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  "body": "\n## Statement\n\nLet $\\mu$ be a positive Borel [[measure|PR11BZQQPG]] on $\\R$ such that every [[moment|01XAHPADBA]] $m_k = \\int x^k \\, d\\mu(x)$, $k \\ge 0$, is finite.\nFor $n \\ge 0$, let $H_n = (m_{j+k})_{j,k=0}^{n}$ be the $(n+1) \\times (n+1)$ [[Hankel matrix|PQ1D8FQB3E]] of the moments.\nThen:\n\n1. $H_n$ is [[positive semidefinite|01MMVW8CKC]] for every $n \\ge 0$.\n2. If the [[support|FQTXV6BQZG]] of $\\mu$ is an infinite set, then $H_n$ is [[positive definite|0KX66CZMBB]] for every $n \\ge 0$, so $\\det H_n \u003e 0$.\n\n## Remarks\n\nFor real $c = (c_0, \\ldots, c_n)$, let $p(x) = \\sum_{j=0}^{n} c_j x^j$.\nThen $c^{T} H_n c = \\sum_{j,k} c_j c_k \\int x^{j+k} \\, d\\mu = \\int p(x)^2 \\, d\\mu(x) \\ge 0$, which is clause 1.\nIf $c \\ne 0$, then $p$ is a nonzero polynomial with at most $n$ real roots.\nIf the support is infinite, then $p^2 \u003e 0$ on a set of positive measure, so the integral is positive, which is clause 2.\nExample: for Lebesgue measure on $[0, 1]$, $m_k = 1/(k+1)$, and $H_n$ is the Hilbert matrix $(1/(j + k + 1))$, which is positive definite.\nFor $\\mu = \\delta_0$, the support is finite, $H_n$ has $m_0 = 1$ in one corner and zeros elsewhere, and it is only semidefinite for $n \\ge 1$.\nThe converse of clause 1 is Hamburger's theorem, part of the [[Hamburger moment problem|TBNCFXTMMW]].\nIt says that a real sequence with all $H_n$ positive semidefinite is the moment sequence of a positive measure on $\\R$.\n\n## Explanations\n\n### L1 human\n\nFrom the moments of a mass distribution, build square tables whose entry in row $j$ and column $k$ is the moment of order $j + k$.\nThese tables always behave like sums of squares: they never give a negative value.\nThe reason is that each value is the average of the square of a polynomial.\n\n### L2 human\n\nThis is the necessary condition in the Hamburger moment problem, and Hamburger showed that it is also sufficient.\nFor the Stieltjes problem on $[0, \\infty)$ one adds the shifted Hankel matrices $(m_{j+k+1})$, whose positivity reflects $x \\ge 0$.\nThe positive definite case gives an inner product $\\ip{p}{q} = \\int p q \\, d\\mu$ on polynomials, and Gram–Schmidt then yields the orthogonal polynomials of $\\mu$.\n",
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