Does the function $G(u)=1+2u^2$ satisfy this property involving the supremum?

Question

Let $G:(x, u)\in\mathbb{R}^N\times\mathbb{R}\mapsto G(x, u)\in\mathbb{R}$ be a function such that for any $r>0$, it is $$(S)\qquad\qquad\sup_{|u|\le r} |G(\cdot, u)|\in L^{\infty}(\mathbb{R}^N).$$

My question is: if I take into account the function $G(u):\mathbb{R}\to\mathbb{R}$ such that $G(u) = 1+2u^2$, does this function satisfy the property $(S)$?

I am in trouble about that. As it is definied $G(u)$ I'd say that the dependence on $x$ is "hidden" in $u$ (I mean, possibly it is $G(u):x\in\mathbb{R}^N\mapsto G(u)(x)=1+2u^2(x)$). Anyway, I don't understand if the property $(S)$ is satisfied in this case.

Could someone please help me?

Thank you in advance!

Answer 1

I believe you have to assume that the real $G$ is $G(x,u) = G(u)\;\forall x\in \mathbb R^N$.

In that case $$\sup_{|u|\le r} |G(x, u)| = \sup_{|u|\le r} |G(u)|$$

The function is then constant (or $+\infty$) and in the former case in $L^\infty(\mathbb R^N)$.

Must be a trivial example given by your professor.