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  "body": "\n## Definition\n\nLet $\\mathbb{D} = \\set{z \\in \\C : \\abs{z} \u003c 1}$ be the [[unit disc|8G5ABXRK85]] and $\\mathbb{T} = \\set{\\zeta \\in \\C : \\abs{\\zeta} = 1}$ its boundary circle.\nThe *Poisson kernel* of $\\mathbb{D}$ is the function $P \\colon \\mathbb{D} \\times \\mathbb{T} \\to \\R$ given by\n$$P(z, \\zeta) = \\frac{1 - \\abs{z}^2}{\\abs{\\zeta - z}^2}.$$\nIn polar form, for $0 \\le r \u003c 1$ and real $\\theta$, one writes $P_r(\\theta) = P(r e^{i\\theta}, 1) = \\dfrac{1 - r^2}{1 - 2r\\cos\\theta + r^2}$.\n\n## Remarks\n\nThe kernel is positive, and $P(z, \\zeta) = \\operatorname{Re} \\dfrac{\\zeta + z}{\\zeta - z}$, so $z \\mapsto P(z, \\zeta)$ is [[harmonic|0DEB0JMN0W]] on $\\mathbb{D}$.\nIt has the expansion $P_r(\\theta) = \\sum_{n \\in \\Z} r^{\\abs{n}} e^{i n\\theta}$, from which $\\frac{1}{2\\pi}\\int_0^{2\\pi} P_r(\\theta)\\, d\\theta = 1$.\nExample: at $z = 0$ one gets $P(0, \\zeta) = 1$ for every $\\zeta$, so the Poisson integral at the centre is the plain mean over the circle.\nFor a continuous function $\\varphi$ on $\\mathbb{T}$, the Poisson integral $u(z) = \\frac{1}{2\\pi}\\int_0^{2\\pi} P(z, e^{it}) \\varphi(e^{it})\\, dt$ is harmonic on $\\mathbb{D}$ and extends continuously to $\\varphi$ on $\\mathbb{T}$.\nNon-example: the function $(z, \\zeta) \\mapsto 1/\\abs{\\zeta - z}^2$ is not the Poisson kernel; it lacks the factor $1 - \\abs{z}^2$, it is not harmonic in $z$, and its integral over the circle is $1/(1 - \\abs{z}^2)$, not $1$.\nFor the upper half-plane, the Poisson kernel is $\\dfrac{1}{\\pi}\\dfrac{y}{(x - t)^2 + y^2}$ for $x + iy$ with $y \u003e 0$ and real $t$.\n",
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