{
  "id": "1802.00350",
  "version": "v1",
  "published": "2018-02-01T15:36:15.000Z",
  "updated": "2018-02-01T15:36:15.000Z",
  "title": "An $L^2$-identity and pinned distance problem",
  "authors": [
    "Bochen Liu"
  ],
  "comment": "10 pages",
  "categories": [
    "math.CA",
    "math.AP",
    "math.CO",
    "math.MG"
  ],
  "abstract": "We prove that if the Hausdorff dimension of $E\\subset\\mathbb{R}^d$, $d\\geq 2$ is greater than $\\frac{d}{2}+\\frac{1}{3}$, there exists $x\\in E$ such that the pinned distance set $$\\Delta_x(E)=\\{|y-x|: y\\in E \\}$$ has positive Lebesgue measure. This improves a result of Peres and Schlag. The key new ingredient in our proof is the following identity. Using a group action argument, we show that for any Schwartz function $f$ on $\\mathbb{R}^d$ and any $x\\in\\mathbb{R}^d$, $$\\int_0^\\infty |\\omega_t*f(x)|^2\\,t^{d-1}dt\\,=\\int_0^\\infty|\\widehat{\\omega_r}*f(x)|^2\\,r^{d-1}dr,$$ where $\\omega_r$ is the normalized surface measure on $r S^{d-1}$. An interesting remark is that the right hand side can be easily seen equals $$c_d\\int_0^\\infty\\left|D_x^{-\\frac{d-1}{2}}e^{-2\\pi i t\\sqrt{-\\Delta}}f(x)\\right|^2\\,dt=c_d'\\int_0^\\infty\\left|D_x^{-\\frac{d-2}{2}}e^{2\\pi i t\\Delta}f(x)\\right|^2\\,dt,$$ where $\\Delta$ is the standard Laplacian and $D_x^\\alpha=(-\\Delta)^{\\frac{\\alpha}{2}}$.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2018-02-01T15:36:15.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "28A75",
      "42B20"
    ],
    "keywords": [
      "pinned distance problem",
      "right hand side",
      "group action argument",
      "pinned distance set",
      "positive lebesgue measure"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 10,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}