Then for $|x|>4$, there exists $c_{38}>1$ such that $$\label{5.8}
\frac1{c_{38}}|x|^{-N-2\alpha}\le -(-\Delta)^\alpha\bar w(x)\le c_{38}|x|^{-N-2\alpha},\quad x\in B_4^c(0).$$

[**Proof.**]{} For $|x|>4$, we have that $$\arraycolsep=1pt
\begin{array}{lll}
  -(-\Delta)^\alpha\bar w(x) =\int_{B_2(0)}\frac{\bar w(z)}{|z-x|^{N+2\alpha}}dz  \le (|x|-2)^{-N-2\alpha}  \int_{B_2(0)}\bar w(z)dz
\end{array}$$ and $$\arraycolsep=1pt
\begin{array}{lll}
  -(-\Delta)^\alpha\bar w(x)  \ge (|x|+2)^{-N-2\alpha}  \int_{B_2(0)}\bar w(z)dz,
\end{array}$$ which, together with $\int_{B_2(0)}\bar w(z)dz<+\infty$, imply (\[5.8\]). $\Box$

Now we are in the position to prove Theorem \[th 1\].

[**Proof of Theorem \[th 1\].**]{} For $p\in(1,\frac{N}{N-2\alpha})$, we denote $$\label{tau_p}
\tau_p=\left\{
\arraycolsep=1pt
\begin{array}{lll}
 -\frac{2\alpha}{p-1}\qquad   &{\rm for}\quad p\in[1+\frac{2\alpha}{N},\frac{N}{N-2\alpha}),\\[2mm]
  \phantom{    }
 -\frac{N+2\alpha}{p}&{\rm for}\quad p\in(1,1+\frac{2\alpha}{N}).
\end{array}
 \right.$$ We note that $\tau_p$ is continuous and strictly increasing with respect to $p$, $\tau_p=-N$ if $p=1+\frac{2\alpha}{N}$ and $\lim_{p\to0^+}\tau_p=-\infty$.

[*Lower bound.*]{} Since $\lim_{|x|\to0}u_1(x)=\infty$ and $u_1$ is continuous and positive in ${\mathbb{R}}^N\setminus\{0\}$, then there exists $c_{39}>0$ such that $$\label{5.9}
 c_{39}w_{\tau_p}\le u_1\quad {\rm in}\ \ \bar B_R(0)\setminus\{0\},$$ where $R>4$ is from Lemma \[lm 5.1\].

We note that for $p\in(1,1+\frac{2\alpha}{N})$, $\tau_pp=-N-2\alpha$; for $p=1+\frac{2\alpha}{N}$, $\tau_p=-N$, $\gamma_0+1=p\gamma_0$ and for $p\in(1+\frac{2\alpha}{N},\frac{N}{N-2\alpha})$, $\tau_p-2\alpha=p\tau_p$. By Lemma \[lm 5.1\], there exists $t_0\in(0,1)$ such that $$(-\Delta)^\alpha (t_0c_{39}w_{\tau_p})+(t_0c_{39}w_{\tau_p})^p\le 0\quad {\rm in }\quad B_R^c(0).$$

We claim that $u_1\ge t_0c_{39}w_{\tau_p}$ in $B_R^c(0)$. In fact, if not, there would exist $x_0\in B_R^c$ such that $$\begin{aligned}
 (u_1-t_0c_{39}w_{\tau_p})(x_0) &=& \min_{x\in B_R^c(0)}(u_1-t_0c_{39}w_{\tau_p})(x) \\
   &=&\min_{x\in {\mathbb{R}}^N\setminus\{0\}}(u_1-t_0c_{39}w_{\tau_p})(x)<0,\end{aligned}$$ since $u_1-t_0c_{39}w_{\tau_p}\ge0$ in $\bar B_R(0)$ and $\lim_{|x|\to\infty}(u_1-t_0c_{39}w_{\tau_p})(x)=0$. Then 