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

机译:平移绝对连续性和傅立叶框架的单数措施

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A finite Borel measure mu in 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 * lambda + delta(t) * 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-0 with theta not equal +/-pi/2, the two-dimensional measure rho(0)(.) := (mu(4) x delta(0))(.) + (delta(0) x mu(16))(R-theta(-1).), 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 lambda,x )}(lambda is an element of Lambda theta), such that it forms a Fourier frame for rho(theta) with frame bounds independent of theta. Nonetheless, rho(+/-pi/2) does not admit any Fourier frame. (C) 2017 Elsevier Inc. All rights reserved.
机译:R-D中的有限硼测量MU被称为帧谱度量,如果它承认L-2(MU)的指数帧(或傅里叶帧)。已经猜测帧谱测量必须是完全连续的,这是描述其支持措施的局部均匀性的标准。在本文中,我们表明,如果没有原子的任何措施的措施,那么Nu * lambda + delta(t)* nu是翻译奇异的,它不承认任何傅立叶帧。特别是,我们表明第四个和第十六令牌的总和测量mu(4)+ mu(16)不承认任何傅立叶帧。我们还在借助于LEV和我们所研究的措施中插入混合型帧谱措施。在这样做时,我们证明了一个不连续性行为:对于任何逆时针旋转映射映射R-0不等于+/- Pi / 2,二维测量Rho(0)(。):=(mu(4)x delta (0))(。)+(Δ(0)x mu(16))(R-Theta(-1)。),支持X轴和Y =(COT THETA)X的联盟,始终承认傅里叶框架。此外,我们可以找到{e(2 pi i& x&)}(lambda是lambdaθ的一个元素),使得它形成rho(θ)的傅里叶框架,与θf的帧界无关。尽管如此,rho(+/pi / 2)不承认任何傅立叶帧。 (c)2017年Elsevier Inc.保留所有权利。

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