Vrftsjtryk

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 $gin C(overline{Omega}times Bbb {R},Bbb {R})$ such that there exist constants $r,sge 1$ and $a_1,a_2ge 0$ such that for all $x in overline{Omega}, yin Bbb{R}$ $$|g(x,y)|le a_1 + a_2|y|^{r/s}$$
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 $xinoverline{Omega}$ 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 $(frac{a+b}{2})^{frac{r}{s}}leqfrac{a^frac{r}{s}+b^frac{r}{s}}{2}$ which holds by the convexity of $xmapsto x^{frac{r}{s}}$ 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.