A pde via the Ekeland Variational Principle The following is a problem from a text on Critical Point Theory I am reading. Below is the beginning of an attempt, but I got stuck. Any hints will be the most appreciated. Thanks in advance and kind regards.

Use the Ekeland Variational Principle to solve the following sublinear problem:
  $$
(P) \quad
\begin{cases}
-\Delta_p u + |u|^{p - 2}u = h(x)|u|^{q - 2}u \quad \text{ in }\Bbb{R}^N \\
u \in W^{1, p}(\Bbb{R}^N)
\end{cases}
$$
  where $\Delta_p$ is the $p$-laplace operator, $N \geq 3$, $2 \leq p < N$, $p - 1 < q < p$, $h \in L^{\frac{p^*}{p^* - q}}(\Bbb{R}^N) \cap L^\infty (\Bbb{R}^N)$, $h \geq 0$ and $h \neq 0$.

Weak solutions to the problem $(P)$ are critical points of the functional
\begin{align*}
I(u) & = \frac1p \int_{\Bbb{R}^N} |\nabla u|^p \ dx+ \frac1p \int_{\Bbb{R}^N} |u|^p \ dx - \frac1q \int_{\Bbb{R}^N} h(x) |u|^q \ dx \\
& = \frac1p ||u||^p - \frac1q \int_{\Bbb{R}^N} h(x)|u|^q \ dx,  \quad u \in W^{1, p} (\Bbb{R}^N)
\end{align*}
which is of class $C^1$, with 
$$
I'(u)v = \int_{\Bbb{R}^N} |\nabla u|^{p - 2} \nabla u \cdot \nabla v \ dx + \int_{\Bbb{R}^N} |u|^{p - 2}uv \ dx - \int_{\Bbb{R}^N} h(x) |u|^{q - 2}uv \ dx, 
$$
for $u, v \in W^{1, p}(\Bbb{R}^N)$. The functional $I$ is also weakly lower semicontinuous and coercive, and hence bounded from below. Therefore, if $c = \inf_{W^{1, p}(\Bbb{R}^N)}I$, by the Ekeland Variational Principle there exists a Palais-Smale sequence at the level $c$.
Let $(u_n)$ be a $(PS)_c$ sequence for $I$. Then $(u_n)$ is bounded, since $I$ is coercive. Since $W^{1, p}(\Bbb{R}^N)$ is reflexive, there exists $u \in W^{1, p} (\Bbb{R}^N)$ such that $u_n \rightharpoonup u$.
Let $\phi \in C_c^\infty (\Bbb{R}^N)$ and let $\Omega = \text{supp} \phi$. Then 
$$
u_n|_\Omega \rightharpoonup u|_\Omega \quad \text{ in } W^{1, p}(\Omega)
$$ 
and therefore, by the compact Sobolev embeddings, 
$$
u_n|_\Omega \to u|_\Omega \quad \text{ in } L^s(\Omega)
$$
for $s \in [1, p^*)$, up to a subsequence. It is can be shown that
$$
\int_{\Bbb{R}^N} h(x) |u_n|^{q - 2} u_n \phi \ dx \to \int_{\Bbb{R}^N} h(x) |u|^{q - 2} u \phi \ dx, 
$$
as well as that
$$
\int_{\Bbb{R}^N} |u_n|^{p - 2} u_n \phi \ dx \to \int_{\Bbb{R}^N} |u|^{p - 2} u \phi \ dx,
$$
which holds for all $\phi \in C_c^\infty(\Bbb{R}^N)$. It remains to show that 
\begin{align*}
\int_{\Bbb{R}^N} |\nabla u_n|^{p - 2} \nabla u_n \cdot \nabla \phi \ dx  \to  \int_{\Bbb{R}^N} |\nabla u|^{p - 2} \nabla u \cdot \nabla \phi \ dx
\end{align*}
 A: I found a way to prove the desired convergence. It is as follows. I would appreciate any critique and comments.
Choose $\phi \in C_c^\infty(\Bbb{R}^N)$. Let $\psi \in C_c^\infty(\Bbb{R}^N)$ be such that $0 \leq \psi \leq 1$ and 
$$
\psi(x) = 
\begin{cases}
1, \quad x \in B_1(0) \\
0, \quad x \in \Bbb{R}^N \setminus B_2(0)
\end{cases}
$$
For each $\rho > 0$, let 
$$
\psi_\rho = \psi \left(\frac x\rho \right).
$$
Then
$$
\psi_\rho = 
\begin{cases}
1, \quad x \in B_\rho(0) \\
0, \quad x \in \Bbb{R}^N \setminus B_{2\rho}(0)
\end{cases}.
$$
Defining
$$
P_n(x) =  (|\nabla u_n|^{p-2} \nabla u_n - |\nabla u|^{p - 2} \nabla u) \cdot (\nabla u_n - \nabla u)
$$
we have that 
\begin{align*}
0 & \leq C_p \int_{B\rho(0)} |\nabla u_n - \nabla u|^p \ dx \\
& \leq \int_{B\rho(0)} P_n(x) \ dx \\
& \leq \int_{B\rho(0)} P_n(x) \psi_\rho(x) \ dx \\
& \leq \int_{\Bbb{R}^N} P_n(x) \psi_\rho(x) \ dx.
\end{align*}
Therefore
\begin{align*}
0 & \leq C_p \int_{B\rho(0)} |\nabla u_n - \nabla u|^p \ dx \\
& \leq \int_{\Bbb{R}^N}|\nabla u_n|^p \psi_\rho \ dx - \int_{\Bbb{R}^N} |\nabla u_n|^{p - 2} \nabla u_n \cdot \nabla u \psi_\rho \ dx - \int_{\Bbb{R}^N} |\nabla u|^{p - 2} \nabla u \cdot \nabla u_n \psi_\rho \ dx \\
& \quad + \int_{\Bbb{R}^N} |\nabla u|^p \psi_\rho \ dx \\
& = J_1(u_n) - J_2(u_n) + J_3(u_n) + J_4(u_n) + J_5(u_n), 
\end{align*}
where
$$
J_1(u_n) = \int_{\Bbb{R}^N} |\nabla u_n|^p \psi_\rho \ dx + \int_{\Bbb{R}^N} |u_n|^p \psi_\rho - \int_{\Bbb{R}^N} h(x) |u_n|^q \psi_\rho \ dx,
$$
\begin{align*}
J_2(u_n) = & \int_{\Bbb{R}^N} |\nabla u_n|^{p - 2} \nabla u_n \cdot \nabla u \psi_\rho \ dx + \int_{\Bbb{R}^N} |u_n|^{p - 2} u_n u \psi_\rho \ dx \\
& - \int_{\Bbb{R}^N} |u_n|^{q - 2} u_n u \psi_\rho \ dx,
\end{align*}
$$
J_3(u_n) = - \int_{\Bbb{R}^N} |\nabla u|^{p - 2} \nabla u \cdot \nabla u_n \psi_\rho \ dx + \int_{\Bbb{R}^N} |\nabla u|^p \psi_\rho \ dx, 
$$
$$
J_4(u_n) = \int_{\Bbb{R}^N} |u_n|^{p - 2} u_n u \psi_\rho \ dx - \int_{\Bbb{R}^N} |u_n|^p \psi_\rho \ dx
$$
and
$$
J_5(u_n) = \int_{\Bbb{R}^N} |u_n|^q \psi_\rho \ dx - \int_{\Bbb{R}^N} |u_n|^{q - 2}u_nu \psi_\rho \ dx.
$$
We begin by noting that 
$$
J_1(u_n) = I'(u_n)(u_n \psi_\rho) - \int_{\Bbb{R}^N} |\nabla u_n|^{p - 2} \nabla u_n \cdot \nabla \psi_\rho u_n \ dx
$$
and also that 
\begin{align*}
||u_n \psi_\rho||^p & = \int_{\Bbb{R}^N} |\nabla u_n \psi_\rho|^p \ dx + \int_{\Bbb{R}^N}|u_n \psi_\rho|^p \ dx \\
& \leq C||u_n||^p \\
& \leq C_1
\end{align*}
for some $C_1 > 0$, since the sequence $(u_n)$ is bounded. But then, since $I'(u_n) \to 0$, 
$$
J_1(u_n) = o_n(1) - \int_{\Bbb{R}^N} |\nabla u_n|^{p - 2} \nabla u_n \cdot \nabla \psi_\rho u_n \ dx.
$$
On the other hand, note that
\begin{align*}
\left|\int_{\Bbb{R}^N} |\nabla u_n|^{p-2} \nabla u_n \cdot \nabla \psi_\rho u_n \ dx \right| & \leq \int_{\Bbb{R}^N} |\nabla u_n|^{p - 1} |\nabla \psi_\rho| |u_n| \ dx \\
& \leq \left(\int_{\Bbb{R}^N}|\nabla u_n|^p \ dx\right)^{\frac{p-1}{p}} \left(\int_{\Bbb{R}^N} |\nabla \psi_\rho|^p |u_n|^p \ dx \right)^{\frac1p} \\
& \leq C_1 \left(\int_{B_{2\rho}(0) \setminus B_\rho(0)} |\nabla \psi_\rho|^p |u_n|^p \ dx \right)^{\frac1p}
\end{align*}
where the first inequality follows from Cauchy-Schwarz, the second from Hölder's Inequality with exponents $p/(p - 1)$ and $p$, and the third by the boundedness of $(u_n)$. Now, note that $u_n \to u$ in $L^p(B_{2\rho}(0) \setminus B_\rho(0))$. Then, applying Vainberg's Theorem and the Dominated Convergence Theorem in sequence yields 
$$
\limsup_{n \to \infty} \left|\int_{\Bbb{R}^N} |\nabla u_n|^{p-2} \nabla u_n \cdot \nabla \psi_\rho u_n \ dx \right| \leq C_1 \left(\int_{B_{2\rho}(0) \setminus B_\rho(0)} |\nabla \psi_\rho|^p |u|^p \ dx \right)^{\frac1p}.
$$ 
From Hölder's Inequality with exponents $N/(N - p)$ and $N/p$ it follows that 
\begin{align*}
\limsup_{n \to \infty} & \left|\int_{\Bbb{R}^N} |\nabla u_n|^{p-2} \nabla u_n \cdot \nabla \psi_\rho u_n \ dx \right| \\
& \leq C_1 \left[ \left(\int_{B_{2\rho}(0) \setminus B_\rho(0)} |u|^{p^*} \ dx \right)^{\frac{N - p}{p}} \left(\int_{B_{2\rho}(0) \setminus B_\rho(0)} |\nabla \psi_\rho|^N \right)^{\frac Np} \right]^{\frac1p} \\
& \leq C_1 \left[ \left(\int_{B_{2\rho}(0) \setminus B_\rho(0)} |u|^{p^*} \ dx \right)^{\frac{N - p}{p}} \left(\int_{\Bbb{R}^N} |\nabla \psi|^N \right)^{\frac Np} \right]^{\frac1p}.
\end{align*}
Then, by the Dominated Convergence Theorem, 
$$
\lim_{\rho \to 0} \limsup_{n \to \infty} \left|\int_{\Bbb{R}^N} |\nabla u_n|^{p-2} \nabla u_n \cdot \nabla \psi_\rho u_n \ dx \right| = 0
$$
and therefore
$$
J_1(u_n) = o_n(1) + o_\rho(1).
$$
By an analogous argument, 
$$
J_2(u_n) = o_n(1) + o_\rho(1).
$$
By the weak convergence, 
$$
J_3(u_n) = o_n(1).
$$
Since $u_n \to u$ in $L^s_{\text{loc}}$ and $\psi_\rho$ has compact support, the Dominated Convergence Theorem yields
$$
J_4(u_n) = o_n(1)
$$
and 
$$
J_5(u_n) = o_n(1).
$$
It therefore follows that 
$$
\frac{\partial u_n}{x_i} \to \frac{\partial u}{x_i} \quad \text{ in } L^p_{\text{loc}} (\Bbb{R}^N)
$$
for all $i \in \{1, \ldots, N\}$. In particular, 
$$
\left. \frac{\partial u_n}{x_i}\right|_{B_R(0)} \to \left. \frac{\partial u}{x_i}\right|_{B_R(0)} \quad \text{ in } L^p(B_R(0)) \ \forall R > 0.
$$
By Vainberg's Theorem, there exists a subsequence $(u_{1n}) \subset (u_n)$ such that 
$$
\frac{\partial u_{1n}}{x_i} (x) \to \frac{\partial u}{x_i}(x) \quad \text{ a.e. in } B_1(0). 
$$
Now, by the compact Sobolev embedding on the sequence $(u_{1n})$ there exists a subsequence $(u_{2n})$ such that 
$$
\frac{\partial u_{2n}}{x_i} (x) \to \frac{\partial u}{x_i}(x) \quad \text{ a.e. in } B_2(0). 
$$
Proceeding in an analogous manner, for every $k \in \Bbb{N}$ there exists $(u_{kn}) \subset (u_n)$ such that
$$
\frac{\partial u_{kn}}{x_i} (x) \to \frac{\partial u}{x_i}(x) \quad \text{ a.e. in } B_k(0). 
$$
We claim that $(u_{jj})$ is such that 
$$
\frac{\partial u_{jj}}{\partial x_i}(x) \to \frac{\partial u}{\partial x_i}(x) \quad \text{ a.e in } \Bbb{R}^N.
$$
Let
$$
S_k = \left\{x \in B_k(0) \ : \ \frac{\partial u_{kn}}{x_i} (x) \not\to \frac{\partial u}{x_i}(x) \right\}
$$
and $S = \cap_k S_k$. It is clear that $|S| = 0$, since it is a countable union of sets of measure $0$. Let $x \in \Bbb{R}^N \setminus S$ and $j_0 \in \Bbb{N}$ such that $x \in B_{j_0}(0)$. Then $x \in B_j(0)$ for all $j \geq j_0$. Moreover, 
$$
\frac{\partial u_{j_0n}}{\partial x_i}(x) \to \frac{\partial u}{\partial x_i}(x) \quad \text{ a.e in } B_{j_0}(0).
$$
Since $(u_{jj})$ is a subsequence of $(u_{j_0n})$, the claim follows. Therefore it holds that 
$$
|\nabla u_n|^{p - 2}\nabla u_n \to |\nabla u|^{p - 2} \nabla u \quad \text{ a.e. in } \Bbb{R}^N.
$$
Moreover, the sequence $(|\nabla u_n|^{p - 2}\nabla u_n)$ is bounded in $L^{\frac{p}{p - 1}}$. Hence, by the Brezis-Lieb Lemma, 
$$
\int_{\Bbb{R}^N} |\nabla u_n|^{p - 2} \nabla u_n \cdot \nabla \phi \ dx \to  \int_{\Bbb{R}^N} |\nabla u|^{p - 2} \nabla u \cdot \nabla \phi \ dx.
$$
