The following theorem appears in Appendix B of Rabinowitz' book Minimax Methods in Critical Point Theory:
Let $\Omega \subset \Bbb{R}^n$ be bounded and $g\in C(\overline{\Omega}\times \Bbb {R},\Bbb {R})$ such that there exist constants $r,s\ge 1$ and $a_1,a_2\ge 0$ such that for all $x \in \overline{\Omega}, y\in \Bbb{R}$ $$|g(x,y)|\le a_1 + a_2|y|^{r/s}$$ Then the map $\varphi(x)\mapsto g(x,\varphi(x))$ belongs to $C(L^r(\Omega),L^s(\Omega))$.
In the proof, he says "To prove the continuity of this map, observe that it is continuous at $\varphi$ if and only if $f(x,z(x)) = g(x,z(x)+\varphi(x))-g(x,\varphi(x))$ is continuous at $z=0$. Therefore we can assume $\varphi = 0$ and $g(x,0)=0$."
I don't understand how this assumpion can be made without loss of generality, and was unable to finish the proof without it. Any help would be appreciated.
Edit: I have found a partial answer in a different thread: Continuity proof of a function between $L^p$ spaces
In the post it says: Using the growth estimate, one can derive a similar estimate for $f$ of the form: $$ |f(x,z(x))|\leq A_1+A_2 |\phi_0(x)|^{r/p}+A_3|z(x)|^{r/p} $$
This would solve my problem, since the former two constants don't depend on $z$ and can be thrown together, leaving the case that was already proven. However, I couldn't derive this estimate.
edit2: I was wrong in assuming this solves the problem since, as pointed out by the users supinf and Peter Melech, the constant may not depend on x.
For $x\in\overline{\Omega}$ and $\frac{r}{s}\geq 1$ You get $$|f(x,z(x))|=|g(x,z(x)+\varphi(x))-g(x,\phi(x))|\leq a_1+a_2|z(x)+\varphi|^{\frac{r}{s}}+a_1+a_2|\varphi(x)|^{\frac{r}{s}}\leq$$ $$2a_1+a_22^{\frac{r}{s}-1}(|z(x)|^{\frac{r}{s}}+|\varphi(x)|^{\frac{r}{s}})+a_2|\varphi(x)|^{\frac{r}{s}}\leq $$ $$\underbrace{2a_1}_{=A_1}+\underbrace{(a_22^{\frac{r}{s}-1}+a_2)}_{=A_2}|\varphi(x)|^{\frac{r}{s}}+\underbrace{a_22^{\frac{r}{s}-1}}_{=A_3}|z(x)|^\frac{r}{s}$$ where You used that $(\frac{a+b}{2})^{\frac{r}{s}}\leq\frac{a^\frac{r}{s}+b^\frac{r}{s}}{2}$ which holds by the convexity of $x\mapsto x^{\frac{r}{s}}$ and Jensen's inequality. In the case $\frac{r}{s}<1$ You can proceed analogously using $(a+b)^{\frac{r}{s}}\leq a^{\frac{r}{s}}+b^{\frac{r}{s}}$ valid in this range.