}\ B_R^c(0).
\end{array}$$ By Lemma \[teo CP\], we have that $$\label{4.2}
\tilde u_n\le u_{n,R}\le u_n \quad {\rm in}\quad {\mathbb{R}}^N.$$

It is known that $u_R:=\lim_{n\to\infty} u_{n,R}$ is a weak solution of (\[3.11\]), since $\{k\eta_n+u_n^p-u^p\ast\eta_n\}$ converges to $k \delta_0$ in the distribution sense as $n\to\infty$. Hence, (\[3.1.7\]) follows by (\[4.2\]) and $\tilde u_n\to u-m_R$ in ${\mathbb{R}}^N$ as $n\to\infty$. $\Box$

With the help of Lemma \[lm 3.1\], we show next the uniqueness of weak solution to (\[eq1.3\]).

\[pr 3.1\] Assume that $k>0$ and $0<p<\frac{N}{N-2\alpha}$. Then (\[eq1.3\]) admits a unique weak solution $u_k$.

[**Proof.**]{} [*Existence.*]{} By Theorem \[teo 1\], there exists at least one weak solution $u_k$ to $$(-\Delta)^\alpha  u+u^p=k\delta_0\quad {\rm in}\ {\mathbb{R}}^N$$ such that $0\le u_k\le k\mathbb{G}_{{\mathbb{R}}^N}[\delta_0]$ a.e. in ${\mathbb{R}}^N$. We observe that $$\mathbb{G}_{{\mathbb{R}}^N}[\delta_0](x)=\frac{c_{N,\alpha}}{|x|^{N-2\alpha}},\quad x\in{\mathbb{R}}^N\setminus\{0\},$$ then $$\lim_{|x|\to\infty}u_k(x)=0.$$ Thus $u_k$ is a weak solution of (\[eq1.3\]).

[*Uniqueness.*]{} We assume that $u_k$, $\tilde u_k$ are two different weak solutions of (\[eq1.3\]) and $$A_0:=\min\{1,\limsup_{x\to0}|\tilde u_k-u_k|(x)\}.$$ We claim that $A_0>0$. In fact, if not, then $\lim_{x\to0}|\tilde u_k-u_k|(x)=0$. Now we may assume that there exists $x_0\in{\mathbb{R}}^N\setminus\{0\}$ such that $$(\tilde u_k-u_k)(x_0)=\sup_{x\in{\mathbb{R}}^N\setminus\{0\}}(\tilde u_k-u_k)(x)>0,$$ which implies that $$(-\Delta)^\alpha(\tilde u_k-u_k)(x_0)\ge0.$$ Then we obtain a contradiction by the fact that $\tilde u_k$ and $u_k$ are classical solutions of (\[eq1.01\]) by Lemma \[lm 3.1\]. 