\begin{aligned}
 & \int_{{\mathbb R}^n\times {\mathbb R}^n}\big(u_j(x)-u_j(y)\big)\big(\varphi(x)-\varphi(y)\big) K(x-y)\,dx\,dy \to \\
& \qquad \qquad
\int_{{\mathbb R}^n\times {\mathbb R}^n}\big(u_\infty(x)-u_\infty(y)\big)\big(\varphi(x)-\varphi(y)\big)
K(x-y)\,dx\,dy  \quad \mbox{for any}\,\, \varphi\in X_0\,,
\end{aligned}$$ while, by [@svmountain Lemma 8], up to a subsequence, $$\label{convergenze0bis}
\begin{aligned}
& u_j \to u_\infty \quad \mbox{in}\,\, L^\nu({\mathbb R}^n)\quad \mbox{for any}\,\,\nu\in [1, 2^*) \\
& u_j \to u_\infty \quad \mbox{a.e. in}\,\, {\mathbb R}^n
\end{aligned}$$ as $j\to +\infty$. Finally, by [@brezis Theorem IV.9] we know that for any $\nu\in [1, 2^*)$ there exists $\ell_\nu\in L^\nu({\mathbb R}^n)$ such that $$\label{dominata20}
|u_j(x)|\leq \ell_\nu(x) \quad \mbox{a.e. in}\,\, {\mathbb R}^n\,\quad \mbox{for any}\,\,j\in {\mathbb N}\,.$$

By , , , the fact that the map $t\mapsto f(\cdot, t)$ is continuous in $t\in {\mathbb R}$ (see ) and the Dominated Convergence Theorem, we get $$\label{convf0}
\int_\Omega f(x, u_j(x))u_j(x)\,dx \to \int_\Omega f(x, u_\infty(x))u_\infty(x)\,dx$$ and $$\label{convfu0}
\int_\Omega f(x, u_j(x))u_\infty(x)\,dx \to \int_\Omega f(x, u_\infty(x))u_\infty(x)\,dx$$ as $j\to +\infty$. Furthermore, by and we have that $$\begin{aligned}
 0\leftarrow \langle \mathcal J'_\lambda(u_j), u_j\rangle & = \int_{{\mathbb R}^n\times {\mathbb R}^n}|u_j(x)-u_j(y)|^2 K(x-y)\,dx\,dy -\lambda \int_\Omega |u_j(x)|^2\,dx\\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad - \int_\Omega f(x, u_j(x))u_j(x)\,dx
\end{aligned}$$ so that, by and we deduce that $$\label{convnorma10}
\int_{{\mathbb R}^n\times {\mathbb R}^n}|u_j(x)-u_j(y)|^2 K(x-y)\,dx\,dy\to \lambda \int_\Omega |u_\infty(x)|^2\,dx+\int_\Omega f(x, u_\infty(x))u_\infty(x)\,dx$$ as $j\to +\infty$, while, by , (both with 