Let consider a function $f\colon \overline{\Omega} \times \mathbb{R} \mapsto \mathbb{R}$, where $\overline{\Omega}$ is subset of $\mathbb{R}^n$. $\overline{\Omega}$ can be either bounded or unbounded. The function $f$ satisfies the Carathéodory conditions:
- $g \colon \mathbb{R} \ni u \mapsto f(\mathbf{x}, u) \in \mathbb{R}$ is a continuous function for each fixed $\mathbf{x} \in \overline{\Omega}$;
- $h \colon \mathbb{R}^n \ni \mathbf{x} \mapsto f(\mathbf{x}, u) \in \mathbb{R}$ is a measurable function for each fixed $u \in \mathbb{R}$;
and the Lipschitz-Carathéodory condition:
- there exists a function $l \in L^1_{\mathrm{loc}}(\overline{\Omega})$ such that $|f(\mathbf{x}, z_1) - f(\mathbf{x}, z_2)| \leqslant l(\mathbf{x}) |z_1 - z_2|$
Define a new operator $$ N[u](\mathbf{x}) = f(\mathbf{x}, u(\mathbf{x})), $$ which acts on continuous functions: $u \in C(\overline{\Omega})$.
The question arises: when an image $N[u]$ of a continuous function $u$ is good enough: continuous or at least locally Lebesgue integrable.
Here I can say the following:
i) The image $N[u]$ is a measurable function.
ii) If the function $f$ were continuous, then the image $N[u]$ would also be continuous. But I do not have it.
iii) Assume $\overline{\Omega}$ is bounded. If the condition $|f(\mathbf{x}, u)| \leqslant b |u| + \alpha(\mathbf{x})$, where $b$ is some constant, $\alpha \in L_1(\overline{\Omega})$, holds then $N \colon L_1(\overline{\Omega}) \mapsto L_1(\overline{\Omega})$ is a continuous operator. (This condition is necessary and sufficient for the continuity of the operator $N \colon L_1(\overline{\Omega}) \mapsto L_1(\overline{\Omega})$.) This implies $N[u] \in L_1(\overline{\Omega})$.
iiii) Arguing similarly, we can conclude that under the condition $|f(\mathbf{x}, u)| \leqslant b |u| + \alpha(\mathbf{x})$, where $b$ is some constant, $\alpha \in L^1_{\mathrm{loc}}(\overline{\Omega})$, $N[u]$ belongs to the class $L^1_{\mathrm{loc}}(\overline{\Omega})$.
But (ii) - (iiii) not the same conditions that I have.
Main question. Is it possible somehow, under the conditions that I have for the function $f$, to show that the image $N[u]$ of a continuous function $u$ is good enough (continuous or at least locally Lebesgue integrable). Or is it necessary to impose stronger conditions (e. g. $f$ is a continuous function)?