On the distance sets of AD-regular sets

I prove that if $\emptyset \neq K \subset \mathbb{R}^{2}$ is a compact $s$-Ahlfors-David regular set with $s \geq 1$, 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 $\emptyset \neq K \subset \mathbb{R}^{2}$ is a compact $s$-Ahlfors-David regular set with $s \geq 1$, 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 $s \geq 1/2$, 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.

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