Let $\Omega$ be an open bounded domain and consider $(W_0^{1, p}(\Omega), \|\cdot\|)$ with $p>1$. Let $(u_n)_n\subset W_0^{1, p}(\Omega)$ be a ${\bf bounded}$ sequence in $ W_0^{1, p}(\Omega)$ and consider the equation
$$\|u_n\|^a-\int_{\Omega}u_nf(u_n) dx =o(\|u_n\|),$$
where $a\in\mathbb{N}$ and $f:\mathbb{R}\to\mathbb{R}$ denotes a continuous function.
I am trying to understand if it is possible to deduce by the above equality that
$$\sup_n \int_{\Omega} u_nf(u_n) dx <+\infty.$$
${\bf MY \,ATTEMPT:}$
By exploiting the definition of $o(\|u_n\|)$, the above equality means that
$$ \lim_{n\to +\infty}\frac{\|u_n\|^a-\int_{\Omega} u_nf(u_n) dx}{\|u_n\|}=0,$$
which means, in particular, that $n_k\in\mathbb{N}$ exists such that
$$\frac{\|u_n\|^a-\int_{\Omega} u_nf(u_n) dx}{\|u_n\|}<1 \quad\forall n\geq n_k.$$
Thus it follows that
$$ -\int_{\Omega} u_nf(u_n) dx<\|u_n\| -\|u_n\|^a<M$$
for a constant $M$ since $(u_n)$ is bounded in $W_0^{1,p}(\Omega)$.
How could I deduce from that
$$\sup_n \int_{\Omega} u_nf(u_n) dx <+\infty?$$
Could anyone please help me?
Thank you in advance!
Best Answer
You don't just have $$\frac{\|u_n\|^a-\int_{\Omega} u_nf(u_n) dx}{\|u_n\|}<1 \quad\forall n\geq n_k.$$ You also have $$\frac{\|u_n\|^a-\int_{\Omega} u_nf(u_n) dx}{\|u_n\|}> -1 \quad\forall n\geq n_k.$$ Hence $\int_\Omega u_n f(u_n)\,dx < \|u_n\|^a + \|u_n\| < M'$ for another constant $M'$. Taking the supremum over $n$ then gives the result.