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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?)

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  • $\begingroup$ A simple sufficient condition for the continuity is that $f$ is Lipschitz with respect to the second argument, i.e., $|f(x,y)-f(x,z)|\le c|y-z|$ for some constant independent of $x,y,z$. $\endgroup$ – Jochen Wengenroth Jul 21 at 8:07
  • $\begingroup$ What about when $L^1$ is replaced by $C(\mathbb{R})$? I don't this works then... we probably need some integrability contstraint also? $\endgroup$ – James_T Jul 21 at 8:51
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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].

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  • $\begingroup$ Meanwhile the OP deleted his comment to which my answer was a reaction. $\endgroup$ – Jochen Wengenroth Jul 21 at 8:53
  • $\begingroup$ Thank you for posting this and especially for the writing tip. I appreciate it. $\endgroup$ – James_T Jul 21 at 9:00
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Since $\nu$ does not appear anywhere else in the question, I suppose that $L^1(X)=L^1(\nu)$.

In order that the functional be defined, one should then assume (probably without loss of generality) that $\mu$ is absolutely continuous w.r.t. $\nu$.

In this case, the two conditions in the question are more than sufficient, and far from being necessary.

To prove that they are sufficient, it suffices to show that for each sequence $x_n\to x$ in $L_1(\nu)$ there is a subsequence with $F_f(x_{n_k})\to F_f(x)$. Since $x_n\to x$ in $L_1(\nu)$, there is a subsequence with $x_{n_k}\to x$ $\nu$-a.e., hence by hypothesis $\mu$-a.e. Since $f$ is Carathéodory, it follows that $g_{n_k}(t)=f(t,x_{n_k}(t))\to g(t)=f(t,x(t))$ for $\mu$-a.e. $t$. It remains to apply Lebesgue's dominated convergence theorem with the dominating function being $\sup_y f(\cdot,y)$.

Using Vitali's instead of Lebesgue's dominated convergence theorem, one can replace the strong integral hypotheses by various sorts of growth conditions, depending on $\mu/\nu$. For instance, in case $\mu=\nu$ the growth condition $f(t,y)\le a(t)+C|y|$ with a constant $C$ and some $\mu$-integrable $a$ is sufficient.

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  • $\begingroup$ Edited the last condition so that the result becomes more general, and the growth condition is of Krasnosel'skij type. $\endgroup$ – Martin Väth Sep 28 at 17:17

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