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?

  • $\begingroup$ 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

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 $d_H(\text{supp}(\mu), \text{supp}(\mu_0)) \to 0$.


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