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  "body": "\n## Definition\n\nLet $\\mathbb{D}$ be the open [[unit disc|8G5ABXRK85]].\nLet $\\varphi$ be a nonnegative measurable function on the unit circle such that $\\log \\varphi$ is integrable with respect to [[Lebesgue measure|2Z2QVWZ262]].\nLet $c \\in \\C$ with $\\abs{c} = 1$.\nThe *outer function* with modulus $\\varphi$ and constant $c$ is the function $F \\colon \\mathbb{D} \\to \\C$ given by\n$$F(z) = c \\exp\\left(\\frac{1}{2\\pi} \\int_0^{2\\pi} \\frac{e^{it} + z}{e^{it} - z} \\log \\varphi(e^{it}) \\, dt\\right).$$\nA holomorphic function on $\\mathbb{D}$ is *outer* if it is of this form for some $\\varphi$ and $c$.\n\n## Remarks\n\n$F$ is [[holomorphic|0DG0Z72CQD]] and has no zeros, and $\\log \\abs{F}$ is the [[Poisson integral|TCYSWCTYYM]] of $\\log \\varphi$.\nIts radial limits satisfy $\\abs{F^*} = \\varphi$ almost everywhere.\nFor $0 \u003c p \\le \\infty$, $F$ lies in the Hardy space $H^p$ if and only if $\\varphi \\in L^p$ ([[Lp space|SASP14MA1Z]]).\n\nExample: $F(z) = 2 + z$ is outer.\nIt is holomorphic and has no zeros on a neighbourhood of the closed disc, so $\\log \\abs{F}$ is harmonic and continuous up to the circle.\nHence $\\log \\abs{F}$ equals the Poisson integral of $\\log \\varphi$, where $\\varphi = \\abs{2 + e^{it}}$.\nSo $F$ and the outer function with modulus $\\varphi$ have the same modulus and no zeros; their quotient is a holomorphic function of constant modulus $1$, hence a constant.\n\nNon-example: $f(z) = z$ is not outer.\nIts boundary modulus is $\\varphi = 1$, and the outer functions with this modulus are the constants of modulus $1$, which differ from $z$.\n",
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