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  "body": "\n## Definition\n\nLet $X$ be a nonempty set and let $\\mathbb{K}$ be $\\R$ or $\\C$.\nLet $H$ be a vector space of functions $X \\to \\mathbb{K}$, with pointwise operations, that is a [[Hilbert space|HM2HHE3J51]] for some inner product.\nThe space $H$ is a *reproducing kernel Hilbert space* if the following holds.\nFor every $x \\in X$, the evaluation map $f \\mapsto f(x)$ is a [[continuous linear map|03TM135G0N]] from $H$ to $\\mathbb{K}$.\n\n## Remarks\n\nFor each $x$ there is then a unique $k_x \\in H$ with $f(x) = \\ip{f}{k_x}$ for all $f \\in H$.\nThe function $K(x, y) = k_y(x) = \\ip{k_y}{k_x}$ on $X \\times X$ is the *reproducing kernel* of $H$.\n\nExample: $H = \\ell^2(\\N)$, seen as functions on $X = \\N$.\nEvaluation at $n$ is $f \\mapsto f(n)$, and $\\abs{f(n)} \\le \\norm{f}_2$, so it is continuous.\nHere $k_n$ is the unit vector $e_n$ and $K(m, n) = 1$ if $m = n$ and $0$ otherwise.\n\nNon-example: $L^2([0, 1])$ is not a reproducing kernel Hilbert space.\nIts elements are classes of functions equal almost everywhere, so $f \\mapsto f(x)$ is not even defined on it.\nThe continuous functions on $[0, 1]$ with the $L^2$ inner product are not one either, because this space is not complete.\n\nThe Hardy space of the unit disc is a reproducing kernel Hilbert space with kernel $K(z, w) = 1/(1 - z \\bar{w})$.\n",
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