\mu\right)\Big(\|u_j\|_{X_0}^2-\lambda \|u_j\|_{L^2(\Omega)}^2\Big)\\
& \qquad +a_3\left(\frac q \mu-1\right)\|u_j\|_{L^q(\Omega)}^q-a_4\left(\frac q \mu-1\right)\,|\Omega|\,.
\end{aligned}$$ Note that for any $\varepsilon>0$ the Young inequality gives $$\label{youngPS}
\|u_j\|_{L^2(\Omega)}^2\leq \frac{2\varepsilon}{q}\,\|u_j\|_{L^q(\Omega)}^q+\frac{q-2}{q}\,\varepsilon^{-2/(q-2)}\,|\Omega|\,,$$ so that, by and , we obtain that $$\label{young2}
\begin{aligned}
\mathcal J_\lambda(u_j)-\frac 1 \mu \langle \mathcal J_\lambda'(u_j), u_j\rangle &
\geq \left(\frac 1 2 -\frac 1 \mu\right)\|u_j\|_{X_0}^2-\lambda\left(\frac 1 2 -\frac 1 \mu\right) \frac{2\varepsilon}{q}\,\|u_j\|_{L^q(\Omega)}^q\\
& \qquad -\lambda\left(\frac 1 2 -\frac 1 \mu\right) \frac{q-2}{q}\,\varepsilon^{-2/(q-2)}\,|\Omega|\\
& \qquad +a_3\left(\frac q \mu-1\right)\|u_j\|_{L^q(\Omega)}^q-a_4\left(\frac q \mu-1\right)\,|\Omega|\\
& = \left(\frac 1 2 -\frac 1 \mu\right)\|u_j\|_{X_0}^2\\
& \qquad +\Big[a_3\left(\frac q \mu-1\right)-\lambda\left(\frac 1 2 -\frac 1 \mu\right)\frac{2\varepsilon}{q}\Big]\|u_j\|_{L^q(\Omega)}^q-C_\varepsilon\,,
\end{aligned}$$ where $C_\varepsilon$ is a constant such that $C_\varepsilon\to +\infty$ as $\varepsilon\to 0$, being $q>\mu>2$.

Now, choosing $\varepsilon$ so small that $$a_3\left(\frac q \mu-1\right)-\lambda\left(\frac 1 2 -\frac 1 \mu\right)\frac{2\varepsilon}{q}>0\,,$$ by we get $$\label{jj'20}
\mathcal J_\lambda(u_j)-\frac 1 \mu \langle \mathcal J_\lambda'(u_j), u_j\rangle \geq \left(\frac 1 2 -\frac 1 \mu\right)\|u_j\|_{X_0}^2-C_\varepsilon\,.$$

Finally, by and for any $j\in {\mathbb N}$ $$\|u_j\|_{X_0}^2 \leq \kappa_*\left(1+\|u_j\|_{X_0}\right)$$ for a suitable positive constant $\kappa_*$. Hence, assertion  is proved.

Now, let us finish the proof of the Palais–Smale condition for $\mathcal J_\lambda$. Since $\{u_j\}_{j\in {\mathbb N}}$ is bounded in $X_0$ and $X_0$ is a reflexive space, up to a subsequence, still denoted by $u_j$, there exists $u_\infty \in X_0$ such that $$\label{convergenze0}
