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

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.