How can i prove this Convergence?

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

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I think i have the answer, please verify if it is correct.

First of all, as i have noted in the post, $F$ is convex (in fact it is striclty convex), to see this just derivate it two times and verify that the operator $f''(u)$ is striclty positive for all $u$.

Assumption 1: Suppose $u$ is the strictly local minimum of $F$ in $K$, then for $\delta>0$ there exist a constant $C_\delta>0$ such that $$F(v)>C_\delta+F(u),\ \forall\ v\in K'=K\backslash (B(u,\delta)\cap K)$$ where $B(u,\delta)$ is the ball with center $u$ and ratio $\delta$.

Proof of Assumption 1: Let $I'=\displaystyle\inf_{v\in K'} F(v)$. As the minimum is unique in $K$ we have that $F(u)<I'$ hence, \begin{eqnarray} F(v) &\geq& I' \nonumber \\ &=& (I'-F(u))+F(u), \forall v\in K' \nonumber \end{eqnarray}

So $C_\delta=I'-F(u)$ and this finish the proof os Assumption 1.

Now suppose ad absurdum that $u_n$ does not converge to $u$ in $W^{1,p}(\Omega)$. Therefore we can find some $\delta>0$ and a subsequence of $u_n$ (not relabeled) such that $u_n\in K'$ for all $n$ (I'm using the above notation). By Assumption 1 we have $$F(u_n)\geq C_\delta+F(u)$$

As $C_\delta$ is a positive constant that does not dependes on $n$, we conclude by passing the limit in the laste inequality that $$F(u)\geq C_\delta +F(u)$$

which is an absurd...

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