support. Moreover, since $q_i>1$ for all $i\in\mathcal{I}_1\cup\mathcal{I}_2$, we get $\eta^{{\left(}\beta+p_-{\right)}/p_+}\in C^1{\left(}{\mathbb{R}}^n{\right)}$ and $$\label{Lem4Eq3}
\left|\partial_{x_i}\eta{\left(}x{\right)}\right|^{p_i}\eta{\left(}x{\right)}^{\beta+p_--p_i}\le{\left(}\frac{4q_i}{\lambda'-\lambda}\max{\left(}1,\frac{p_+}{\beta+p_-}{\right)}{\right)}^{p_i}R_{\delta_i}^{-\frac{p_i}{q_i}}$$ for all $x\in\operatorname{supp}{\left(}\eta{\right)}$, where $\delta_i:=1$ if $i\in\mathcal{I}_1$, $\delta_i:=2$ if $i\in\mathcal{I}_2$. By applying Lemma \[Lem5\] with $v=\left|u\right|$ and $\eta$ as in , and using , we obtain $$\begin{gathered}
\label{Lem4Eq4}
{\left(}\int_{\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda{\right)}}u^\gamma dx{\right)}^{\frac{n-p}{n}}\le C{\left(}\left|\beta\right|^p+1{\right)}\bigg(\Lambda{\left(}\beta+1{\right)}^{-1}\int_{\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda'{\right)}}u^{\beta+p^*}dx\\
+\sum_{i\in\mathcal{I}_1\cup\mathcal{I}_2}\min{\left(}1,\beta+1{\right)}^{-p_i}{\left(}\lambda'-\lambda{\right)}^{-p_i}q_i^{p_i}R_{\delta_i}^{-\frac{p_i}{q_i}}\int_{\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda'{\right)}}u^{\beta+p_i}dx\bigg)\end{gathered}$$ for some constant $C=C{\left(}n,\operatorname{\overrightarrow{p}}{\right)}$.

Now, we estimate the first integral in the right-hand side of . We claim that there exists a constant $C'=C{\left(}n,\operatorname{\overrightarrow{p}},K_0{\right)}$ such that $$\label{Lem4Eq5}
u{\left(}x{\right)}^{p^*-p_{i_0}}\le C'R_2^{-\frac{p_{i_0}}{q_{i_0}}}\quad\text{for all }x\in\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda'{\right)},$$ where $K_0$ is the constant given by Lemma \[Lem3\] and $i_0\in\mathcal{I}_2$ is such that $$\label{Lem4Eq6}
\frac{q_{i_0}{\left(}p^*-p_{i_0}{\right)}}{p_{i_0}}=\max_{i\in\mathcal{I}_2}{\left(}\frac{q_i{\left(}p^*-p_i{\right)}}{p_i}{\right)}\,.$$ We prove this claim. For any $x\in\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda'{\right)}$, since $\lambda'\le1/2$, we obtain $$\label{Lem4Eq7}
\frac{R_2}{2}\le\sum_{i\in\mathcal{I}_2}\left|x_i\right|^{q_i}\le n\cdot|x_{i{\left(}x{\right)}}|^{q_{i{\left(}x{\right)}}}\le n\cdot d_{\operatorname{\overrightarrow{p}}}{\left(}x,0{\right)}^{\frac{q_{i{\left(}x{\right)}}(p^*-p_{i{\left(}x{\right)}})}{\delta p_{i{\left(}x{\right)}}}},$$ where $i{\left(}x{\right)}\in\mathcal{I}_2$ is such that $|x_{i{\left(}x{\right)}}|^{q_{i{\left(}x{\right)}}}=\max{\left(}\left\{\left|x_i\right|^{q_i}:\,i\in\mathcal{I}_2\right\}{\right)}$, and the distance function $d_{\operatorname{\overrightarrow{p}}}$ and the real number $\delta$ are as in . Since $\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda'{\right)}\cap B_{\operatorname{\overrightarrow{p}}}{\left(}0,\max{\left(}r,1{\right)}{\right)}=\emptyset$ by assumption, follows from , , and Lemma \[Lem3\]. In particular, implies $$\label{Lem4Eq9}
\int_{\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda'{\right)}}u^{\beta+p^*}dx\le C'R_2^{-\frac{p_{i_0}}{q_{i_0}}}\int_{\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda'{\right)}}u^{\beta+p_{i_0}}dx\,.$$ Finally, follows from , , and the fact that $\beta+1=\frac{n-p}{n}{\left(}\gamma-p_*+1{\right)}$ and $\beta+p_i=\gamma_i$. This ends the proof of Lemma \[Lem4\].

The iteration scheme {#Sec5}
====================

In this section, we describe the iteration scheme which leads to the proofs of our main results.

Let $\mathcal{I}_1$ and $\mathcal{I}_2$ be two disjoint subsets of $\left\{1,\dotsc,n\right\}$, $\mathcal{I}_2\ne\emptyset$, 