The Experts below are selected from a list of 66 Experts worldwide ranked by ideXlab platform

Chunkit Lai - One of the best experts on this subject based on the ideXlab platform.

  • translational absolute continuity and fourier frames on a sum of singular Measures
    Journal of Functional Analysis, 2017
    Co-Authors: Chunkit Lai
    Abstract:

    Abstract A Finite Borel Measure μ in R d is called a frame-spectral Measure if it admits an exponential frame (or Fourier frame) for L 2 ( μ ) . 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 ν and λ without atoms whose supports form a packing pair, then ν ⁎ λ + δ t ⁎ ν 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 μ 4 + μ 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 θ with θ ≠ ± π / 2 , the two-dimensional Measure ρ θ ( ⋅ ) : = ( μ 4 × δ 0 ) ( ⋅ ) + ( δ 0 × μ 16 ) ( R θ − 1 ⋅ ) , supported on the union of x-axis and y = ( cot ⁡ θ ) x , always admit a Fourier frame. Furthermore, we can find { e 2 π i 〈 λ , x 〉 } λ ∈ Λ θ such that it forms a Fourier frame for ρ θ with frame bounds independent of θ. Nonetheless, ρ ± π / 2 does not admit any Fourier frame.

  • translational absolute continuity and fourier frames on a sum of singular Measures
    arXiv: Functional Analysis, 2017
    Co-Authors: Chunkit Lai
    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.

Tamás Keleti - One of the best experts on this subject based on the ideXlab platform.

  • is lebesgue Measure the only sigma Finite invariant Borel Measure
    arXiv: Classical Analysis and ODEs, 2011
    Co-Authors: Márton Elekes, Tamás Keleti
    Abstract:

    R.D.Mauldin asked if every translation invariant $\sigma$-Finite Borel Measure on $\RR^d$ is a constant multiple of Lebesgue Measure. The aim of this paper is to show that the answer is "yes and no", since surprisingly the answer depends on what we mean by Borel Measure and by constant. We present Mauldin's proof of what he called a folklore result, stating that if the Measure is only defined for Borel sets then the answer is affirmative. Then we show that if the Measure is defined on a $\sigma$-algebra \emph{containing} the Borel sets then the answer is negative. However, if we allow the multiplicative constant to be infinity, then the answer is affirmative in this case as well. Moreover, our construction also shows that an isometry invariant $\sigma$-Finite Borel Measure (in the wider sense) on $\RR^d$ can be non-$\sigma$-Finite when we restrict it to the Borel sets.

  • Is Lebesgue Measure the only σ-Finite invariant Borel Measure?
    Journal of Mathematical Analysis and Applications, 2006
    Co-Authors: Márton Elekes, Tamás Keleti
    Abstract:

    AbstractS. Saks and recently R.D. Mauldin asked if every translation invariant σ-Finite Borel Measure on Rd is a constant multiple of Lebesgue Measure. The aim of this paper is to investigate the versions of this question, since surprisingly the answer is “yes and no,” depending on what we mean by Borel Measure and by constant. According to a folklore result, if the Measure is only defined for Borel sets, then the answer is affirmative. We show that if the Measure is defined on a σ-algebra containing the Borel sets, then the answer is negative. However, if we allow the multiplicative constant to be infinity, then the answer is affirmative in this case as well. Moreover, our construction also shows that an isometry invariant σ-Finite Borel Measure (in the wider sense) on Rd can be non-σ-Finite when we restrict it to the Borel sets

Márton Elekes - One of the best experts on this subject based on the ideXlab platform.

  • is lebesgue Measure the only sigma Finite invariant Borel Measure
    arXiv: Classical Analysis and ODEs, 2011
    Co-Authors: Márton Elekes, Tamás Keleti
    Abstract:

    R.D.Mauldin asked if every translation invariant $\sigma$-Finite Borel Measure on $\RR^d$ is a constant multiple of Lebesgue Measure. The aim of this paper is to show that the answer is "yes and no", since surprisingly the answer depends on what we mean by Borel Measure and by constant. We present Mauldin's proof of what he called a folklore result, stating that if the Measure is only defined for Borel sets then the answer is affirmative. Then we show that if the Measure is defined on a $\sigma$-algebra \emph{containing} the Borel sets then the answer is negative. However, if we allow the multiplicative constant to be infinity, then the answer is affirmative in this case as well. Moreover, our construction also shows that an isometry invariant $\sigma$-Finite Borel Measure (in the wider sense) on $\RR^d$ can be non-$\sigma$-Finite when we restrict it to the Borel sets.

  • Is Lebesgue Measure the only σ-Finite invariant Borel Measure?
    Journal of Mathematical Analysis and Applications, 2006
    Co-Authors: Márton Elekes, Tamás Keleti
    Abstract:

    AbstractS. Saks and recently R.D. Mauldin asked if every translation invariant σ-Finite Borel Measure on Rd is a constant multiple of Lebesgue Measure. The aim of this paper is to investigate the versions of this question, since surprisingly the answer is “yes and no,” depending on what we mean by Borel Measure and by constant. According to a folklore result, if the Measure is only defined for Borel sets, then the answer is affirmative. We show that if the Measure is defined on a σ-algebra containing the Borel sets, then the answer is negative. However, if we allow the multiplicative constant to be infinity, then the answer is affirmative in this case as well. Moreover, our construction also shows that an isometry invariant σ-Finite Borel Measure (in the wider sense) on Rd can be non-σ-Finite when we restrict it to the Borel sets

Lining Tong - One of the best experts on this subject based on the ideXlab platform.

Tong Lining - One of the best experts on this subject based on the ideXlab platform.