minimizes the functional to solve a pde I am trying to do this exercise:
Let $\Omega$ a open bounded domain in $R^n$. Consider the Dirichlet problem
$$ \left\{ \begin{array}{ccccccc}
 -\Delta u  = \lambda \sin (u) + f ,  \  \text{in $\Omega$}   \\ \ u=0 , \ \text{on $\partial\Omega$} \\ \end{array} \right. $$
where $\lambda >0$ and $f \in L^{2}(\Omega)$. Show that for $ \lambda < \lambda_1 $ ($\lambda_1$ is the first eigenvalue of the Laplacean), the above problem has a weak solution minimizing the functional
$$ \varphi(u) = \displaystyle\int_{\Omega} \frac{1}{2}|\nabla u|^2 + \lambda (\cos u -1) - fu \ dx.$$
I know that if this functional is coercive (that is $\varphi(u) \rightarrow \infty$ as $\|u\| \rightarrow \infty$) and satisfies $\varphi(u) \leq \liminf_n \varphi(u_n)$ for $u_n$ converging weakly to $u$ then the problem is done. 
I know a proof for the the coercivity(see Show that this functional is coercive - variational methods , when i posted in the above link i  did a part of the solution, and my only problem was the coercivity, but after a time concluded that the other part of my solution was worng), but i dont know how to prove the $\liminf$ condition (the cossine is the problem).
Someone can give me a help with the $\liminf$ condition ?
thanks in advance
 A: Let $g:\mathbb{R}\to \mathbb{R}$ be a $L^\infty$ function and define $G(x)=\int_0^x g(t)dt$. Consider the functional $$I(u)=\frac{1}{2}\|\nabla u\|_2^2+\lambda\int_\Omega G(u)-\int_\Omega fu,\ \forall u\in H_0^1(\Omega)$$
I will assume here that you know, that the first term in he above sum is weakly sequentially lower semi continuous (w.s.l.s.c.). Let's study the other terms. First the term $$\int_\Omega fu$$
Suppose that $u_n\rightharpoonup u$ in $H_0^1(\Omega)$ and assume without loss of generality that $u_n\to u$ in $L^2(\Omega)$, $u_n\to u$ a.e. in $\Omega$ and $|u_n|\le v_1\in L^2$. Conclude that $$\int_\Omega fu_n\to \int_\Omega fu\tag{1}$$
Now the term $$\int_\Omega G(u)$$
Suppose that $u_n\rightharpoonup u$ in $H_0^1(\Omega)$ and assume without loss of generality that $u_n\to u$ in $L^2(\Omega)$, $u_n\to u$ a.e. in $\Omega$ and $|u_n|\le v_1\in L^2$. Note that 
\begin{eqnarray}
 |G(u_n)| &\le& \int_0^{u_n} |g(t)|dt      \nonumber \\
   &\le& \|g\|_\infty |u_n| \nonumber \\
   &\le& \|g\|_\infty v_1 \in L^2(\Omega) \tag{2}
\end{eqnarray}
Moreover, because $G$ is continuous we have that $$G(u_n(x))\to G(u(x))\ \mbox{a.e. in} \ \Omega\tag{3}$$
Conclude from $(2)$ and $(3)$ that $$\int_\Omega G(u_n)\to \int_\Omega G(u)\tag{4}$$
To finish the proof first note that $$\liminf I(u_n)\ge \liminf \frac{1}{2}\|\nabla u_n\|_2^2+\lim\lambda\int_\Omega G(u_n)-\lim \int_\Omega fu_n \tag{5}$$
From $(1)$, $(4)$ and $(5)$ we see that $I$ is w.s.l.s.c. Now, choose $g(x)=\sin(x)$.
