The following theorem appears in the appendix of P.H. Rabinowitz monograph on Critical Point Theory:
Let $\Omega \subset \mathbb R^n$ be bounded. Let $g$ be such that
(i) $g \in C(\overline{\Omega} \times \mathbb R, \mathbb R)$
(ii) there exist constants $r,s \geq 1$ and $a_1,a_2 \geq 0$ such that
$$|g(x,\eta)| \leq a_1 + a_2|\eta|^{\frac{r}{s}}$$
then the map $\phi(x) \to g(x,\phi(x))$ is in $C\big(L^r(\Omega),L^s(\Omega)\big)$.
I suppose I'm a bit confused in regards to two things:
(i) How does a map from $\phi(x) \to g(x,\phi(x))$ define a map $L^r(\Omega) \to L^s(\Omega)$ if we're considering only one such $\phi$ in the equivalence class?
(ii) How is $g(x,\phi(x)) \in L^s(\Omega)$ if $g$ is defined on $\overline{\Omega} \times \mathbb{R}$? I understand that $g$ takes $\phi$ to some function involving $\phi(x)$, but the fact that everything is being done point-wise is throwing me off.
I understand the proof, it's fairly straightforward, since $L^\infty \subset L^r$ in this case.
Any comments appreciated.
(i) if $\phi_1(x)=\phi(x)$ except on a set of measure zero $A$, then $g(x,\phi(x))=g(x,\phi_1(x))$ except on that same set of measure zero. Thus two functions which are members of the same a.e. equivalence class in $L^r$ give two functions equivalent in $L^s$.
(ii) holds because $x\mapsto g(x,\phi(x))$ is a real function of $x$ defined for $x\in \Omega$ provided $\phi$'s domain is (or includes) $\Omega$, as is the case for $\phi\in L^r$. The theorem proves that the function is in $L^s$ and the map is continuous.