Proof. For any fixed $x\in \mathbb {R}^{n}$ and $R > > |x|+1$, let

(1.1)\begin{align} & (-\Delta)^{\sigma}_{p}u(x)-(-\Delta)^{\sigma}_{p}u_{i}(x)\nonumber\\ & = c_{n,\sigma p}\int_{B_{R}(0)}\frac{|u(x)-u(y)|^{p-2}(u(x)\!-\!u(y))\!-\!|u_{i}(x)\!-\!u_{i}(y)|^{p-2}(u_{i}(x)-u_{i}(y))}{|x-y|^{n+\sigma p}}\,{\rm d}y\nonumber\\ & \quad+c_{n,\sigma p}\int_{B_{R}^{c}(0)}\frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))}{|x-y|^{n+\sigma p}}\,{\rm d}y\nonumber\\ & \quad+c_{n,\sigma p}\int_{B_{R}^{c}(0)}\frac{-|u_{i}(x)-u_{i}(y)|^{p-2}(u_{i}(x)-u_{i}(y))}{|x-y|^{n+\sigma p}}\,{\rm d}y\nonumber\\ & = :\Phi_{i}(x,R)+\mathfrak{G}(x,R)+\Psi_{i}(x,R). \end{align}

In light of the fact that $u_{i}\rightarrow u$ in $C^{\sigma p+\alpha }(B_{2R}(0))$, we obtain that for each $0<\varepsilon <1$, there exists an integer $N>0$ such that for every $i>N$,

(1.2)\begin{equation} \|u_{i}-u\|_{C^{\sigma p+\alpha}(B_{2R}(0))}\leq \varepsilon^{\frac{p}{\min\{1,p-2\}}},\quad \|u_{i}\|_{C^{\sigma p+\alpha}(B_{2R}(0))}\leq\|u\|_{C^{\sigma p+\alpha}(B_{2R}(0))}+1. \end{equation}

Define

\[ \Phi_{i}(x,R\setminus\varepsilon):=\Phi_{i}(x,R)-\Phi_{i}(x,\varepsilon),\quad\mathcal{M}:=\|u\|_{C^{\sigma p+\alpha}(B_{2R}(0))}+1, \]

where $\Phi _{i}(x,\varepsilon )$ denotes the integral in $\Phi _{i}(x,R)$ with the domain $B_{R}(0)$ replaced by $B_{\varepsilon }(x)$. Using (1.2), we deduce that for $x,y\in B_{2R}(0)$, $i>N$,

\begin{align*} & \left||u(x)-u(y)|^{p-2}(u(x)-u(y))-|u_{i}(x)-u_{i}(y)|^{p-2}(u_{i}(x)-u_{i}(y))\right|\\ & \leq|u(x)-u(y)|^{p-2}|(u-u_{i})(x)-(u-u_{i})(y)|\\ & \quad+\left||u(x)-u(y)|^{p-2}-|u_{i}(x)-u_{i}(y)|^{p-2}\right||u_{i}(x)-u_{i}(y)|\\ & \leq C(p,\mathcal{M})\|u_{i}-u\|^{\min\{1,p-2\}}_{L^{\infty}(B_{2R}(0))}\leq C(p,\mathcal{M})\varepsilon^{p}, \end{align*}

which yields that

(1.3)\begin{equation} \left|\Phi_{i}(x,R\setminus\varepsilon)\right|\leq C(p,\mathcal{M})\varepsilon^{p}\int_{B_{2R}(x)\setminus B_{\varepsilon}(x)}\frac{{\rm d}y}{|x-y|^{n+\sigma p}}\leq C(p,n,\sigma,\mathcal{M})\varepsilon^{(1-\sigma)p}. \end{equation}

On the other hand, if $\sigma p+\alpha \in (0,1]$, then it follows from (1.2) that

(1.4)\begin{align} \left|\Phi_{i}(x,\varepsilon)\right|& \leq C(p,\sigma,\alpha,\mathcal{M})\int_{B_{\varepsilon}(x)}\frac{|x-y|^{(\sigma p+\alpha)(p-1)}}{|x-y|^{n+\sigma p}}\nonumber\\ & \leq C(p,n,\sigma,\alpha,\mathcal{M})\varepsilon^{(\sigma p+\alpha)(p-2)+\alpha}. \end{align}

When $\sigma p+\alpha \in (1,\infty )$, utilizing (1.2) again, it follows from Taylor expansion that

\begin{align*} & \left||u_{i}(x)-u_{i}(y)|^{p-2}(u_{i}(x)-u_{i}(y))-|\nabla u_{i}(x)(x-y)|^{p-2}\nabla u_{i}(x)(x-y)\right|\\ & \leq C(p,\sigma,\alpha,\mathcal{M})\left(|\nabla u_{i}(x)(x-y)|^{p-2}+|x-y|^{\min\{2,\sigma p+\alpha\}(p-2)}\right)|x-y|^{\min\{2,\sigma p+\alpha\}}\\ & \leq C(p,\sigma,\alpha,\mathcal{M})|x-y|^{\min\{p,(\sigma+1)p+\alpha-2\}}, \end{align*}

where we utilized the following element inequality:

\[ \left||a|^{p-2}a-|b|^{p-2}b\right|\leq C(p)|a-b|\left(|a-b|^{p-2}+|b|^{p-2}\right),\quad\text{for}\ a,b\in\mathbb{R}^{n}. \]

By the same argument, we have

\begin{align*} & \left||u(x)-u(y)|^{p-2}(u(x)-u(y))-|\nabla u(x)(x-y)|^{p-2}\nabla u(x)(x-y)\right|\\ & \leq C(p,\sigma,\alpha,\mathcal{M})|x-y|^{\min\{p,(\sigma+1)p+\alpha-2\}}. \end{align*}

Therefore, we obtain that if $\sigma p+\alpha \in (1,\infty )$,

(1.5)\begin{align} \left|\Phi_{i}(x,\varepsilon)\right|& \leq C(p,\sigma,\alpha,\mathcal{M})\int_{B_{\varepsilon}(x)}\frac{|x-y|^{\min\{p,(\sigma+1)p+\alpha-2\}}}{|x-y|^{n+\sigma p}}\,{\rm d}y\nonumber\\ & \leq C(p,n,\sigma,\alpha,\mathcal{M})\varepsilon^{\min\{(1-\sigma)p,p+\alpha-2\}}, \end{align}

where we utilized the anti-symmetry of $\nabla u(x)(x-y)$ and $\nabla u_{i}(x)(x-y)$ with regard to the centre $x$. Consequently, combining (1.3)–(1.5), we deduce that for every $i>N$,

\[ \left|\Phi_{i}(x,R)\right|\leq C(p,n,\sigma,\alpha,\mathcal{M}) \begin{cases} \varepsilon^{\min\{(1-\sigma)p,(\sigma p+\alpha)(p-2)+\alpha\}}, & \text{if}\ \sigma p+\alpha\in(0,1],\\ \varepsilon^{\min\{(1-\sigma)p,p+\alpha-2\}}, & \text{if}\ \sigma p+\alpha\in(1,\infty), \end{cases} \]

which implies that

(1.6)\begin{equation} \lim_{i\rightarrow\infty}\Phi_{i}(x,R)=0. \end{equation}

Note that $\{(-\Delta )^{\sigma }_{p}u_{i}\}$ is a pointwise convergent sequence, we then deduce from (1.1) and (1.6) that

(1.7)\begin{equation} \lim\limits_{i\rightarrow\infty}\Psi_{i}(x,R)\;\,\text{exists and is finite}. \end{equation}

Since $u\in \mathcal {L}_{\sigma p}(\mathbb {R}^{n})$ and $R > > |x|+1$, then

\begin{align*} & \limsup_{R\rightarrow\infty}\int_{B_{R}^{c}(0)}\frac{|u(x)-u(y)|^{p-1}}{|x-y|^{n+\sigma p}}\,{\rm d}y\\ & \leq \limsup_{R\rightarrow\infty}\left(\frac{R}{R-|x|}\right)^{n+\sigma p}\int_{B_{R}^{c}(0)}\frac{C(p)(u^{p-1}(x)+u^{p-1}(y))}{|y|^{n+\sigma p}}\,{\rm d}y=0, \end{align*}

which yields that

\[ \lim_{R\rightarrow\infty}\mathfrak{G}(x,R)=0. \]

This, in combination with (1.1) and (1.6)–(1.7), leads to that $\lim \limits _{R\rightarrow \infty }\lim \limits _{i\rightarrow \infty }\Psi _{i}(x,R)$ exists and is finite,

(1.8)\begin{equation} (-\Delta)^{\sigma}_{p}u(x)-\lim_{i\rightarrow\infty}(-\Delta)^{\sigma}_{p}u_{i}(x)=\lim_{R\rightarrow\infty}\lim_{i\rightarrow\infty}\Psi_{i}(x,R). \end{equation}

Denote

\begin{align*} \mathcal{K}_{1}& :={-}u_{i}^{p-2}(y)u_{i}(x),\\ \mathcal{K}_{2}& :=\left(u_{i}^{p-2}(y)-|u_{i}(x)-u_{i}(y)|^{p-2}\right)u_{i}(x),\\ \mathcal{K}_{3}& :={-}\left(u_{i}^{p-2}(y)-|u_{i}(x)-u_{i}(y)|^{p-2}\right)u_{i}(y),\\ \Theta& :={-}|u_{i}(x)-u_{i}(y)|^{p-2}(u_{i}(x)-u_{i}(y)). \end{align*}

Then we have

(1.9)\begin{equation} u_{i}^{p-1}(y)-\sum^{3}_{j=1}|\mathcal{K}_{j}|\leq\Theta=u_{i}^{p-1}(y)+\sum^{3}_{j=1}\mathcal{K}_{j}\leq u_{i}^{p-1}(y)+\sum^{3}_{j=2}|\mathcal{K}_{j}|. \end{equation}

For any given $\varepsilon >0$, it follows from Young's inequality that

(1.10)\begin{equation} |\mathcal{K}_{1}|\leq \varepsilon u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{p-2}}u^{p-1}_{i}(x). \end{equation}

We now divide into three cases to estimate $\mathcal {K}_{2}$ and $\mathcal {K}_{3}$ in the following.

Case 1. Consider $2< p\leq 3$. Since

\begin{align*} & u_{i}^{p-2}(y)\leq (|u_{i}(y)-u_{i}(x)|+u_{i}(x))^{p-2}\leq|u_{i}(y)-u_{i}(x)|^{p-2}+u_{i}^{p-2}(x),\\ & |u_{i}(y)-u_{i}(x)|^{p-2}\leq u_{i}^{p-2}(y)+u_{i}^{p-2}(x), \end{align*}

then

\[ \left|u_{i}^{p-2}(y)-|u_{i}(x)-u_{i}(y)|^{p-2}\right|\leq u_{i}^{p-2}(x). \]

Hence it follows from Young's inequality that

(1.11)\begin{equation} |\mathcal{K}_{2}|\leq u^{p-1}_{i}(x),\quad|\mathcal{K}_{3}|\leq \varepsilon u_{i}^{p-1}(y)+\frac{C(p)}{\varepsilon^{\frac{1}{p-2}}}u^{p-1}_{i}(x). \end{equation}

Substituting (1.10)–(1.11) into (1.9), we derive

(1.12)\begin{equation} (1-2\varepsilon)u^{p-1}_{i}(y)-\frac{C(p)}{\varepsilon^{p-2}}u^{p-1}_{i}(x)\leq\Theta\leq(1+\varepsilon)u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{\frac{1}{p-2}}}u^{p-1}_{i}(x). \end{equation}

Case 2. Consider the case when $p>3$ is an integer. From the binomial theorem and Young's inequality, we have

(1.13)\begin{align} (a+b)^{p-2}& =a^{p-2}+\sum^{p-2}_{j=1}C_{p-2}^{j}a^{p-2-j}b^{j}\leq(1+\varepsilon)a^{p-2}+C(p)b^{p-2}\sum^{p-2}_{j=1}\varepsilon^{-\frac{p-2-k}{k}}\nonumber\\ & \leq(1+\varepsilon)a^{p-2}+\frac{C(p)}{\varepsilon^{p-3}}b^{p-2},\quad \text{for any }a,b\geq0. \end{align}

Using (1.13), we deduce

\[ u_{i}^{p-2}(y)\leq(|u_{i}(y)-u_{i}(x)|+u_{i}(x))^{p-2}\leq(1+\varepsilon)|u_{i}(x)-u_{i}(y)|^{p-2}+\frac{C(p)}{\varepsilon^{p-3}}u_{i}^{p-2}(x), \]

which implies that

\begin{align*} u_{i}^{p-2}(y)-|u_{i}(y)-u_{i}(x)|^{p-2}& \leq\varepsilon|u_{i}(x)-u_{i}(y)|^{p-2}+\frac{C(p)}{\varepsilon^{p-3}}u_{i}^{p-2}(x)\\ & \leq \varepsilon(1+\varepsilon)u^{p-2}_{i}(y)+\frac{C(p)}{\varepsilon^{p-3}}u^{p-2}_{i}(x). \end{align*}

Analogously,

\[ |u_{i}(y)-u_{i}(x)|^{p-2}-u_{i}^{p-2}(y)\leq\varepsilon u^{p-2}_{i}(y)+\frac{C(p)}{\varepsilon^{p-3}}u^{p-2}_{i}(x). \]

Hence, we have

(1.14)\begin{equation} \left|u_{i}^{p-2}(y)-|u_{i}(y)-u_{i}(x)|^{p-2}\right|\leq\varepsilon(1+\varepsilon)u^{p-2}_{i}(y)+\frac{C(p)}{\varepsilon^{p-3}}u^{p-2}_{i}(x). \end{equation}

Utilizing (1.14) and Young's inequality, we obtain

\begin{align*} |\mathcal{K}_{2}|& \leq\varepsilon(1+\varepsilon)u^{p-2}_{i}(y)u_{i}(x)+\frac{C(p)}{\varepsilon^{p-3}}u^{p-1}_{i}(x)\leq\varepsilon u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{p-3}}u^{p-1}_{i}(x),\\ |\mathcal{K}_{3}|& \leq\varepsilon(1+\varepsilon)u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{p-3}}u^{p-2}_{i}(x)u_{i}(y)\leq\varepsilon(2+\varepsilon) u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{p-2}}u^{p-1}_{i}(x), \end{align*}

which, in combination with (1.9)–(1.10), gives that

(1.15)\begin{align} \Theta& \leq(1+3\varepsilon+\varepsilon^{2})u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{p-2}}u^{p-1}_{i}(x), \end{align}

(1.16)\begin{align} \Theta& \geq(1-4\varepsilon-\varepsilon^{2})u^{p-1}_{i}(y)-\frac{C(p)}{\varepsilon^{p-2}}u^{p-1}_{i}(x). \end{align}

Case 3. Consider the case when $p>3$ is not an integer. On one hand, making use of (1.13), we obtain

(1.17)\begin{align} |u_{i}(x)-u_{i}(y)|^{p-2}& \leq(u_{i}(x)+u_{i}(y))^{[p-2]+(p-[p])}\nonumber\\ & \leq\left((1+\varepsilon)u^{[p-2]}_{i}(y)+\frac{C(p)}{\varepsilon^{[p-3]}}u^{[p-2]}_{i}(x)\right)\left(u^{p-[p]}_{i}(y)+u_{i}^{p-[p]}(x)\right)\nonumber\\ & = (1+\varepsilon)u^{p-2}_{i}(y)+\frac{C(p)}{\varepsilon^{[p-3]}}u^{p-[p]}_{i}(y)u^{[p-2]}_{i}(x)\nonumber\\ & \quad+(1+\varepsilon)u^{[p-2]}_{i}(y)u^{p-[p]}_{i}(x)+\frac{C(p)}{\varepsilon^{[p-3]}}u^{p-2}_{i}(x). \end{align}

From Young's inequality, we deduce

(1.18)\begin{align} & \frac{C(p)}{\varepsilon^{[p-3]}}u^{p-[p]}_{i}(y)u^{[p-2]}_{i}(x)\leq\varepsilon u^{p-2}_{i}(y)+\frac{C(p)}{\varepsilon^{p-3}}u^{p-2}_{i}(x), \end{align}

(1.19)\begin{align} & (1+\varepsilon)u^{[p-2]}_{i}(y)u^{p-[p]}_{i}(x)\leq\varepsilon u^{p-2}_{i}(y)+\frac{C(p)}{\varepsilon^{\frac{[p-2]}{p-[p]}}}u^{p-2}_{i}(x). \end{align}

Substituting (1.18)–(1.19) into (1.17), it follows that

(1.20)\begin{equation} |u_{i}(x)-u_{i}(y)|^{p-2}-u^{p-2}_{i}(y)\leq3\varepsilon u^{p-2}_{i}(y)+\frac{C(p)}{\varepsilon^{\frac{[p-2]}{p-[p]}}}u^{p-2}_{i}(x). \end{equation}

On the other hand, using (1.13) again, we have

(1.21)\begin{align} u_{i}^{p-2}(y)& \leq(|u_{i}(y)-u_{i}(x)|+u_{i}(x))^{[p-2]+(p-[p])}\nonumber\\ & \leq\left((1+\varepsilon)|u_{i}(x)-u_{i}(y)|^{[p-2]}+\frac{C(p)}{\varepsilon^{[p-3]}}u^{[p-2]}_{i}(x)\right)\nonumber\\ & \quad\cdot\left(|u_{i}(x)-u_{i}(y)|^{p-[p]}+u^{p-[p]}_{i}(x)\right)\nonumber\\ & = (1+\varepsilon)|u_{i}(x)-u_{i}(y)|^{p-2}+\frac{C(p)}{\varepsilon^{[p-3]}}u^{[p-2]}_{i}(x)|u_{i}(x)-u_{i}(y)|^{p-[p]}\nonumber\\ & \quad+(1+\varepsilon)u^{p-[p]}_{i}(x)|u_{i}(x)-u_{i}(y)|^{[p-2]}+\frac{C(p)}{\varepsilon^{[p-3]}}u^{p-2}_{i}(x). \end{align}

It follows from Young's inequality that

(1.22)\begin{align} & \frac{C(p)}{\varepsilon^{[p-3]}}u^{[p-2]}_{i}(x)|u_{i}(x)-u_{i}(y)|^{p-[p]}\leq\varepsilon|u_{i}(x)-u_{i}(y)|^{p-2}+\frac{C(p)}{\varepsilon^{p-3}}u^{p-2}_{i}(x), \end{align}

(1.23)\begin{align} & (1+\varepsilon)u^{p-[p]}_{i}(x)|u_{i}(x)-u_{i}(y)|^{[p-2]}\leq\varepsilon|u_{i}(x)-u_{i}(y)|^{p-2}+\frac{C(p)}{\varepsilon^{\frac{[p-2]}{p-[p]}}}u^{p-2}_{i}(x). \end{align}

Combining (1.20)–(1.23), we deduce

\begin{align*} u_{i}^{p-2}(y)-|u_{i}(x)-u_{i}(y)|^{p-2}& \leq3\varepsilon|u_{i}(x)-u_{i}(y)|^{p-2}+\frac{C(p)}{\varepsilon^{\frac{[p-2]}{p-[p]}}}u^{p-2}_{i}(x)\nonumber\\ & \leq 3\varepsilon(1+3\varepsilon)u^{p-2}_{i}(y)+\frac{C(p)}{\varepsilon^{\frac{[p-2]}{p-[p]}}}u^{p-2}_{i}(x). \end{align*}

This, together with (1.20) again, gives that

(1.24)\begin{equation} \left|u_{i}^{p-2}(y)-|u_{i}(x)-u_{i}(y)|^{p-2}\right|\leq3\varepsilon(1+3\varepsilon)u^{p-2}_{i}(y)+\frac{C(p)}{\varepsilon^{\frac{[p-2]}{p-[p]}}}u^{p-2}_{i}(x). \end{equation}

In light of (1.24), it follows from Young's inequality that

(1.25)\begin{align} |\mathcal{K}_{2}|& \leq3\varepsilon(1+3\varepsilon)u^{p-2}_{i}(y)u_{i}(x)+\frac{C(p)}{\varepsilon^{\frac{[p-2]}{p-[p]}}}u^{p-1}_{i}(x)\nonumber\\ & \leq\varepsilon u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{\frac{[p-2]}{p-[p]}}}u^{p-1}_{i}(x),\end{align}

(1.26)\begin{align} |\mathcal{K}_{3}|& \leq3\varepsilon(1+3\varepsilon)u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{\frac{[p-2]}{p-[p]}}}u^{p-2}_{i}(x)u_{i}(y)\nonumber\\ & \leq\varepsilon(4+9\varepsilon) u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{\frac{[p-1]}{p-[p]}}}u^{p-1}_{i}(x). \end{align}

Therefore, substituting (1.10) and (1.25)–(1.26) into (1.9), we derive

(1.27)\begin{align} \Theta& \leq(1+5\varepsilon+9\varepsilon^{2})u^{p-1}_{i}(y)+\frac{C(p)}{\varepsilon^{\frac{[p-1]}{p-[p]}}}u^{p-1}_{i}(x), \end{align}

(1.28)\begin{align} \Theta& \geq(1-6\varepsilon-9\varepsilon^{2})u^{p-1}_{i}(y)-\frac{C(p)}{\varepsilon^{\frac{[p-1]}{p-[p]}}}u^{p-1}_{i}(x). \end{align}

Observe that

(1.29)\begin{align} \lim_{R\rightarrow\infty}\lim_{i\rightarrow\infty}\int_{B_{R}^{c}(0)}\frac{u_{i}^{p-1}(x)}{|x-y|^{n+\sigma p}}\,{\rm d}y& =u^{p-1}(x)\lim_{R\rightarrow\infty}\int_{B_{R}^{c}(0)}\frac{{\rm d}y}{|x-y|^{n+\sigma p}}\nonumber\\ & \leq u^{p-1}(x)\lim_{R\rightarrow\infty}\int_{B_{R-|x|}^{c}(x)}\frac{{\rm d}y}{|x-y|^{n+\sigma p}}=0. \end{align}

Since $\lim \limits _{R\rightarrow \infty }\lim \limits _{i\rightarrow \infty }\Psi _{i}(x,R)$ exists and is finite, it follows from (1.12), (1.15)–(1.16) and (1.27)–(1.29) that

\begin{align*} \lim_{R\rightarrow\infty}\lim_{i\rightarrow\infty}\Psi_{i}(x,R)& \leq c_{n,\sigma p}(1+\varepsilon_{p}^{(1)})\liminf_{R\rightarrow\infty}\liminf_{i\rightarrow\infty}\int_{B^{c}_{R}(0)}\frac{u^{p-1}_{i}(y)}{|x-y|^{n+\sigma p}}\,{\rm d}y,\\ \lim_{R\rightarrow\infty}\lim_{i\rightarrow\infty}\Psi_{i}(x,R)& \geq c_{n,\sigma p}(1-\varepsilon_{p}^{(2)})\limsup_{R\rightarrow\infty}\limsup_{i\rightarrow\infty}\int_{B^{c}_{R}(0)}\frac{u^{p-1}_{i}(y)}{|x-y|^{n+\sigma p}}\,{\rm d}y, \end{align*}

where

\begin{align*} \varepsilon_{p}^{(1)}& =\begin{cases} \varepsilon, & \text{if}\,2< p\leq3,\\ \varepsilon(3+\varepsilon), & \text{if}\,p>3\;\text{is an integer},\\ \varepsilon(5+9\varepsilon), & \text{if}\,p>3\;\text{is not an integer}, \end{cases}\\ \varepsilon^{(2)}_{p}& = \begin{cases} 2\varepsilon, & \text{if }\,2< p\leq3,\\ \varepsilon(4+\varepsilon), & \text{if }\,p>3\;\text{is an integer},\\ 3\varepsilon(2+3\varepsilon), & \text{if }\,p>3\;\text{is not an integer}. \end{cases} \end{align*}

Due to the fact that $R > > |x|+1$, we have

\[ \frac{(R-|x|)|y|}{R}\leq|y-x|\leq\frac{(R+|x|)|y|}{R},\quad\text{for }y\in B_{R}^{c}(0). \]

Hence, we deduce

\begin{align*} \lim_{R\rightarrow\infty}\lim_{i\rightarrow\infty}\Psi_{i}(x,R)& \leq c_{n,\sigma p}(1+\varepsilon_{p}^{(1)})\liminf_{R\rightarrow\infty}\liminf_{i\rightarrow\infty}\int_{B^{c}_{R}(0)}\frac{u^{p-1}_{i}(y)}{|y|^{n+\sigma p}}\,{\rm d}y,\\ \lim_{R\rightarrow\infty}\lim_{i\rightarrow\infty}\Psi_{i}(x,R)& \geq c_{n,\sigma p}(1-\varepsilon_{p}^{(2)})\limsup_{R\rightarrow\infty}\limsup_{i\rightarrow\infty}\int_{B^{c}_{R}(0)}\frac{u^{p-1}_{i}(y)}{|y|^{n+\sigma p}}\,{\rm d}y. \end{align*}

By virtue of the arbitrariness of $\varepsilon$ and $\{u_{i}\}$ is nonnegative, we obtain

\[ \lim_{R\rightarrow\infty}\lim_{i\rightarrow\infty}\Psi_{i}(x,R)=c_{n,\sigma p}\lim_{R\rightarrow\infty}\lim_{i\rightarrow\infty}\int_{B^{c}_{R}(0)}\frac{u^{p-1}_{i}(y)}{|y|^{n+\sigma p}}\,{\rm d}y\geq0. \]

This, together with (1.8), yields that theorem 1.1 holds.

Proof. It is easily seen from (1.31) that $v_{j}\in C_{c}^{\infty }(\mathbb {R}^{n})$, $v_{j}\geq 0$ in $\mathbb {R}^{n}$, $v_{j}=1$ in $B_{R_{j}}$, and $\|v_{j}-1\|_{C^{2}_{loc}}\rightarrow 0$, as $i\rightarrow \infty$. A direct computation gives that

(1.33)\begin{equation} (-\Delta)^{\sigma}_{p}v_{j}(x)=j^{{-}s(p-1)}R_{j}^{-\sigma p}(-\Delta)^{\sigma}_{p}w_{j}(R_{j}^{{-}1}x),\quad\text{for $x\in B_{R_{j}}$}. \end{equation}

For any fixed $x\in \mathbb {R}^{n}$, we have

\begin{align*} & j^{{-}t(p-1)}(-\Delta)^{\sigma}_{p}w_{j}(R_{j}^{{-}1}x)\\ & ={-}c_{n,\sigma p}\int_{B_{4}\setminus B_{3}}\frac{\phi^{p-1}(y)}{|R_{j}^{{-}1}x-y|^{n+\sigma p}}\,{\rm d}y-c_{n,\sigma p}\int_{B_{6}\setminus B_{4}}\frac{{\rm d}y}{|R_{j}^{{-}1}x-y|^{n+\sigma p}}\\ & \quad+\frac{c_{n,\sigma p}}{j^{t-s}}\int_{B_{6}^{c}}\frac{\psi(y)|\psi(y)-1+\psi(y)j^{-(t-s)}|^{p-2}}{|R_{j}^{{-}1}x-y|^{n+\sigma p}}\,{\rm d}y\\ & \quad-c_{n,\sigma p}\int_{B_{6}^{c}}\frac{(1-\psi(y))|\psi(y)-1+\psi(y)j^{-(t-s)}|^{p-2}}{|R_{j}^{{-}1}x-y|^{n+\sigma p}}\,{\rm d}y\\ & \quad\rightarrow-\beta,\quad\text{as}\ j\ {\rm goes\ to}\ \infty, \end{align*}

where $\beta$ is defined by (1.32). This, together with (1.33), gives that

\[ \lim_{j\rightarrow\infty}(-\Delta)^{\sigma}_{p}v_{j}(x)=\beta^{{-}1}\lim_{j\rightarrow\infty}j^{{-}t(p-1)}(-\Delta)^{\sigma}_{p}w_{j}(R_{j}^{{-}1}x)={-}1,\quad\text{in $\mathbb{R}^{n}$}. \]

The proof is complete.

Proof. Let $\eta$ and $\phi$ be defined in (1.30) and (1.31). For $q\in \mathbb {R}$ and $s>-\frac {\sigma p}{p-1}$, let

\[ u_{j}(x)=\begin{cases} j+j^{q}\phi(x), & \mathrm{in}\;B_{R},\\ (1-\varphi(x))(j+j^{q})+\varphi(x)|x|^{{-}s}, & \mathrm{in}\;B_{R}^{c}, \end{cases} \]

where $\varphi (x)=\eta (|x|-R)$ and $R=R(n,p,q,\sigma,s,j)>9$ is a sufficiently large constant to be determined later. Then $u_{j}\in C^{\infty }(\mathbb {R}^{n})\cap \mathcal {L}_{\sigma p}(\mathbb {R}^{n})$ and $u_{j}>0$ in $\mathbb {R}^{n}$. Denote

\[ K_{j}(x):=\frac{(-\Delta)^{\sigma}_{p}u_{j}(x)}{u_{j}^{q(p-1)}(x)},\quad\text{in}\ \mathbb{R}^{n}. \]

Then $K_{j}\in C^{\infty }(\mathbb {R}^{n})$. Moreover, $\{K_{j}\}$ satisfies the following properties: there exists four positive constants $C_{i}:=C_{i}(n,\sigma,p)$, $i=1,2,3,4$, such that for every $j\geq 1$,

1. (K1) $-C_{1}\leq K_{j}(x)\leq -C_{2}$, and $\sum \limits ^{3}_{i=1}|\nabla ^{i}K_{j}(x)|\leq C_{3}$ in $B_{2}$;

2. (K2) $\nabla ^{2}K_{j}(0)\leq -C_{4}\mathbf {I}_{n}$, where $\mathbf {I}_{n}$ denotes $n\times n$ identity matrix.

We first prove $(\mathbf {K1})$. Observe that

(2.2)\begin{align} c_{n,\sigma p}^{{-}1}K_{j}(x)& ={-}\int_{B_{4}\setminus B_{3}}\frac{\phi^{p-1}(y)}{|x-y|^{n+\sigma p}}\,{\rm d}y-\int_{B_{R}\setminus B_{4}}\frac{{\rm d}y}{|x-y|^{n+\sigma p}}\nonumber\\ & \quad+\int_{B_{R}^{c}}\frac{|\mathcal{A}_{\varphi}(y)|^{p-2}\mathcal{A}_{\varphi}(y)}{|x-y|^{n+\sigma p}}\,{\rm d}y:=\sum^{3}_{i=1}J_{i}, \end{align}

where $\mathcal {A}_{\varphi }(y):=\varphi (y)-1+j^{1-q}\varphi (y)-j^{-q}|y|^{-s}\varphi (y)$. For simplicity, let

\begin{align*} \gamma& :=\gamma(n,\sigma,p)=\int_{B_{1}^{c}}\frac{{\rm d}y}{|y|^{n+\sigma p}}=\frac{|\mathbb{S}^{n-1}|}{\sigma p},\\ \tau:=& \tau(n,\sigma,p,s)=\int_{B_{1}^{c}}\frac{{\rm d}y}{|y|^{n+\sigma p+s(p-1)}}=\frac{|\mathbb{S}^{n-1}|}{\sigma p+s(p-1)}. \end{align*}

A straightforward computation yields that

\[ 0\geq J_{1}\geq{-}\int_{B_{1}(x)^{c}}\frac{{\rm d}y}{|x-y|^{n+\sigma p}}={-}\gamma, \]

and

\[ -\gamma\leq J_{2}\leq{-}\int_{B_{R-2}\setminus B_{6}}\frac{{\rm d}y}{|y|^{n+\sigma p}}={-}(6^{-\sigma p}-(R-2)^{-\sigma p})\gamma. \]

For $x\in B_{2}$, $y\in B_{R}^{c}$, we have $|x-y|\geq |y|/2$ in virtue of $R>9$. Then

\begin{align*} |J_{3}|& \leq2^{(\sigma+1)p+n-2}\int_{B_{R}^{c}}\frac{(1+j^{1-q})^{p-1}+j^{{-}q(p-1)}|y|^{{-}s(p-1)}}{|y|^{n+\sigma p}}\,{\rm d}y\\ & = 2^{(\sigma+1)p+n-2}\gamma(1+j^{1-q})^{p-1}R^{-\sigma p}+2^{(\sigma+1)p+n-2}\tau j^{{-}q(p-1)}R^{-\sigma p-s(p-1)}. \end{align*}

For a sufficiently large $R>9$, we have

\[ (R-2)^{-\sigma p}\gamma+2^{(\sigma+1)p+n-2}R^{-\sigma p}\left(\gamma(1+j^{1-q})^{p-1}+\tau j^{{-}q(p-1)}R^{{-}s(p-1)}\right)\leq\frac{\gamma6^{-\sigma p}}{2}, \]

which implies that

\[{-}3c_{n,\sigma p}\gamma\leq K_{j}(x)\leq{-}\frac{c_{n,\sigma p}\gamma6^{-\sigma p}}{2},\quad\forall\;x\in B_{2},\;j\geq1. \]

Furthermore, after differentiating (2.2), it follows from a similar calculation that

\[ \sum^{3}_{i=1}|\nabla^{i}K_{j}(x)|\leq C(n,\sigma,p),\quad\text{for}\ x\in B_{2},\;j\geq1. \]

We proceed to verify property $(\bf {K2})$. A simple calculation shows that for $y\in B_{3}^{c}$,

(2.3)\begin{equation} \partial_{x_{k}x_{l}}^{2}\left(\frac{1}{|x-y|^{n+\sigma p}}\right)(0)=\frac{(n+\sigma p)[(n+\sigma p+2)y_{k}y_{l}-\delta_{kl}|y|^{2}]}{|y|^{n+\sigma p+4}}. \end{equation}

Since the integral domain is symmetric, then we see from (2.2) to (2.3) that

\[ \partial^{2}_{x_{k}x_{l}}K_{j}(0)=0,\quad\text{for}\ k\neq l. \]

If $k=l$, it follows from the radial symmetry of $\phi$ and $\varphi$ that

\begin{align*} & [(n+\sigma p)c_{n,\sigma p}]^{{-}1}\partial^{2}_{x_{k}x_{k}}K_{j}(0)\\ & ={-}\frac{\sigma p+2}{n}\left(\int_{B_{4}\setminus B_{3}}\frac{\phi^{p-1}(y)}{|y|^{n+\sigma p+2}}\,{\rm d}y+\int_{B_{R}^{c}}\frac{|\mathcal{A}_{\varphi}(y)|^{p-2}(j^{1-q}\varphi(y)-\mathcal{A}_{\varphi}(y))}{|y|^{n+\sigma p+2}}\,{\rm d}y\right)\\ & \quad-\frac{\sigma p+2}{n}\left(\int_{B_{R}\setminus B_{4}}\frac{{\rm d}y}{|y|^{n+\sigma p+2}}-j^{1-q}\int_{B_{R}^{c}}\frac{|\mathcal{A}_{\varphi}(y)|^{p-2}\varphi(y)}{|y|^{n+\sigma p+2}}\,{\rm d}y\right)\\ & \leq{-}|B_{1}|\left(4^{-(\sigma p+2)}-R^{-(\sigma p+2)}-3j^{1-q}R^{-(\sigma p+2)}\right)\\ & \leq{-}|B_{1}|4^{-(\sigma p+3)},\quad\text{for a sufficiently large}\ R>9, \end{align*}

where we used the fact that $|\mathcal {A}_{\varphi }(y)|^{p-2}\varphi (y)\leq 3$ in $B_{R}^{c}$. That is, property $(\bf {K2})$ holds.

From the radial symmetry of $u_{j}$ with respect to the origin, we know that $K_{j}$ is also radially symmetric. Then we have

\[ \nabla K_{j}(0)=0, \]

which, together with $(\bf {K1})$–$(\bf {K2})$, leads to that for $j\geq 1$,

(2.4)\begin{equation} |\nabla K_{j}(x)|\geq c_{1},\quad\text{in}\ B_{2\varepsilon_{0}}(4\varepsilon_{0}e_{1}), \end{equation}

where $e_{1}=(1,0,...,0)\in \mathbb {R}^{n}$, $\varepsilon _{0}:=\varepsilon _{0}(n,p,\sigma )\in (0,1/4)$ is a small constant and $c_{1}:=c_{1}(n,p,\sigma )$ is a positive constant.

Define

\[ \bar{u}_{j}(x):=\varepsilon_{0}^{s}u_{j}(\varepsilon_{0}(x+4e_{1})),\quad\mathrm{and}\ \bar{K}_{j}(x):=\varepsilon_{0}^{\sigma p-s(p-1)(q-1)}K_{j}(\varepsilon_{0}(x+4e_{1})). \]

Therefore,

\[ (-\Delta)^{\sigma}_{p}\bar{u}_{j}=\bar{K}_{j}(x)\bar{u}_{j}^{q(p-1)},\quad\text{for}\ x\in\mathbb{R}^{n}. \]

Then combining $(\bf {K1})$ and (2.4), we obtain

\[ -\bar{C}\leq \bar{K}_{j}(x)\leq{-}\bar{c},\;\,\bar{c}\leq|\nabla \bar{K}_{j}(x)|\leq \bar{C},\ \text{and}\ |\nabla^{2}\bar{K}_{j}(x)|\leq \bar{C},\quad\text{in}\ B_{2}, \]

where $\bar {c}=\bar {c}(n,\sigma,p,q,s)$ and $\bar {C}=\bar {C}(n,\sigma,p,q,s)$. Moreover, recalling the definition of $u_{j}$, we have

\[ \lim_{|x|\rightarrow\infty}|x|^{s}\bar{u}_{j}=1,\quad\text{and}\ \min_{\overline{B}_{1}}\bar{u}_{j}=\varepsilon_{0}^{s}j\rightarrow\infty,\quad\text{as}\ j\rightarrow\infty. \]

The proof is finished.