Nemytskii operator

Remarks

A theorem of Krasnosel'skii describes the $L^p$ spaces case. Let $\Omega \subseteq \R^m$ be measurable with Lebesgue measure, and let $1 \le p, q < \infty$. Then $N_g$ 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 $u$. In that case $N_g \colon L^p(\Omega) \to L^q(\Omega)$ is continuous and maps bounded sets to bounded sets.

Example: $\Omega = [0, 1]$ with Lebesgue measure and $g(x, u) = u^2$. Both 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 $\int_0^1 (u^2)^2 = \int_0^1 u^4 < \infty$.

Non-example: the map $u \mapsto \bigl(x \mapsto \int_0^1 u(t)\, dt\bigr)$ is not a Nemytskii operator. The value at $x$ depends on all values of $u$, not only on $u(x)$.

Other names are superposition operator and, for $g$ independent of $x$, composition operator.