{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Statement\n\nLet $(a_n)_{n \\ge 0}$ be real numbers and let $(b_n)_{n \\ge 1}$ be positive real numbers.\nDefine polynomials by $p_{-1} = 0$, $p_0 = 1$ and the [[recurrence|0FYX44GP92]]\n$$p_{n+1}(x) = (x - a_n)\\, p_n(x) - b_n\\, p_{n-1}(x) \\qquad (n \\ge 0).$$\nThen there is a positive [[Borel measure|DYR1QFNGNJ]] $\\mu$ on $\\R$ with infinite support and all moments finite such that\n$$\\int_\\R p_m(x)\\, p_n(x)\\, d\\mu(x) = 0 \\quad \\text{for } m \\ne n.$$\n\n## Remarks\n\nThe proof defines a linear functional $L$ on polynomials by $L(1) = 1$ and $L(p_n) = 0$ for $n \\ge 1$, which is possible since $p_n$ is monic of degree $n$.\nThe recurrence gives $L(p_m p_n) = 0$ for $m \\ne n$ and $L(p_n^2) = b_1 b_2 \\cdots b_n \u003e 0$, so $L$ is positive on nonzero squares of polynomials.\nThen $L(f) \\ge 0$ for every polynomial $f \\ge 0$ on $\\R$, since such $f$ is a sum of two squares, and the Hamburger moment problem gives a measure $\\mu$ with $L(f) = \\int f\\, d\\mu$.\nThe measure need not be unique.\nIt is unique when the moment problem is determinate, for instance when $(a_n)$ and $(b_n)$ are bounded, and then $\\mu$ has compact support.\nExample: $a_n = 0$ and $b_n = \\frac{1}{4}$ give the monic Chebyshev polynomials of the second kind, and $\\mu$ is the semicircle measure $\\frac{2}{\\pi} \\sqrt{1 - x^2}\\, dx$ on $[-1, 1]$.\nThe hypothesis $b_n \u003e 0$ is needed.\nFor orthogonal polynomials of a positive measure with infinite support, $b_n = \\frac{\\int p_n^2\\, d\\mu}{\\int p_{n-1}^2\\, d\\mu} \u003e 0$, so a recurrence with some $b_n \\le 0$ cannot come from such a measure.\n\n## Explanations\n\n### L1 human\n\nOrthogonal polynomials always satisfy a three-term recurrence.\nFavard's theorem says the converse: any polynomials built by such a recurrence, with positive coefficients in the right place, are orthogonal for some weight.\nSo one can produce orthogonal polynomials just by choosing two sequences of numbers.\n\n### L2 human\n\nThe theorem identifies orthogonal polynomial sequences on the real line with pairs of sequences $(a_n)$, $(b_n)$ with $b_n \u003e 0$, that is, with Jacobi matrices.\nIn operator language, $\\mu$ is a spectral measure of the Jacobi operator with diagonal $a_n$ and off-diagonal $\\sqrt{b_n}$.\nIt reverses the [[three-term recurrence|YRYWCP9TYV]], and its proof reduces to the Hamburger moment problem.\nIt is used to show that families defined by recurrences, such as many in the Askey scheme, are orthogonal before the weight is known.\n",
  "commit": "ff8a40749dc16545e3d2fc637f6b0196e81d0bfa",
  "front": {
    "about": [
      "0FYX44GP92"
    ],
    "kind": "statement",
    "msc": [
      "42C05",
      "44A60"
    ],
    "name": "Favard's theorem",
    "role": "theorem",
    "same_as": {
      "wikidata": "Q5438430",
      "wikipedia": "Favard's theorem"
    },
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Favard%27s_theorem"
      }
    ],
    "tag": "1YADG8ESHY",
    "uses": [
      "DYR1QFNGNJ",
      "0FYX44GP92"
    ]
  },
  "kind": "statement",
  "link": "https://tamnd.github.io/math2/node/1YADG8ESHY/",
  "links": [
    {
      "field": "about",
      "target": "0FYX44GP92"
    },
    {
      "field": "uses",
      "target": "DYR1QFNGNJ"
    },
    {
      "field": "uses",
      "target": "0FYX44GP92"
    }
  ],
  "markdown": "https://tamnd.github.io/math2/node/1YADG8ESHY.md",
  "msc": [
    "42C05",
    "44A60"
  ],
  "name": "Favard's theorem",
  "prov": null,
  "role": "theorem",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/1YA/1YADG8ESHY.md",
  "status": null,
  "tag": "1YADG8ESHY",
  "uri": "m2://1YADG8ESHY",
  "url": "https://tamnd.github.io/math2/wiki/favards-theorem/",
  "writing": {
    "tag": "1YADG8ESHY",
    "kind": "statement",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}
