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  "body": "\n## Definition\n\nLet $\\Omega \\subseteq \\C^n$ be a domain, and let $L^2(\\Omega)$ be the [[Lebesgue space|SASP14MA1Z]] of square-integrable functions for Lebesgue measure $dV$.\nLet $A^2(\\Omega) \\subseteq L^2(\\Omega)$ be the [[Bergman space|2D619WTPYC]] of holomorphic functions in $L^2(\\Omega)$, which is a closed subspace.\nThe *Bergman projection* of $\\Omega$ is the [[orthogonal projection|0B8KZRT28M]]\n$$P \\colon L^2(\\Omega) \\to A^2(\\Omega).$$\nIt is given by\n$$P f(z) = \\int_\\Omega K(z, w) \\, f(w) \\, dV(w),$$\nwhere $K$ is the [[Bergman kernel|8RWDHW0CXC]] of $\\Omega$.\n\n## Remarks\n\nThe integral formula follows from the reproducing property: for $f \\in L^2(\\Omega)$, $Pf(z) = \\ip{Pf}{K(\\cdot, z)} = \\ip{f}{K(\\cdot, z)}$, since $K(\\cdot, z) \\in A^2(\\Omega)$ and $f - Pf$ is orthogonal to $A^2(\\Omega)$.\nExample: on the unit disc, the functions $z^k$, $k \\ge 0$, form an orthogonal basis of $A^2$, and $\\ip{\\bar z}{z^k} = \\int_{\\mathbb{D}} \\bar z \\, \\bar z^k \\, dV = 0$ for every $k$, so $P \\bar z = 0$.\nSimilarly $P \\abs{z}^2 = \\frac{1}{2}$, since $\\ip{\\abs{z}^2}{1} = \\int_{\\mathbb{D}} \\abs{z}^2 dV = \\frac{\\pi}{2}$ and $\\norm{1}^2 = \\pi$, and $\\abs{z}^2$ is orthogonal to $z^k$ for $k \\ge 1$.\nNon-example: the orthogonal projection of $L^2(\\Omega)$ onto the harmonic functions in $L^2(\\Omega)$ is a different operator, the harmonic Bergman projection; on the unit disc it fixes $\\bar z$.\n$P$ is self-adjoint, $P^2 = P$, and $\\norm{P} = 1$ when $A^2(\\Omega) \\ne \\set{0}$.\nOn smoothly bounded strongly pseudoconvex domains, $P$ maps $C^\\infty(\\bar\\Omega)$ into itself.\n",
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