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  "body": "\n## Statement\n\nLet $\\mu$ be a probability measure on the unit circle $\\partial\\mathbb{D}$ with infinite support, and let $\\Phi_n$ be its monic [[orthogonal polynomials|3F9G3DRQ62]].\nFor a polynomial $p$ of degree at most $n$, let $p^*(z) = z^n \\, \\overline{p(1/\\bar z)}$ be its reversed polynomial of order $n$; so $\\Phi_n^*$ has constant term $1$.\n\n1. There are complex numbers $\\alpha_0, \\alpha_1, \\alpha_2, \\dots$ with $\\abs{\\alpha_n} \u003c 1$ such that for every $n \\ge 0$\n$$\\Phi_{n+1}(z) = z \\, \\Phi_n(z) - \\overline{\\alpha_n} \\, \\Phi_n^*(z).$$\nThey are given by $\\alpha_n = -\\overline{\\Phi_{n+1}(0)}$ and are called the *Verblunsky coefficients* of $\\mu$.\n2. The norms satisfy $\\norm{\\Phi_{n+1}}^2 = (1 - \\abs{\\alpha_n}^2) \\norm{\\Phi_n}^2$, so $\\norm{\\Phi_n}^2 = \\prod_{j=0}^{n-1} (1 - \\abs{\\alpha_j}^2)$.\n3. (Verblunsky's theorem) Let $\\mathbb{D}$ be the open unit disk. The map $\\mu \\mapsto (\\alpha_n)_{n \\ge 0}$ is a bijection from the probability measures on $\\partial\\mathbb{D}$ with infinite support onto the sequences in $\\mathbb{D}$.\n\n## Remarks\n\nFor part 1, the polynomial $\\Phi_{n+1} - z\\Phi_n$ has degree at most $n$.\nIt is orthogonal to $z, z^2, \\dots, z^n$, because $\\Phi_{n+1}$ is, and multiplication by $z$ is unitary on $L^2(\\mu)$ and maps $\\Phi_n \\perp \\set{1, \\dots, z^{n-1}}$ onto $z\\Phi_n \\perp \\set{z, \\dots, z^n}$.\nThe polynomials of degree at most $n$ orthogonal to $z, \\dots, z^n$ form a one-dimensional space spanned by $\\Phi_n^*$, which gives the recurrence.\nTaking norms, with $\\norm{z\\Phi_n} = \\norm{\\Phi_n^*} = \\norm{\\Phi_n}$, gives part 2, and $\\norm{\\Phi_{n+1}} \u003e 0$ forces $\\abs{\\alpha_n} \u003c 1$.\nPart 3 builds $\\mu$ from any sequence in $\\mathbb{D}$ by running the recurrence and taking a weak limit of the measures $\\abs{\\varphi_n(e^{i\\theta})}^{-2} d\\theta/2\\pi$.\n\nExample: for $d\\mu = d\\theta/2\\pi$, $\\Phi_n(z) = z^n$ and $\\Phi_n^* = 1$, so $\\Phi_{n+1} = z\\Phi_n$ and all $\\alpha_n = 0$.\nConversely, by part 3, the uniform measure is the only one with all Verblunsky coefficients zero.\n\nInfinite support is needed.\nIf $\\mu$ is supported on $N$ points, then $L^2(\\mu)$ has dimension $N$ and $\\Phi_N$ has norm $0$; the recurrence ends with $\\abs{\\alpha_{N-1}} = 1$.\n\n## Explanations\n\n### L1 human\n\nOn the real line, orthogonal polynomials are built step by step by a rule with two numbers per step.\nOn the unit circle the rule is simpler: one complex number $\\alpha_n$ per step, of modulus less than $1$.\nEvery measure on the circle gives such a sequence, and every such sequence comes from exactly one measure.\nSo the sequence of Verblunsky coefficients is a complete, free set of coordinates for measures on the circle.\n\n### L2 human\n\nThe recurrence turns questions about the measure $\\mu$ into questions about the sequence $(\\alpha_n)$.\nFor example, [[Szegő's theorem|SZ8ARMDEMY]] says that $\\sum \\abs{\\alpha_n}^2 \u003c \\infty$ exactly when $\\log$ of the density of $\\mu$ is integrable.\nThe $\\alpha_n$ are the entries of the CMV matrix, the unitary analogue of the Jacobi matrix, which represents multiplication by $z$ on $L^2(\\mu)$.\nIn prediction theory and signal processing the same numbers appear as reflection coefficients, computed by the Levinson algorithm from the moments of $\\mu$.\n",
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