A simple proof of the Mattila-Sjolin theorem

The Mattila-Sjolin theorem says that if the Hausdorff dimension of $E \subset {[0,1]}^d$ is greater than $\frac{d+1}{2}$ , the the distance set $\Delta(E)=\{|x-y|: x,y \in E \}$ contains an interval. Mihalis Mourgoglou, Krystal Taylor and recently extended this result to distance sets $\Delta_B(E)=\{ {||x-y||}_B: x,y \in E \}$ , where ${|| \cdot ||}_B$ is the distance generated by a symmetric convex body $B$ with a smooth boundary and everywhere non-vanishing Gaussian curvature. We also prove other things, but my goal here is to describe our proof of the Mattila-Sjolin result extended to smooth well-curved metrics.

Define the measure $\nu$ on $\Delta_B(E)$ by the relation

$\int f(t) d\nu(t)=\int \int f({||x-y||}_B) d\mu(x) d\mu(y)$ , where $d\mu$ is a Frostman measure on $E$ . We prove that

$\frac{\mu((t-\epsilon, t+\epsilon))}{\epsilon}=M(t)+R^{\epsilon}(t)$ , where $M(t)$ is continuous away from the origin and $\lim_{\epsilon \to 0} R^{\epsilon}(t)=0$ . Indeed, in place of $\frac{\mu((t-\epsilon, t+\epsilon))}{\epsilon}$ we may consider

$\int \int \sigma_t^{\epsilon}(x-y) d\mu(x) d\mu(y)$ , where $\sigma_t$ is the Lebesgue measure on $t\partial B$ and $\sigma^{epsilon}$ is $\sigma_t$ convolved with the approximation to the identity at level $\epsilon$ , i.e $\sigma_t^{\epsilon}(x)=\sigma_t* \rho(\cdot /\epsilon) \epsilon^{-d}(x)$ .

By Plancherel, this expression equals

$\int {|\widehat{\mu}(\xi)|}^2 \widehat{\sigma}(t\xi) \widehat{\rho}(\epsilon \xi) d\xi$

$=\int {|\widehat{\mu}(\xi)|}^2 \widehat{\sigma}(t\xi) d\xi+\int {|\widehat{\mu}(\xi)|}^2 \widehat{\sigma}(t\xi) (1-\widehat{\rho}(\epsilon \xi) d\xi$

$M(t)+R^{\epsilon}(t)$ .

The function $M(t)$ is continuous away fro $t=0$ by the dominated convergence theorem since

$|\widehat{\sigma}(t \xi)| \lesssim {|t\xi|}^{-\frac{d-1}{2}}$ by the method of stationary phase and

$\int {|\widehat{\mu}(\xi)|}^2 {|\xi|}^{-\frac{d-1}{2}} d\xi<\infty$ since $\mu$ is a Frostman measure on $E$ of Hausdorff dimension greater than $\frac{d+1}{2}$ .

It remains to show that $\lim_{\epsilon \to 0} R^{\epsilon}(t)=0$ . But

$|R^{\epsilon}(t)| \lessapprox \int_{|\xi|>\epsilon^{-1}} {|\widehat{\mu}(\xi)|}^2 {|\xi|}^{-\frac{d-1}{2}} d\xi$

$\lesssim \sum_{2^j>\epsilon^{-1}} 2^{-j \frac{d-1}{2}} \int_{|\xi| \approx 2^j} {|\widehat{\mu}(\xi)|}^2 d\xi$

$\lesssim \sum_{2^j>\epsilon^{-1}} 2^{j(d-s-\frac{d-1}{2})}$

$\sum_{2^j>\epsilon^{-1}} 2^{j(\frac{d+1}{2}-s)} \lesssim \epsilon^{s-\frac{d+1}{2}}$ and this quantity goes to $0$ as $\epsilon \to 0$ if $s>\frac{d+1}{2}$ .