Let $\Omega\subset\mathbb{R}^n$ be a bounded smooth domain. Let $p\in (1,\infty)$ and $K=\{u\in W^{1,p}(\Omega):\ u-g\in W^{1,p}_0(\Omega)\}$, where $g\in W^{1,p}(\Omega)$. Let $\epsilon\geq 0$ and $F:W^{1,p}(\Omega)\rightarrow\mathbb{R}$ defined by $F(u)=\frac{1}{p}\int_\Omega (\epsilon^2+|\nabla u|^2)^\frac{p}{2}$
Let $I=\displaystyle\inf_{u\in K} F(u)$ and note that $I\geq 0$, therefore, there exists a sequence $u_n\in K$ such that $F(u_n)\rightarrow I$.
By using the monotonicity of the function $|\ |^\frac{p}{2}$ and the Poincaré inequality, we can show that $\|u_n\|_{1,p}$ is bounded and then extract a subsequence (not relabeled) such that $u_n \rightharpoonup u$ for some $u\in K$. By using the fact that $F$ is convex and lower semi continuous, we can show that $u$ is the "unique" solution of the problem.
Now my problem is: I want to show that in fact $u_n\rightarrow u$ in $W^{1,p}(\Omega)$.
Beyond the facts stated above, we have that $\int_\Omega (\epsilon^2+|\nabla u|^2)^\frac{p-2}{2}\nabla u\nabla v=0,\ \forall\ v\in W^{1,p}_0(\Omega)$
where $u$ is the solution of the problem. The last equality appear when we derivate the functional $F$ in the point $u$.
Remark: Suppose $\epsilon=0$, hence in this case, we have that $F(u_n)=\frac{1}{p}\|\nabla u_n\|_p\rightarrow \frac{1}{p}\|\nabla u\|_p=F(u)$. By using the compact immersion $W^{1,p}(\Omega)\hookrightarrow L^p(\Omega)$ we have (up to a seubsequence ) that $\|u_n\|_p\rightarrow \|u\|_p$ and hence $\|u_n\|_{1,p}\rightarrow\|u\|_{1,p}$. As the space $W^{1,p}(\Omega)$ is uniformly convex and $u_n \rightharpoonup u$ we can conclude that $u_n\rightarrow u$ in $W^{1,p}(\Omega)$
In the case $\epsilon>0$ this sill Works: so if we can show that $\|\nabla u_n\|_p\rightarrow \|\nabla u\|_p$, the problem is solved.
Any help is appreciated.
Thanks