Let $(u_n)$ be a bounded sequence in $L^2(\mathbb R)$, i.e. there exists $M>0$ such that $||u_n||_{L^2(\mathbb R)}\le M$.
Let $p>2$ and $F:\mathbb R\times \mathbb R\to\mathbb R$ be a continuous function such that $$|F(x, u)|\le c_1 +c_2 |u|^{p-1}\quad\mbox{ for a.e. } x\in\mathbb R, \ \forall u\in\mathbb R,$$ for some $c_1, c_2>0$.
Under these assumptions, is it possible to conclude that $$\sup_{n\in\mathbb N} \int_{\mathbb R}|F(x, u_n)|^2 dx \le M_2,$$ for a constant $M_2>0$?
I'd say yes, as $$|F(x, u_n)|\le c_1 +c_2 |u_n|^{p-1}$$ and $(u_n)_n$ is bounded, but I am not sure about that.
Anyone could please me in proving (or disproving) that?
$\bf{EDIT:}$ By the answer of @Matt it seems that the statement is false. Anyone could please let me understand if it holds true replacing $\mathbb R$ with any compact subset $K$ of $\mathbb R$? Namely, if $$\sup_{n\in\mathbb N} \int_{K}|F(x, u_n)|^2 dx<\infty$$
Without knowing anything further about $F$ this is in general not possible.
For example let $F(x,u) = 1$ then clearly $|F(x,u)| \leq c_1 + c_2|u|^{p-1}$ with $c_1 = 1$ and any $c_2 > 0$. Therefore this $F$ satisfies the conditions on $F$.
Also let $u_n = 0$ for every $n$ so that the sequence $(u_n)$ satisfies its conditions.
However for every $n$ we have $$ \int_\mathbb{R} F(x,u_n)^2 dx = \int_\mathbb{R} 1^2 dx = \infty $$ which shows that there is no finite upper bound.