On the distance sets of AD-regular sets

Abstract

I prove that if $\varnothing \ne K\subset {\mathbb{R}}^2$ is a compact $s$-Ahlfors-David regular set with $s\ge 1$, then $${\dim }_{\mathrm{p}}D(K)=1,$$ where $D(K):=\{|x-y|:x,y\in K\}$ is the distance set of $K$, and ${\dim }_{\mathrm{p}}$ stands for packing dimension. The same proof strategy applies to other problems of similar nature. For instance, one can show that if $\varnothing \ne K\subset {\mathbb{R}}^2$ is a compact $s$-Ahlfors-David regular set with $s\ge 1$, then there exists a point $x_0\in K$ such that ${\dim }_{\mathrm{p}}K\cdot (K-x_0)=1$. Specialising to product sets, one derives the following sum-product corollary: if $A\subset \mathbb{R}$ is a non-empty compact $s$-Ahlfors-David regular set with $s\ge 1/2$, then $${\dim }_{\mathrm{p}}[A(A-a_1)+A(A-a_1)]=1$$ for some $a_1,a_2\in A$. In particular, ${\dim }_{\mathrm{p}}[AA+AA-AA-AA]=1$. In all of the results mentioned above, compactness can be relaxed to boundedness and ${\mathcal{H}}^s$-measurability, if packing dimension is replaced by upper box dimension.

12 pages. v3: The proof of the claimed "further results" in v2 contained a gap. The statements of these results have been downgraded accordingly

References in corpus (2)

• Projections of planar sets in well-separated directions
• Micromeasure distributions and applications for conformally generated fractals

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