p_0$ for all $j=1,\dotsc,k$, we obtain $$\label{Lem7Eq12}
\mathcal{D}_{i_1,\dotsc,i_k,\gamma,\varepsilon}\le2^{\frac{1}{\gamma}\overset{k}{\underset{j=1}{\sum}}p_{i_j}{\left(}\frac{n}{n-p}{\right)}^j{\left(}k_{\gamma,\varepsilon}^+-j+4{\right)}}\le2^{\frac{p_0}{\gamma}\overset{k_{\gamma,\varepsilon}^+}{\underset{j=1}{\sum}}{\left(}\frac{n}{n-p}{\right)}^j{\left(}k_{\gamma,\varepsilon}^+-j+4{\right)}}.$$ We find $$\begin{aligned}
\overset{k_{\gamma,\varepsilon}^+}{\underset{j=1}{\sum}}{\left(}\frac{n}{n-p}{\right)}^j{\left(}k_{\gamma,\varepsilon}^+-j+4{\right)}&=\frac{n}{p}{\left[}{\left(}\frac{n}{n-p}{\right)}^{k_{\gamma,\varepsilon}^+}{\left(}3+\frac{n}{p}{\right)}-k_{\gamma,\varepsilon}^+-3-\frac{n}{p}{\right]}\\
&<\frac{n}{p}{\left(}\frac{n}{n-p}{\right)}^{k_{\gamma,\varepsilon}^+}{\left(}3+\frac{n}{p}{\right)},\end{aligned}$$ and hence by , we obtain $$\label{Lem7Eq13}
\overset{k_{\gamma,\varepsilon}^+}{\underset{j=1}{\sum}}{\left(}\frac{n}{n-p}{\right)}^j{\left(}k_{\gamma,\varepsilon}^+-j+4{\right)}<\frac{\gamma}{\varepsilon p_0}\cdot\frac{n}{n-p}{\left(}3+\frac{n}{p}{\right)}.$$ It follows from and that $$\label{Lem7Eq14}
\mathcal{D}_{i_1,\dotsc,i_k,\gamma,\varepsilon}\le2^{\frac{1}{\varepsilon}\cdot\frac{n}{n-p}{\left(}3+\frac{n}{p}{\right)}}.$$

The estimate follows from , , , , and .

The vanishing result {#Sec6}
====================

In this section, we prove a vanishing result which will give Point (i) in Theorem \[Th3\]. We define $$\label{Eq15}
\overline{p}_0:=\max{\left(}p_*,\left\{p_i\in\operatorname{\overrightarrow{p}}\,:\,i\in\Theta\right\}{\right)},$$ where $p_*$ is as in and $\Theta$ is the set of all indices $i$ such that $$\label{Eq16}
\big(p_i-p_--\frac{n}{p}{\left(}p_i-p_*{\right)}\big)\sum_{j=1}^n\max{\left(}\frac{p_i-p_j}{p_j},0{\right)}\ge{\left(}p_*-1{\right)}{\left(}p_i-p_-{\right)}$$ with $p_-:=\min{\left(}\left\{p_i\in\operatorname{\overrightarrow{p}}\right\}{\right)}$. We define $\mathcal{I}_0$ as the set of all indices $i$ such that $p_i>\overline{p}_0$.

When $p_+>p_*$, one easily sees that the condition does not hold true for $p_i=p_+$, and hence we have $\overline{p}_0<p_+$ and $\mathcal{I}_0\ne\emptyset$.

We prove the following result.

\[Th6\] Assume that $p_*<p_+<p^*$. Let $f:{\mathbb{R}}^n\times{\mathbb{R}}\to{\mathbb{R}}$ be a Caratheodory function such that holds true and $u$ be a solution of . Then there exists a constant $R_0=R_0{\left(}n,\operatorname{\overrightarrow{p}},\Lambda,u{\right)}$ such that $u{\left(}x{\right)}=0$ for all $x\in{\mathbb{R}}^n$ such that $\sum_{i\in \mathcal{I}_0}\left|x_i\right|\ge R_0$.

The proof of Theorem \[Th6\] is based on the following result, which we obtain by applying the iteration scheme in Section \[Sec5\].

\[Lem8\] Assume that $p_*<p_+<p^*$. Let $f:{\mathbb{R}}^n\times{\mathbb{R}}\to{\mathbb{R}}$ be a Caratheodory function such that holds true, $u$ be a 