Hausdorff Dimension and Hölder Continuity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:59:08Z http://mathoverflow.net/feeds/question/16959 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16959/hausdorff-dimension-and-holder-continuity Hausdorff Dimension and Hölder Continuity Brent Werness 2010-03-03T07:46:45Z 2012-05-29T14:53:54Z <p>Suppose we have a curve &gamma; : [0,1] -> &#x211D;<sup>n</sup>. It is well known that if this curve is Hölder continuous for some exponent &alpha; then the Hausdorff dimension of &gamma;[0,1] is bounded above by 1/&alpha;.</p> <p>My question is: Is there a partial converse of the following form. Suppose that the curve &gamma; is such that &gamma;[0,1] has Hausdorff dimension d, then is it true that for any &alpha; &lt; 1/d we have that there exists some <em>reparameterization</em> of &gamma; so that &gamma; is Hölder continuous with exponent &alpha;?</p> <p>I would like it to be true, although I doubt it is, however I have not been able to find a counterexample. Proofs, references, or counterexamples would be appreciated.</p> http://mathoverflow.net/questions/16959/hausdorff-dimension-and-holder-continuity/16962#16962 Answer by Lasse Rempe-Gillen for Hausdorff Dimension and Hölder Continuity Lasse Rempe-Gillen 2010-03-03T09:45:47Z 2010-03-03T15:08:06Z <p>Edit: The short answer is that there are planar curves that cannot be parametrized in a Holder continuous manner. Thus any such curve provides a counterexample for some $d \le 2$.</p> <p>My original answer, showing that the curve can be chosen even of Hausdorff dimension $d=1$, follows.</p> <p>I believe the answer to your question is negative. Indeed, I would think you can construct a curve in $\mathbb{R}^2$ that has Hausdorff dimension 1 but cannot be parametrized by a Holder-continuous curve.</p> <p>The basic idea is that at smaller and smaller scales, the curve oscillates rather wildly in the horizontal direction, which can be done to ensure that the curve cannot be parametrized in a Holder-continuous way. However, if we make sure that the vertical extent of these oscillations is much, much smaller at each step, then the Hausdorff dimension should still be one. (Of course the linear measure of such a curve will be infinite.)</p> <p>Let me try to be a bit more precise. I hope the idea is suitably clear.</p> <p>Let us suppose that we construct the curve $\gamma_k$ at successive scales $\delta_k = 2^{-k}$. The curve $\gamma_k$ will consist of a number $M_k$ of horizontal segments, each of length $\delta_k$, together with some vertical segments. In the construction, we may ensure that the vertical segments have bounded total length, so we can just ignore them. </p> <p>To obtain $\gamma_{k+1}$, we split each horizontal segment in two segments of length $\delta_{k+1}$, and then replace each of these segments by a curve that oscillates, consisting of some number $m_{k+1}$ of horizontal segments of length $\delta_{k+1}$. However, the vertical extent of this oscillation should be some very small number $\varepsilon_{k+1}$. (Which in particular we may make so small that the total length of vertical segments added in this step is smaller than, say $2^{-k}$, so the total length of vertical segments stays bounded at all times.) </p> <p>Now the total number of segments at stage $k+1$ is $M_{k+1} = 2m_{k+1} M_k$. Any parametrization of this curve by the unit interval must contain two points that are at distance at most $1/M_{k+1}$ but have points whose images are at distance at least $\delta_{k+1}$. So if we choose $m_{k+1}$ sufficiently large, we can ensure that the limit curve cannot be Holder-parametrized.</p> <p>On the other hand, the curve can be covered by about</p> <p>$M_{k+1}\cdot\frac{\delta_{k+1}}{\varepsilon_{k+1}}$</p> <p>sets of diameter $\varepsilon_{k+1}$. If we make $\varepsilon_{k+1}$ small enough, this number will be smaller than, say $\varepsilon_{k+1}^{-(1+1/k)}$, so the Hausdorff dimension of the limiting curve will be equal to one.</p>