# Subset of $\mathbb R$ with equal Fourier, Hausdorff and Minkowski dimensions

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?

• Can you remind us of the definition of Fourier dimension? (Or give a link)? Sep 18 '20 at 15:58
• 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}. Sep 18 '20 at 20:13
• 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 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 $$d_H(\text{supp}(\mu), \text{supp}(\mu_0)) \to 0$$.