What conditions are needed to have $F(f,g) \in L^\infty$ for $f, g \in L^\infty$?

Question

Let $\Omega \subset \mathbb{R}$ and $f, g \in L^\infty(\Omega)$. Define $F: \Omega \times \Omega \rightarrow \mathbb{R}$ by $F(x) = \tilde{f}(f(x), g(x))$ for some $\tilde{f}$. I am interested in what conditions can be placed on $\tilde{f}$ so that $F \in L^\infty(\Omega)$.

I am thinking of two extreme cases: $\tilde{f} \in C^0$ and $\tilde{f} \in C^\infty$. If $\Omega$ was bounded this seems to be trivial since we know $f, g \in L^\infty$ so we can take their maximum and use that to bound $\tilde{f}$. However I'm not sure about the case where $\Omega$ is unbounded.

Answer 1

In relation to my comment on the previous answer, it is sufficient to choose $\tilde{f} \in L^\infty_{\mathrm{loc}}(\mathbb{R}^2)$. Since $f,g$ are bounded and measurable the set $D=(f(\mathbb{R}), g(\mathbb{R}))$ is bounded as a subset of $\mathbb{R}^2$ and so \begin{align} \mathrm{ess}\sup_{(x,y)\in\mathbb{R}^2} F(x,y)\leq \mathrm{ess}\sup_{(x,y)\in\bar{D}} \tilde{f}(x,y) <\infty \, . \end{align} For a given $f,g$ you can make the condition sharper by requiring only $\tilde{f} \in L^\infty(D)$ where $D$ is as above.

Answer 2

You need a condition to ensure that $F$ is measurable, e.g., $\tilde f$ continuous.

This is sufficient to get $F \in L^\infty$: Take $c \ge \max(\|f\|_{L^\infty}, \|g\|_{L^\infty})$. Then $\tilde f$ is bounded on the compact set $[-c,+c]^2$, i.e., there is $M>0$ such that $|\tilde f(s,t)| \le M$ for all $(s,t) \in [-c,+c]^2$. Then $|F(x)| \le M$ for almost all $x$.