Let $X$ be a metric space, $\nu,\mu$ be Borel measures on $X$, $f:X\times \mathbb{R}\rightarrow [0,\infty)$ be a measurable function. Under what conditions is the integral functional $F\_f$, defined by:
$$
\begin{aligned}
F\_f: L^1(X) & \rightarrow [0,\infty]
\\
g&\mapsto \int\_{x \in X} f(x,g(x))d\mu(x)
\end{aligned}
$$
continuous?


I expect that $f$ should at-least be Carath'{e}odory; i.e.: measurable in its first argument and continuous in its second, and probably some sort of growth condition such as $\int\_{x\in X} \sup\_{y \in \mathbb{R}}f(x,y)d\mu(x)<\infty$ to ensure that $F\_f$ is finite-valued...


I'm particularly interested in the case where $X$ is $\mathbb{R}^n$ or is a topological manifold modelled thereon.


(Does the situation simplify when $F\_f$ is instead considered on $C(X)$ with the uniform topology and $X$ were a compact space?)



Conditions for continuity of an integral functional
Besides some condition ensuring that the map is well defined a simple and natural condition for the continuity is a Lipschitz condition with respect to the second variable, i.e., $$|f(x,y)-f(x,z)|\le c|y-z|$$ for some constant $c$ independent of $x,y,z$. This is a standard assumption for (a version of) the Picard-Lindelöf theorem for ODEs.


I repeated my comment as an answer to make the following remark more visible: **Please do not try to minimize space but try to maximize the readability.**


I hate to read somthing like


*Using [xx, lemma 2.3] in combination with [yy, theorem 4.17] the reader can easily check that $F\_f$ is continuous whenever $f$ satisfies condition ($\ast$) in [zz, proposition 4].*


