Translational absolute continuity and Fourier frames on a sum of singular measures

TL;DR

This paper investigates conditions under which measures admit Fourier frames, showing that certain singular measures and their sums lack Fourier frames, while some rotated mixed measures do admit frames with bounds independent of rotation angle.

Contribution

It proves that measures supported on packing pairs are translationally singular and lack Fourier frames, and demonstrates a discontinuity in frame existence for rotated measures.

Findings

Sum of certain singular measures does not admit Fourier frames.

Rotated mixed measures admit Fourier frames with uniform bounds for most angles.

Fourier frame existence fails at specific rotation angles like ±π/2.

Abstract

A finite Borel measure $\mu$ in ${\mathbb R}^d$ is called a frame-spectral measure if it admits an exponential frame (or Fourier frame) for $L^2(\mu)$. It has been conjectured that a frame-spectral measure must be translationally absolutely continuous, which is a criterion describing the local uniformity of a measure on its support. In this paper, we show that if any measures $\nu$ and $\lambda$ without atoms whose supports form a packing pair, then $\nu\ast \lambda +\delta_t\ast\nu$ is translationally singular and it does not admit any Fourier frame. In particular, we show that the sum of one-fourth and one-sixteenth Cantor measure $\mu_4+\mu_{16}$ does not admit any Fourier frame. We also interpolate the mixed-type frame-spectral measures studied by Lev and the measure we studied. In doing so, we demonstrate a discontinuity behavior: For any anticlockwise rotation mapping $R_{\theta}$ with $\theta\ne \pm\pi/2$, the two-dimensional measure $\rho_{\theta} (\cdot): = (\mu_4\times\delta_0)(\cdot)+(\delta_0\times\mu_{16})(R_{\theta}^{-1}\cdot)$, supported on the union of $x$-axis and $y=(\cot \theta)x$, always admit a Fourier frame. Furthermore, we can find $\{e^{2\pi i \langle\lambda,x\rangle}\}_{\lambda\in\Lambda_{\theta}}$ such that it forms a Fourier frame for $\rho_{\theta}$ with frame bounds independent of $\theta$. Nonetheless, $\rho_{\pm\pi/2}$ does not admit any Fourier frame.