Let $\Omega$ be an open bounded domain in $\mathbb{R^N}$, $p\geq 1$ and $(u_n)_n\subset W_0^{1,p}(\Omega)$. I would like to show that $(u_n)_n$ is bounded in $W_0^{1,p}(\Omega)$ and the only thing I know is that $$\| u_n\|_{W_0^{1,p}(\Omega)} +\int_{\Omega} f(u_n) dx = c + o(\| u_n\|_{W_0^{1,p}(\Omega)} +1),$$ where $c$ is a positive constant and $f:\Omega\to\mathbb{R}$ is a continuous function.
It is written on my notes that in order to show $(u_n)_n$ is bounded in $W_0^{1,p}(\Omega)$ it is enough to prove that the function $f$ is bounded.
I am trying, but I am not able to understand why this late statement is true, I mean, why it is enough just to show that $f$ is bounded? In addition, bounded from above and\or from below?
Could anyone please help or give some hints?
Thank you in advance!
Dividing the equation $$\| u_n\|_{W_0^{1,p}(\Omega)} +\int_{\Omega} f(u_n) dx = c + o(\| u_n\|_{W_0^{1,p}(\Omega)} +1)$$ by $\| u_n\|_{W_0^{1,p}(\Omega)} +1$, we find that $$ \limsup_{n\to\infty} \left[\frac{\| u_n\|_{W_0^{1,p}(\Omega)}}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1} + \frac{1}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1}\int_{\Omega}f(u_n)~dx\right] = \limsup_{n\to\infty} \frac{c}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1}.$$ Now suppose that there is a subsequence, again denoted by $u_n$, such that $\lim_{n\to\infty} \| u_n\|_{W_0^{1,p}(\Omega)} = \infty.$ Then, as $f$ is bounded we have $$ \limsup_{n\to\infty} \left[ \frac{1}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1}\int_{\Omega}f(u_n)~dx\right] \leq \limsup_{n\to\infty} \frac{\|f\|_{\infty}|\Omega|}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1} = 0. $$ We clearly also have $$ \limsup_{n\to\infty} \frac{c}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1} = 0,$$ and as $$ \limsup_{n\to\infty} \frac{\| u_n\|_{W_0^{1,p}(\Omega)}}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1} = 1, $$ we reach the contradiction $1 = 0$. It follows that there is no subsequence of $(u_n)_{n \geq 1}$ such that $\lim_{n\to\infty} \| u_n\|_{W_0^{1,p}(\Omega)} = \infty$, hence $(u_n)_{n\geq 1}$ is bounded in $W_0^{1,p}(\Omega)$.