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  "body": "\n## Definition\n\nLet $(\\Omega, \\mathcal{A}, m)$ be a $\\sigma$-finite [[measure space|P15VQS5DJ9]] and let $g \\colon \\Omega \\times \\R \\to \\R$ be a *Carathéodory function*:\n\n1. for every $u \\in \\R$, the map $x \\mapsto g(x, u)$ is [[measurable|YYNCJSR0PJ]];\n2. for almost every $x \\in \\Omega$, the map $u \\mapsto g(x, u)$ is continuous.\n\nThe *Nemytskii operator* of $g$ sends each measurable function $u \\colon \\Omega \\to \\R$ to the function $N_g u \\colon \\Omega \\to \\R$ given by\n$$(N_g u)(x) = g\\bigl(x, u(x)\\bigr).$$\nUnder the two conditions, $N_g u$ is again measurable.\n\n## Remarks\n\nA theorem of Krasnosel'skii describes the [[$L^p$ spaces|SASP14MA1Z]] case.\nLet $\\Omega \\subseteq \\R^m$ be measurable with Lebesgue measure, and let $1 \\le p, q \u003c \\infty$.\nThen $N_g$ maps $L^p(\\Omega)$ into $L^q(\\Omega)$ if and only if there are $a \\in L^q(\\Omega)$ and $b \\ge 0$ with $\\abs{g(x, u)} \\le a(x) + b \\abs{u}^{p/q}$ for almost every $x$ and all $u$.\nIn that case $N_g \\colon L^p(\\Omega) \\to L^q(\\Omega)$ is continuous and maps bounded sets to bounded sets.\n\nExample: $\\Omega = [0, 1]$ with Lebesgue measure and $g(x, u) = u^2$.\nBoth conditions hold, and $\\abs{g(x, u)} \\le \\abs{u}^{4/2}$, so $N_g$ maps $L^4[0, 1]$ into $L^2[0, 1]$; indeed $\\int_0^1 (u^2)^2 = \\int_0^1 u^4 \u003c \\infty$.\n\nNon-example: the map $u \\mapsto \\bigl(x \\mapsto \\int_0^1 u(t)\\, dt\\bigr)$ is not a Nemytskii operator.\nThe value at $x$ depends on all values of $u$, not only on $u(x)$.\n\nOther names are superposition operator and, for $g$ independent of $x$, composition operator.\n",
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