On Falconer’s distance set problem in the plane

Abstract

If $E \subset \mathbb {R}^2$ is a compact set of Hausdorff dimension greater than 5 / 4, we prove that there is a point $x \in E$ so that the set of distances $\{ |x-y| \}_{y \in E}$ has positive Lebesgue measure.

Appendix: Discussion of the lower bound on the upper Minkowski dimension of \(\Delta _{x,K}(E)\) in Remark 1.4

Let $\rho$ be a smooth cut-off function supported in the ball of radius 2 and equal to 1 in the ball of radius 1 centered at the origin. Let $\rho _{\delta }(x)=\delta ^{-d} \rho (\delta ^{-1}x)$. Following the argument in (2.5) with $\mu _{1,good}$ replaced by $\mu _{1,good}*\rho _{\delta }$, we see that the Lebesgue measure of the $\delta$-neighborhood of $\Delta _{x,K}(E)$ is bounded from below by

$$\begin{aligned} \frac{\left( 1-\frac{2}{1000}\right) ^2}{\int {|d_{*}^{x}\mu _{1,good}*\rho _{\delta }|}^2}.\end{aligned}$$

Following (5.2) with $\mu _{1,good}$ replaced by $\mu _{1,good}*\rho _{\delta }$, we see that the expression above is bounded from below by $C \delta ^{\frac{5}{3}-\frac{4 \alpha }{3}+\epsilon }$, hence there exists $x \in E$ such that the upper Minkowski dimension of $\Delta _{x,K}(E)$ is bounded from below by

$$\begin{aligned} 1-\left( \frac{5}{3}-\frac{4 \alpha }{3} \right) =\frac{4}{3}\alpha -\frac{2}{3}, \end{aligned}$$

as claimed.

It would be interesting to obtain a lower bound on the Hausdorff dimension of $\Delta _{x,K}(E)$. If $\mu _{1,good}$ were positive, it would be sufficient to show that

$$\begin{aligned} \int _{E_2} I_{\gamma } d^x_*(\mu _{{1, good}}) d\mu _2(x) \end{aligned}$$

(9.1)

is bounded with $\gamma <\frac{4}{3}\alpha -\frac{2}{3}$. This estimate follows from the same argument as in (5.2) above. Unfortunately, in view of the fact that $\mu _{1,good}$ is complex valued, the estimate (9.1) does not appear to be sufficient to draw the desired conclusion.