$\mathcal{I}_2\ne\emptyset$, and ${\overrightarrow{q}}={\left(}q_i{\right)}_{i\in\mathcal{I}_1\cup\mathcal{I}_2}$ be such that $q_i>1$ for all $i\in\mathcal{I}_1\cup\mathcal{I}_2$. Then for any $\varepsilon>0$, $\gamma\ge\frac{n}{p}{\left(}p_\varepsilon-p{\right)}$, $k\in{\mathbb{N}}$, and ${\left(}i_1,\dotsc,i_k{\right)}\in{\left(}\mathcal{I}_1\cup\mathcal{I}_2{\right)}^k$, we have $$\label{Lem6Eq1}
\gamma_{i_1,\dotsc,i_k}>\frac{n}{p}{\left(}p_\varepsilon-p{\right)}\quad\text{if }k<k_{\gamma,\varepsilon}^-\quad\text{and}\quad\gamma_{i_1,\dotsc,i_k}<\frac{n}{p}{\left(}p_\varepsilon-p{\right)}\quad\text{if }k\ge k_{\gamma,\varepsilon}^+\,,$$ where $\gamma_{i_1,\dotsc,i_k}$ is as in , $p_\varepsilon$ is as in , and $k_{\gamma,\varepsilon}^-$ and $k_{\gamma,\varepsilon}^+$ are the smallest and largest natural numbers, respectively, such that $$\label{Lem6Eq2}
\frac{n}{p}{\left(}\frac{n}{n-p}{\right)}^{k_{\gamma,\varepsilon}^+-1}\varepsilon p_0<\gamma<\frac{n}{p}{\left(}\frac{n}{n-p}{\right)}^{k_{\gamma,\varepsilon}^-}{\left(}p_\varepsilon-p_-{\right)},$$ where $p_-:=\min{\left(}\left\{p_i\in\operatorname{\overrightarrow{p}}\right\}{\right)}$. In particular, we have $\Phi_{k,\gamma,\varepsilon}=\emptyset$ for all $k<k_{\gamma,\varepsilon}^-$ and $k>k_{\gamma,\varepsilon}^+$, where $\Phi_{k,\gamma,\varepsilon}$ is as in .

Since $p_-\le p_{i_j}\le p_0$ for all $j=1,\dotsc,k$, it follows from that $$\label{Lem6Eq3}
-\sum_{j=1}^k{\left(}\frac{n-p}{n}{\right)}^{k-j}{\left(}p-p_-{\right)}\le\gamma_{i_1,\dotsc,i_k}-{\left(}\frac{n-p}{n}{\right)}^k\gamma\le\sum_{j=1}^k{\left(}\frac{n-p}{n}{\right)}^{k-j}{\left(}p_0-p{\right)}.$$ Moreover, by a simple calculation, we obtain $$\label{Lem6Eq4}
\sum_{j=1}^k{\left(}\frac{n-p}{n}{\right)}^{k-j}=\frac{n}{p}{\left(}1-{\left(}\frac{n-p}{n}{\right)}^k{\right)}<\frac{n}{p}\,.$$ It follows from and that $$\label{Lem6Eq5}
-\frac{n}{p}{\left(}p-p_-{\right)}<\gamma_{i_1,\dotsc,i_k}-{\left(}\frac{n-p}{n}{\right)}^k\gamma<\frac{n}{p}{\left(}p_0-p{\right)}.$$ Finally, follows from together with the definitions of $k_{\gamma,\varepsilon}^-$ and $k_{\gamma,\varepsilon}^+$.

Now, we can prove the main result of this section.

\[Lem7\] Assume that $p_+<p^*$. Let $f:{\mathbb{R}}^n\times{\mathbb{R}}\to{\mathbb{R}}$ be a Caratheodory function such that holds true, $u$ be a solution of , and $\kappa$, $r$, and $K_0$ be as in Lemma \[Lem3\]. Let $\mathcal{I}_1$ and $\mathcal{I}_2$ be two disjoint subsets of $\left\{1,\dotsc,n\right\}$, $\mathcal{I}_2\ne\emptyset$, and ${\overrightarrow{q}}={\left(}q_i{\right)}_{i\in\mathcal{I}_1\cup\mathcal{I}_2}$ be such that $q_i>1$ for all $i\in\mathcal{I}_1\cup\mathcal{I}_2$. Then there exists a constant $c_1=c_1{\left(}n,\operatorname{\overrightarrow{p}},\Lambda,K_0{\right)}>1$ such that for any $\varepsilon\in{\left(}0,1{\right)}$, $\gamma>\frac{n}{p}{\left(}p_\varepsilon-p{\right)}$, and $R_1,R_2>0$ such that $\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,1/2{\right)}\cap B_{\operatorname{\overrightarrow{p}}}{\left(}0,\max{\left(}r,1{\right)}{\right)}=\emptyset$ and $\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,1/2{\right)}\cap\operatorname{supp}{\left(}u{\right)}$ is bounded, we have $$\begin{gathered}
\label{Lem7Eq1}
\left\|u\right\|_{L^{\gamma}{\left(}\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda_{0,\gamma,\varepsilon}{\right)}{\right)}}\le c_1^{\frac{1}{\varepsilon}}\max{\left(}q_i{\right)}^{\frac{1}{\varepsilon}\cdot\frac{n}{n-p}}{\left(}p_\varepsilon-p_*{\right)}^{-\frac{1}{\varepsilon}\cdot\frac{n}{n-p}}\\
\times\max_{{\left(}i_1,\dotsc,i_k{\right)}\in\Phi_{\gamma,\varepsilon}}{\left(}{\left(}\prod_{j=1}^kR_{\delta_{i_j}}^{-\frac{1}{\gamma}{\left(}\frac{n}{n-p}{\right)}^j\frac{p_{i_j}}{q_{i_j}}}{\right)}\left\|u\right\|_{L^{\gamma_{i_1,\dotsc,i_k}}{\left(}\Omega_{{\overrightarrow{q}}}{\left(}\mathcal{I}_1,R_1,\mathcal{I}_2,R_2,\lambda_{k,\gamma,\varepsilon}{\right)}{\right)}}^{\frac{\gamma_{i_1,\dotsc,i_k}}{\gamma}{\left(}\frac{n}{n-p}{\right)}^k}{\right)},\end{gathered}$$ 