ and $\ell_n&lt;\epsilon$ for any $n$. Set $v_i=f(b_i)-f(a_i)$. Since $f$ is Hölder continuous, $$v_i &lt; C{\cdot}\ell_i^\alpha.\ \ \ \ \ (**)$$ But clearly $$\sum v_i=1$$ which contradicts $( * )$ and $( * * )$. </p> <p>Did I miss something?</p> http://mathoverflow.net/questions/45020/non-holder-continuous-devils-staircases/45026#45026 Answer by Vaughn Climenhaga for Non-Hölder continuous devil's staircases Vaughn Climenhaga 2010-11-06T03:19:32Z 2010-11-06T03:19:32Z <p>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 H&ouml;lder 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. </p> <p>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 H&ouml;lder 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.</p>