6
$\begingroup$

It is a standard fact that for $0\leq s\le1$, there is a compact set $C\subseteq [0,1]$ with Hausdorff and Minkowski dimensions $s$ (by modifying the construction of a Cantor set).

It is also a standard fact that for $0\leq s\le1$, there is a compact set $S\subseteq [0,1]$ with Fourier and Hausdorff dimensions $s$.

My question is: for an arbitrary $0\leq s\le1$, can we find a subset of $\mathbb R$ so that all three dimensions are equal?

$\endgroup$
3
  • $\begingroup$ Can you remind us of the definition of Fourier dimension? (Or give a link)? $\endgroup$ Sep 18 '20 at 15:58
  • $\begingroup$ The Fourier dimension of a bored set in R^n is the supremum of real numbers $s\in [0,n]$ such that there is a Borel probability measure supported on A, with decay of the order |x|^{-s/2}. $\endgroup$ Sep 18 '20 at 20:13
  • 1
    $\begingroup$ Here's a reference: Thomas William Körner, Hausdorff and Fourier dimension Studia Mathematica 206, Issue 1 (2011) pages 37-50, doi.org/10.4064/sm206-1-3 $\endgroup$ Sep 29 '20 at 22:01
2
$\begingroup$

Yes. We just use a Baire Category argument (a similar technique also works in high dimensions). Consider the complete metric space $X$ of pairs $(E,\mu)$, where $\mu$ is a probability measure supported on $E$ such that

$$ \sup_{\xi \in \mathbf{Z}} |\widehat{\mu}(\xi)| |\xi|^{s/2} < \infty, $$

and $E$ is a compact subset of $[0,1]$. We define a distance function

$$ d((E_1,\mu_1),(E_2,\mu_2)) = \max \left( d_H(E_1,E_2), \sup_{\xi \in \mathbf{Z}} |\widehat{\mu_1}(\xi) - \widehat{\mu_2}(\xi)| |\xi|^{s/2} \right) $$

where $d_H$ is the Hausdorff metric between two sets. It is a useful heuristic that a generic set is as `thin as possible' with respect to the Hausdorff metric. It is simple to see that for any $(E,\mu)$ in $X$, the Fourier dimension of $E$ is at least equal to $s$, so we should expect quasi-all elements of $X$ have dimension $s$.

For each $t > s$, $\delta > 0$, and $\varepsilon > 0$, set

$$ A(t,\delta,s) = \{ (E,\mu) \in \mathcal{X} : |E_\delta| < \varepsilon \cdot \delta^s \} $$

where $E_\delta$ is the $\delta$ thickening of $E$. Then $A(t,\delta,s)$ is an open subset of $X$, and

$$ \bigcap_{n = 1}^\infty \bigcap_{m = 1}^\infty \bigcap_{k = 1}^\infty A(s+1/n,1/m,1/k) $$

is the set of all pairs $(E,\mu)$ in $X$ where $E$ has Minkowski dimension $s$. Thus it suffices to argue that $A(t,\delta,\varepsilon)$ is dense in $X$ for all required parameters. It is slightly technical to argue this, but the basic idea is to consider a random construction which, given a pair $(E_0,\mu_0)$, considers the random measure

$$ \mu = \mu_0 \cdot \sum_{k = 1}^K \phi_{\varepsilon_0}(x - x_k) $$

where $x_1,\dots, x_K$ are uniformly distributed on $[0,1]$, $\varepsilon_0 = K^{-1/s}$, and $\phi_{\varepsilon_0}$ is a smooth bump function supported on a ball radius $\varepsilon_0$. One then shows that with high probability that

$$ \sup_{\xi \in \mathbf{Z}} |\widehat{\mu}(\xi) - \widehat{\mu_0}(\xi)| = o(1) $$

as $K \to \infty$, and that $d_H(\text{supp}(\mu), \text{supp}(\mu_0)) \to 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.