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Given a measurable set $K\subset \mathbb{R}^d$ we consider the occupation ratio $$f_r(x)=vol(K\cap B(x,r))/r^d$$ and especially the asymptotics when $r\to 0$. When $K$ has a fractal boundary and $x$ is on the boundary, it is not clear whether most $x$'s have a large occupation ratio.

Has anyone already seen this quantity? Is it studied somewhere?

If you want a precise question, let $K$ be a non-negligible set with self-similar boundary with dimension $s$ and positive lower Minkowski content. Do we have $$\liminf_r\int_{\partial K}f_r(x)d\nu(x)>0$$ where $\nu$ is the $s$-dimensional Hausdorff measure? For the Von Koch flake i think the answer is yes but I don't see a general scheme.

EDIT: I insist on the fact that the set $K$ itself is not fractal, but its boundary is.

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  • $\begingroup$ Why you are interested in $r\to \infty$?, as fractals are usually defined as compact sets (say attractors), that will give you $0$ in the limit for any $K,x$. The obvious direction would be to take $r\to 0$, which leads to something similar to density, and the answer to this question would be basically Frostman's lemma. $\endgroup$
    – Asaf
    Commented Aug 19, 2014 at 18:09
  • $\begingroup$ Right. Presumably the $\infty$ is a typo. All points have the same limit in the $\infty$ direction. $\endgroup$
    – Anthony Quas
    Commented Aug 19, 2014 at 22:14
  • $\begingroup$ Indeed, edited. Thanks for pointing it out. I don't think it is related to Frostman's lemma because I don't look at the measure of the boundary, but the measure of the intersection with the set (and I don't divide by $r^s$ but by $r^d$). $\endgroup$
    – kaleidoscop
    Commented Aug 20, 2014 at 9:14
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    $\begingroup$ It looks as if $f_r(x)$ tends to the derivative of the measure $\mu(A)= vol(K\cap A)$ w.r.t. Lebesgue measure. Lebegue's theorem (see e.g. Rudin's Real and Complex Analysis, chapter 8) tells you that $\lim f_r = 1_K$ almost everywhere which, unfortunately, does not say anything about the integral w.r.t. the Hausdorff measure on $\partial K$. $\endgroup$
    – Jochen Wengenroth
    Commented Aug 20, 2014 at 13:36
  • $\begingroup$ For the sets I am interested in, the topological boundary is negligible, so indeed a result valid a.e. in the plane is not relevant. But I will try to search with these key-words, thanks. $\endgroup$
    – kaleidoscop
    Commented Aug 20, 2014 at 14:21

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I don't have a reference for such quantities, but I think the answer to your precise question is negative.

Consider the Sierpiński triangle $S$ as the boundary of $K$. Then if you define $K$ as the union $S \cup \text{blue triangle}$ (see the picture below), then $$\liminf_r\int_{\partial K}f_r(x)d\nu(x) = 0.$$ This is (for example) because the $\partial (\text{int}K)$ is the boundary of the blue triangle which has lower dimension than the dimension of $S$.

alt text (source)

If you would like to require that $\partial (\text{int}K) = S$ still the answer would be negative. To see this, you can consider $K = S \cup \text{red triangles}$ where we have an infinite number of "removed" (red) triangles as shown in the second picture. By having sufficiently large gaps between the scales at which you attach the removed triangles, you still have $$\liminf_r\int_{\partial K}f_r(x)d\nu(x) = 0.$$

The next question could be what happens if you consider $$\int_{\partial K}\liminf_rf_r(x)d\nu(x),$$ but even for this I think the second example, if correctly tuned, should give zero.

One could go further with inventing question and ask for example what happens if you require the boundary to be a self-similar type Jordan curve. Then the answer should be positive.

As a final remark: I guess you could do the above examples in $\mathbb R$ with a Cantor set, but I preferred a planar case for easier illustration.

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  • $\begingroup$ Fair enough, you answered the question. As you mention at the end my interest is more in the case where the boundary is a Jordan curve, such as in the VK flake, but it proves that the question is indeed tricky. $\endgroup$
    – kaleidoscop
    Commented Aug 21, 2014 at 11:20
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    $\begingroup$ If you consider the Jordan curve case and assume that the iterated function system that produces the boundary (locally) satisfies the open set condition, then the boundary has positive Hausdorff measure with the similitude dimension and the iterated open sets in the OSC contain at least a fixed portion (in Lebesgue measure) of the set $K$ in comparison to the measure of the open set. Thus for small $r$ the quantity $f_r(x)$ is uniformly bounded below by a positive constant and the integral you consider is positive. $\endgroup$
    – Tapio Rajala
    Commented Aug 22, 2014 at 5:59

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