On genericity of non-uniform Dvoretzky coverings of the circle

TL;DR

This paper investigates the conditions under which a sequence of random intervals with non-uniform distribution covers the circle infinitely often, extending classical results to cases with absolutely continuous densities and analyzing the impact of the density's support.

Contribution

It establishes necessary and sufficient conditions for non-uniform Dvoretzky coverings on the circle, including the role of the density's essential infimum and the geometric properties of its support.

Findings

The covering condition depends on the sum of lengths and the density's essential infimum.

If the support of the density has upper box-counting dimension less than 1, the main result holds.

A Menshov type result shows coverage can be achieved by small modifications of the density when the support has measure zero.

Abstract

The classical Dvoretzky covering problem asks for conditions on the sequence of lengths $\{\ell_n\}_{n\in \mathbb{N}}$ so that the random intervals $I_n : = (\omega_n -(\ell_n/2), \omega_n +(\ell_n/2))$ where $\omega_n$ is a sequence of i.i.d. uniformly distributed random variable, covers any point on the circle $\mathbb{T}$ infinitely often. We consider the case when $\omega_n$ are absolutely continuous with a density function $f$. When $m_f=essinf_\mathbb{T}f>0$ and the set $K_f$ of its essential infimum points satisfies $\overline{\dim}_\mathrm{B} K_f<1$, where $\overline{\dim}_\mathrm{B}$ is the upper box-counting dimension, we show that the following condition is necessary and sufficient for $\mathbb{T}$ to be $\mu_f$-Dvoretzky covered \[ \limsup_{n \rightarrow \infty} \left(\frac{\ell_1 + \dots + \ell_n}{\ln n}\right)\geq \frac{1}{m_f}. \] Under more restrictive assumptions on $\{\ell_n\}$ the above result is true if $\dim_H K_f<1$. We next show that as long as $\{\ell_n\}_{n\in \mathbb{N}}$ and $f$ satisfy the above condition and $|K_f|=0$, then a Menshov type result holds, i.e. Dvoretzky covering can be achieved by changing $f$ on a set of arbitrarily small Lebesgue measure. This, however, is not true for the uniform density.