\kappa \lambda_{k+m}} & \leq \frac{\|f(\cdot, u_j(\cdot))\|_{L^{q/(q-1)}(\Omega)}}{\|u_j\|_{X_0}}\\
& \leq \frac{{\displaystyle \left(\int_\Omega \left(2\varepsilon |u_j(x)|+qC_{\varepsilon} |u_j(x)|^{q-1}\right)^{q/(q-1)}\,dx\right)^{(q-1)/q}}}{\|u_j\|_{X_0}}\\
& \leq \frac{\left( 2^{1/(q-1)}\left((2\varepsilon)^{q/(q-1)}\|u_j\|_{L^{q/(q-1)}(\Omega)}^{q/(q-1)}+
(qC_{\varepsilon})^{q/(q-1)}\|u_j\|_{L^q(\Omega)}^q\right)\right)^{(q-1)/q}}{\|u_j\|_{X_0}}\\
& \leq \frac{2\varepsilon\|u_j\|_{L^{q/(q-1)}(\Omega)}+
qC_{\varepsilon}\|u_j\|_{L^q(\Omega)}^{q-1}}{\|u_j\|_{X_0}}\\
& \leq C\left(2\varepsilon+
qC_{\varepsilon}\|u_j\|_{X_0}^{q-2}\right)\,,
\end{aligned}$$ thanks to the continuous embedding $X_0\hookrightarrow L^\nu(\Omega)$ for any $\nu\in [1, 2^*)$, and for some universal positive constant $C$.

By , , , and the Dominated Convergence Theorem, it is easily seen that $$\label{Fdominata}
\int_\Omega F(x, u_j(x))\,dx \to \int_\Omega F(x, u_\infty(x))\,dx$$ and $$\label{fdominata}
\int_\Omega |f(x, u_j(x))|^{q/(q-1)}\,dx \to \int_\Omega |f(x, u_\infty(x))|^{q/(q-1)}\,dx$$ as $j\to +\infty$. Relation , combined with , the fact that $F(x,0)=0$ a.e. $x\in \Omega$, and , yields that $$\label{uinftynabla}
u_\infty \equiv 0\,.$$

Now, two cases can occur. If $$\label{uinfty0}
u_j \to u_\infty\equiv 0\qquad \mbox{strongly in}\,\, X_0$$ as $j\to +\infty$, then, by we get that $$0<\frac{\bar \sigma}{\tilde\kappa \lambda_{k+m}} \leq 2C\varepsilon\,,$$ which gives a contradiction, due to the fact that $\varepsilon$ is arbitrary. Otherwise, there exists $\eta>0$ such that $\|u_j\|_{X_0}\geq \eta$ for $j$ large enough. Then, by this, , , and the fact that $f(x,0)=0$ a.e. $x\in \Omega$ (by ), we get that $$\frac{\bar \sigma\eta}{\tilde\kappa \lambda_{k+m}}\leq 0\,,$$ which is a contradiction. This completes the proof of Lemma \[lemmanabla1\].

The second lemma we need in order to prove the $\nabla$-condition is the following one:

\[lemmanabla2\] 