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 $\theta \mathrm{\ne }\pm \pi \mathrm{/}2$, 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, ${\rho }_{\pm \pi \mathrm{/}2}$ does not admit any Fourier frame.