Let $\Omega$ be a bounded domain of $\mathbb{R}^3$ with smooth boudary, and consider the following nonlinear equation: $$ Au=-\Delta u+cu+u^3=f,\quad \textrm{in}\, \Omega, $$ and $u=0$ on $\partial \Omega$. Since $A$ is a maximal monotone operator with domain $H_0^1(\Omega)\cap L^6(\Omega)$, we know from $A^{-1}$ is well-defined and bounded. I am wondering whether I can show $A^{-1}:f\in L^2\to u_f\in H_0^1$ is compact?
What I have tried:
Consider $\{f_n\}$ be bounded sequence in $L^2$, we have $f_n\rightharpoonup f$ in $L^2$, and I aim to show $u_n=u_{f_n}\to u_f$ in $H_0^1$.
The weak formulation gives me that $$ \int_\Omega \nabla u_n\nabla \phi+c\int_\Omega u_n \phi+\int_\Omega u_n^3 \phi=\int_\Omega f_n \phi,\quad \phi\in H_0^1(\Omega)\cap L^6, \tag{1} $$ set $\phi=u_n$ implies $|\nabla u_n|_{L^2}\le |f_n|_{L^2}$, and hence Poincare inequality implies $|u_n|_{H_0^1}$ and $|u_n|_{L^6}$ are bounded. Hence $u_n\rightharpoonup u$ in $H_0^1$ and $u_n\rightharpoonup u'$ in $L^6$.
In order to pass $n\to \infty$ in $(1)$, I need to show $$ \int_\Omega u_n^3 \phi\to \int_\Omega u^3 \phi, $$ however I am not sure how to prove that.
The Sobolev embedding theorem together with Rellich-Kondrachov shows that the embedding $H_0^1(\Omega) \hookrightarrow L^p(\Omega)$ is compact for all $p < 6$. Thus, $u_n^3 \to u^3$ in $L^{p/3}(\Omega)$ for all $p < 6$ and $\{u_n^3\}$ is bounded in $L^6(\Omega)$. This shows that $u_n^3 \rightharpoonup u^3$ in $L^2(\Omega)$.