I am going back to a question I posted yesterday. This is because I think I have made some steps forward.
Let $(E, \|\cdot\|)$ denote a Hilbert space such that $E$ is continuously embedded in $L^q(\mathbb R)$ for any $q\in [2, 4]$. Let $(u_n)$ be a bounded sequence in $E$, i.e. there exists $M>0$ such that $\|u_n\|_{E}\le M$.
Let $p\in (2, 3)$ and $F:\mathbb R\times \mathbb R\to\mathbb R$ be a continuous function such that
$$|F(x, u)|\le c_1 |u| +c_2 |u|^{p-1}\quad\mbox{ for a.e. } x\in\mathbb R, \ \forall u\in\mathbb R,$$
for some $c_1, c_2>0$.
Under these assumptions, I think that it is possible to conclude that
$$\sup_{n\in\mathbb N} \int_{\mathbb R}|F(x, u_n)|^2 dx \le M_1,$$
for a constant $M_1>0$.
Indeed:
$$
\begin{split}
\int_{\mathbb R}|F(x, u_n)|^2 dx &\le c_1 \int_{\mathbb R} |u_n|^2 dx + c_1 c_2 \int_{\mathbb R} |u_n|^p dx + c_2^2 \int_{\mathbb R} |u_n|^{2(p-1)} dx \\
&\le k_1 \|u_n\|_E \le k_1 M =:M_1.
\end{split}
$$
Hence, passing to the supremum, one has
$$\sup_{n\in\mathbb N} \int_{\mathbb R}|F(x, u_n)|^2 dx \le M_1.$$
Anyone could please tell me if I argued it right? Thank you in advance.
Best Answer
Looks good, but there is one minor mistake. The continuous embedding $E\hookrightarrow L^q(\mathbb{R})$ yields a constant $C>0$ such that $$ \Vert u \Vert_{L^q(\mathbb{R})} \leq C \Vert u \Vert_E$$ for all $q\in[2,4]$. Hence, e.g., $$\int_\mathbb{R} \vert u_n \vert^p\,dx = \Vert u_n \Vert_{L^p(\mathbb{R})}^p \leq C^p \Vert u_n \Vert_E^p,$$ and analogously for the other integrals.