Proof. By $(V_{1})$, lemma 2.1 and the definition of $e(a)$, we have

(3.1)\begin{equation} 0=\inf_{x\in\mathbb{R}^{2}}V(x)\leq e(a)\leq e(0)=\delta_{1}, \quad \text{for }a\in[0, a^{{\ast}}), \end{equation}

where $\delta _{1}$ is the first eigenvalue of $(-\Delta )^{s}+V(x)$ in $E$. Moreover, lemma 2.1 also shows that

(3.2)\begin{equation} \begin{aligned} e(a) & =\int_{\mathbb{R}^{2}}(|(-\Delta)^{\frac{s}{2}}u_{a}|^{2}+V(x)|u_{a}|^{2})\,{\rm d}x-\frac{a}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x\\ & \geq\frac{a^{{\ast}}}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x-\frac{a}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x\\ & =\frac{a^{{\ast}}-a}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x. \end{aligned} \end{equation}

From (3.1)–(3.2), it follows that

(3.3)\begin{equation} \int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x\leq\frac{1+s}{a^{{\ast}}-a}e(a)\leq\frac{1+s}{a^{{\ast}}-a}\delta_{1}\quad \hbox{for}\ a\in[0,\, a^{{\ast}}). \end{equation}

For any $a_{1},\, a_{2}\in [0,\, a^{\ast })$, it is easy to see that

(3.4)\begin{equation} e(a_{1})\geq e(a_{2})+\frac{a_{2}-a_{1}}{1+s}\int_{\mathbb{R}^{2}}|u_{a_{1}}|^{2+2s}\,{\rm d}x,\quad \forall u_{a_{1}}\in K_{a_{1}}, \end{equation}

and

(3.5)\begin{equation} e(a_{2})\geq e(a_{1})+\frac{a_{1}-a_{2}}{1+s}\int_{\mathbb{R}^{2}}|u_{a_{2}}|^{2+2s}\,{\rm d}x,\quad \forall u_{a_{2}}\in K_{a_{2}}. \end{equation}

By (3.4)–(3.5), we get

\[ \lim_{a_{2}\rightarrow a_{1}}e(a_{2})=e(a_{1}), \]

which implies that

(3.6)\begin{equation} e(a)\in C([0, a^{{\ast}}), \mathbb{R}^+). \end{equation}

By using (3.4)–(3.5), for $u_{a_{1}}\in K_{a_{1}}$ and $u_{a_{2}}\in K_{a_{2}}$, we have

(3.7)\begin{equation} \frac{a_{2}-a_{1}}{1+s}\int_{\mathbb{R}^{2}}|u_{a_{1}}|^{2+2s}\,{\rm d}x\leq e(a_{1})-e(a_{2})\leq\frac{a_{2}-a_{1}}{1+s}\int_{\mathbb{R}^{2}}|u_{a_{2}}|^{2+2s}\,{\rm d}x. \end{equation}

Without loss of generality, we set $0< a_{1}< a_{2}< a^{\ast }$, from (3.7), we can see that

(3.8)\begin{equation} -\frac{1}{1+s}\int_{\mathbb{R}^{2}}|u_{a_{2}}|^{2+2s}\,{\rm d}x\leq \frac{e(a_{2})-e(a_{1})}{a_{2}-a_{1}}\leq{-}\frac{1}{1+s}\int_{\mathbb{R}^{2}}|u_{a_{1}}|^{2+2s}\,{\rm d}x, \end{equation}

for $\forall u_{a_{i}}\in K_{a_{i}}$, $i=1,\, 2$. This shows that

(3.9)\begin{equation} -\frac{1}{1+s}\inf_{u_{a_{2}}\in K_{a_{2}}}\int_{\mathbb{R}^{2}}|u_{a_{2}}|^{2+2s}\,{\rm d}x\leq\frac{e(a_{2})-e(a_{1})}{a_{2}-a_{1}}\leq{-}\frac{1}{1+s}\beta_{a_{1}}. \end{equation}

By (3.3) and lemma 2.3, we know that $\{u_{a_{2}}\}$ is bounded in $E$. Up to a subsequence, we may assume that there exists $u\in E$ such that

\[ \left\{ \begin{array}{@{}ll@{}} u_{a_{2}}\rightharpoonup u, & \text{in }E \text{ as }a_{2}\searrow a_{1},\\ u_{a_{2}}\rightarrow u, & \text{in }L^{q}(\mathbb{R}^{2})\text{ for }q\in[2, 2_{s}^{{\ast}}). \end{array} \right. \]

By (3.6), we deduce that

\[ e(a_{1})=\lim_{a_{2}\searrow a_{1}}e(a_{2})=\lim_{a_{2}\searrow a_{1}}J_{a_{2}}(u_{a_{2}})\geq J_{a_{1}}(u)\geq e(a_{1}), \]

which implies that

\[ u_{a_{2}}\rightarrow u\in E \quad \text{and}\quad u\in K_{a_{1}}. \]

Thus, (3.9) shows that

(3.10)\begin{equation} \begin{aligned} & -\frac{1}{1+s}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x\leq\liminf_{a_{2}\searrow a_{1}}\frac{e(a_{2})-e(a_{1})}{a_{2}-a_{1}}\\ & \leq\limsup_{a_{2}\searrow a_{1}}\frac{e(a_{2})-e(a_{1})}{a_{2}-a_{1}}\leq{-}\frac{1}{1+s}\beta_{a_{1}}. \end{aligned} \end{equation}

Moreover, by the definition of $\beta _{a}$, we get

(3.11)\begin{equation} \int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x\leq\beta_{a_{1}}. \end{equation}

In view of (3.10)–(3.11), we can obtain that

\[ e'_+(a_{1})={-}\frac{1}{1+s}\beta_{a_{1}}. \]

Similarly, we also have

\[ e'_{-}(a_{1})={-}\frac{1}{1+s}\alpha_{a_{1}}. \]

Proof of theorem 1.1 By using (3.7), we can see that

\begin{align*} |e(a_{1})-e(a_{2})|& \leq\frac{1}{1+s}|a_{2}-a_{1}|\max\left\{\int_{\mathbb{R}^{2}}|u_{a_{1}}|^{2+2s}\,{\rm d}x, \int_{\mathbb{R}^{2}}|u_{a_{2}}|^{2+2s}\,{\rm d}x\right\}\\ & \leq\frac{1}{1+s}\max\{\beta_{a_{1}}, \beta_{a_{2}}\}|a_{2}-a_{1}|,\quad \forall a_{1}, a_{2}\in[0, a^{{\ast}}), \end{align*}

which implies that $e(a)$ is locally Lipschitz continuous in $[0,\, a^{\ast })$. Thus, by using Rademacher's theorem, we know that $e(a)$ is differentiable for a.e. $a\in [0,\, a^{\ast })$. Moreover, lemma 3.1 shows that $e'(a)$ exists for all $a\in [0,\, a^{\ast \ast })$ and a.e. $a\in [a^{\ast \ast },\, a^{\ast })$ and

(3.12)\begin{equation} e'(a)={-}\frac{1}{1+s}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x, \quad \forall u\in K_{a}. \end{equation}

Thus, we know that all minimizers of $e(a)$ have the same $L^{2+2s}(\mathbb {R}^{2})$-norm. For $a\in [0,\, a^{\ast })$, taking each $u_{a}\in K_{a}$ such that $e'(a)$ satisfies (3.12), then $u_{a}$ satisfies (1.1) for some Lagrange multiplier $\lambda _{a}\in \mathbb {R}$. By (1.1) and (3.12), we have

(3.13)\begin{equation} \lambda_{a}=e(a)-\frac{sa}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x=e(a)+sae'(a), \end{equation}

which implies that all minimizers of $e(a)$ satisfy equation (1.1) with the same Lagrange multiplier $\lambda _{a}$, that is,

(3.14)\begin{equation} (-\Delta)^{s}u_{a}+V(x)u_{a}=\lambda_{a}u_{a}+au^{2s+1}_{a},\quad \text{in }\mathbb{R}^2 \end{equation}

For $\forall \varphi \in E$, by (1.2) and (3.14), we deduce that

\begin{align*} \langle I'_{\lambda_{a}, a}(u_{a}), \varphi\rangle& =\int_{\mathbb{R}^{2}}(-\Delta)^{\frac{s}{2}}u_{a}(-\Delta)^{\frac{s}{2}}\varphi \,{\rm d}x+\int_{\mathbb{R}^{2}}(V(x)-\lambda_{a})u_{a}\varphi \,{\rm d}x\\ & \quad -a\int_{\mathbb{R}^{2}}u_{a}^{1+2s}\varphi \,{\rm d}x=0, \end{align*}

which implies that $u_{a}\in S_{\lambda _{a},\, a}$. Now, we show that $G_{\lambda _{a},\, a}$ is nonempty. In fact, according to the definition of $G_{\lambda _{a},\, a}$, we only need to show that for any $u\in S_{\lambda _{a},\, a}$ one has

(3.15)\begin{equation} I_{\lambda_{a}, a}(u)\geq I_{\lambda_{a}, a}(u_{a}). \end{equation}

Let $u_{a}\in K_{a}$ and $u\in S_{\lambda _{a},\, a}$, we have

(3.16)\begin{equation} \int_{\mathbb{R}^{2}}|(-\Delta)^{\frac{s}{2}}u|^{2}\,{\rm d}x+\int_{\mathbb{R}^{2}}(V(x)-\lambda_{a})|u|^{2}\,{\rm d}x=a\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x, \end{equation}

which implies that

(3.17)\begin{equation} I_{\lambda_{a}, a}(u)=\frac{a}{2}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x-\frac{a}{2+2s}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x=\frac{sa}{2+2s}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x>0. \end{equation}

Similarly, we also have

(3.18)\begin{equation} I_{\lambda_{a}, a}(u_{a})=\frac{sa}{2+2s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x. \end{equation}

Define

\[ \hat{u}=\frac{1}{\sqrt{\sigma}}u \quad \hbox{with}\quad \sigma=\int_{\mathbb{R}^{2}}|u|^{2}\,{\rm d}x. \]

Clearly, $\int _{\mathbb {R}^{2}}|\hat {u}|^{2}\,{\rm d}x=1$. Thus, we get that $J_{a}(\hat {u})\geq J_{a}(u_{a})$. This shows that

(3.19)\begin{equation} I_{\lambda_{a}, a}(\hat{u})=\frac{1}{2}J_{a}(\hat{u})-\frac{\lambda_{a}}{2}\int_{\mathbb{R}^{2}}|\hat{u}|^{2}\,{\rm d}x\geq\frac{1}{2}J_{a}(u_{a})-\frac{\lambda_{a}}{2}\int_{\mathbb{R}^{2}}|u_{a}|^{2}\,{\rm d}x=I_{\lambda_{a}, a}(u_{a}). \end{equation}

Moreover, by using (3.16), we deduce that

(3.20)\begin{align} I_{\lambda_{a}, a}(\hat{u}) & =\frac{1}{2\sigma}\int_{\mathbb{R}^{2}}[|(-\Delta)^{\frac{s}{2}}u|^{2}+(V(x)-\lambda_{a})|u|^{2}]\,{\rm d}x\nonumber\\ & \quad-\frac{a}{(2+2s)\sigma^{1+s}}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x\nonumber\\ & =\frac{a}{2+2s}\frac{1}{\sigma}\left(1+s-\frac{1}{\sigma^{s}}\right)\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x. \end{align}

Let $g(\sigma ):=\frac {1}{\sigma }(1+s-\frac {1}{\sigma ^{s}})$. It is easy to check that $g(\sigma )$ achieves the unique global maximum at $\sigma =1$. By (3.17), (3.19) and (3.20), we deduce that

\begin{align*} I_{\lambda_{a}, a}(u_{a})\leq I_{\lambda_{a}, a}(\hat{u})& =\frac{a}{2+2s}g(\sigma)\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x\\ & \leq\frac{a}{2+2s}g(1)\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x\\ & =\frac{sa}{2+2s}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x=I_{\lambda_{a}, a}(u), \end{align*}

which shows that (3.15) holds. Hence, $G_{\lambda _{a},\, a}$ is nonempty.

Next, we prove that $K_{a}=G_{\lambda _{a},\, a}$. For any given $a\in [0,\, a^{\ast })$, consider any $u_{a}\in K_{a}$ and $u\in G_{\lambda _{a},\, a}$. Clearly, $u\in S_{\lambda _{a},\, a}$. From (3.17)–(3.20), it follows that

(3.21)\begin{align} & \frac{sa}{2+2s}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x=I_{\lambda_{a}, a}(u)\leq I_{\lambda_{a}, a}(u_{a})\nonumber\\ & \leq I_{\lambda_{a}, a}(\hat{u})=\frac{a}{2+2s}\frac{1}{\sigma}\left(1+s-\frac{1}{\sigma^{s}}\right)\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x, \end{align}

which implies that

(3.22)\begin{equation} \sigma^{1+s}-\frac{1+s}{s}\sigma^{s}+\frac{1}{s}\leq0. \end{equation}

Define $h(\sigma ):=\sigma ^{1+s}-\frac {1+s}{s}\sigma ^{s}+\frac {1}{s}$. Taking the derivative of $h(\sigma )$, we have

(3.23)\begin{equation} \left\{ \begin{array}{@{}ll@{}} h'(\sigma)<0, & 0<\sigma<1, \\ h'(\sigma)=0, & \sigma=1, \\ h'(\sigma)>0, & \sigma>1. \end{array} \right. \end{equation}

By (3.22)–(3.23), we know that $\sigma =1$, that is, $\int _{\mathbb {R}^{2}}|u|^{2}\,{\rm d}x=1$. Therefore, (3.21) shows that

\[ I_{\lambda_{a}, a}(u)=I_{\lambda_{a}, a}(u_{a}) \quad \hbox{and}\quad J_{a}(u)=J_{a}(u_{a}). \]

This implies that $u\in K_{a}$ and $u_{a}\in G_{\lambda _{a},\, a}$.

Proof. For any $\kappa >0$ and $u\in E$ with $\|u\|_{2}^{2}=1$, by lemma 2.1, we have

(4.2)\begin{align} J_{a}(u) & \geq\int_{\mathbb{R}^{2}}(|x|-M)^{2}|u|^{2}\,{\rm d}x+\frac{a^{{\ast}}-a}{1+s}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x\nonumber\\ & =\kappa+\int_{\mathbb{R}^{2}}[(|x|-M)^{2}-\kappa]|u|^{2}\,{\rm d}x+\frac{a^{{\ast}}-a}{1+s}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x \nonumber\\ & \geq\kappa-\frac{s}{(1+s)(a^{{\ast}}-a)^{\frac{1}{s}}}\int_{\mathbb{R}^{2}}[\kappa-(|x|-M)^{2}]^{1+\frac{1}{s}}_+\,{\rm d}x, \end{align}

where $[\cdot ]_+=\max \{0,\, \cdot \}$ denotes the positive part. Taking the variable $r=M+\sqrt {\kappa }\sin \theta$ with $-\frac {\pi }{2}\leq \theta \leq \frac {\pi }{2}$, by direct computation, we deduce that

(4.3)\begin{align} & \int_{\mathbb{R}^{2}}[\kappa-(|x|-M)^{2}]^{1+\frac{1}{s}}_+\,{\rm d}x=2\pi\int_{M-\sqrt{\kappa}}^{M+\sqrt{\kappa}}[\kappa-(r-M)^{2}]^{1+\frac{1}{s}}rdr\nonumber\\ & =2\pi\kappa^{1+\frac{1}{s}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\cos\theta)^{2+\frac{2}{s}}(M+\sqrt{\kappa}\sin\theta)\sqrt{\kappa}\cos\theta d\theta\leq C\kappa^{\frac{2+3s}{2s}}, \end{align}

for $\kappa >0$ small enough. Taking $\kappa =(\frac {a^{\ast }-a}{2C^{s}})^{\frac {2}{2+s}}$ in (4.2) and (4.3), we have

(4.4)\begin{align} J_{a}(u) & \geq\left(\frac{a^{{\ast}}-a}{2C^{s}}\right)^{\frac{2}{2+s}}-\frac{sC}{(1+s)(a^{{\ast}}-a)^{\frac{1}{s}}}\left(\frac{a^{{\ast}}-a}{2C^{s}}\right)^{\frac{2+3s}{s(2+s)}}\nonumber\\ & \geq(a^{{\ast}}-a)^{\frac{2}{2+s}}\frac{1}{2^{\frac{2}{2+s}}C^{\frac{2s}{2+s}}}\left(1-\frac{s}{1+s}\frac{1}{2^{\frac{1}{s}}}\right). \end{align}

Clearly, we know that $1-\frac {s}{1+s}\frac {1}{2^{\frac {1}{s}}}>0$ since $s\in (\frac {1}{2},\, 1)$. Hence, (4.4) shows that

\[ e(a)\geq C_{1}(a^{{\ast}}-a)^{\frac{2}{2+s}},\text{as }a\nearrow a^{{\ast}}. \]

On the other hand, set a cut-off function $\eta \in C_{0}^{\infty }(\mathbb {R}^{2})$ such that $\eta (x)=1$ for $|x|\leq 1$, $\eta (x)=0$ for $|x|\geq 2$, $0\leq \eta \leq 1$ and $|\nabla \eta |\leq 2$. Define

(4.5)\begin{equation} u(x):=A_{R, \tau}\frac{\tau}{\|Q\|_{2}}\eta\left(\frac{x-x_{0}}{R}\right)Q(\tau(x-x_{0})), \end{equation}

where $x_{0}\in \mathbb {R}^{2}$, $R,\, \tau >0$ and $A_{R, \tau }>0$ is chosen so that $\int _{\mathbb {R}^{2}}|u|^{2}\,{\rm d}x=1$. First, we show that $\lim _{R\tau \rightarrow \infty }A_{R, \tau }=1$. In fact, by (4.5) and lemma 2.1, we can see that

(4.6)\begin{equation} \frac{1}{A_{R, \tau}^{2}}=\frac{1}{\|Q\|_{2}^{2}}\int_{\mathbb{R}^{2}}\eta^{2}\left(\frac{x}{R\tau}\right)Q^{2}(x)\,{\rm d}x=1+O((R\tau)^{{-}2-4s}), \text{as }R\tau\rightarrow\infty. \end{equation}

Now, taking $R=1$ in (4.5), from lemma 2.1, lemma 2.2 and (4.6), it follows that

(4.7)\begin{align} & \int_{\mathbb{R}^{2}}|(-\Delta)^{\frac{s}{2}}u|^{2}\,{\rm d}x-\frac{a}{1+s}\int_{\mathbb{R}^{2}}|u|^{2+2s}\,{\rm d}x\nonumber\\ & \leq\frac{\tau^{2s}}{\|Q\|_{2}^{2}}\left[\int_{\mathbb{R}^{2}}|(-\Delta)^{\frac{s}{2}}Q|^{2}\,{\rm d}x-\frac{a}{(1+s)\|Q\|_{2}^{2s}}\int_{\mathbb{R}^{2}}|Q|^{2+2s}\,{\rm d}x+O(\tau^{{-}4s})\right]\notag\\ & =\frac{\tau^{2s}}{(1+s)\|Q\|_{2}^{2}}\left[\left(1-\frac{a}{\|Q\|_{2}^{2s}}\right)\int_{\mathbb{R}^{2}}|Q|^{2+2s}\,{\rm d}x+O(\tau^{{-}4s})\right]. \end{align}

Moreover, from lemma 2.1, by direct computation, we get

(4.8)\begin{equation} \int_{\mathbb{R}^{2}}(|x|-M)^{2}|u|^{2}\,{\rm d}x\leq\frac{C}{\tau^{2}}\int_{\mathbb{R}^{2}}|x|^{2}|Q|^{2}\,{\rm d}x\leq\frac{C}{\tau^{2}}. \end{equation}

In view of (4.7) and (4.8), we can deduce that

(4.9)\begin{equation} e(a)\leq C\tau^{2s}(a^{{\ast}}-a)+\frac{C}{\tau^{2}}+O(\tau^{{-}4s}). \end{equation}

Taking $\tau =(a^{\ast }-a)^{-\frac {1}{2+2s}}$ in (4.9), we get

\[ e(a)\leq C_{2}(a^{{\ast}}-a)^{\frac{1}{1+s}},\text{as }a\nearrow a^{{\ast}}. \]

Proof. From (4.2), it follows that

\[ e(a)=J_{a}(u_{a})\geq\frac{a^{{\ast}}-a}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x, \]

which implies that the upper bounded of (4.10) since lemma 4.1.

Moreover, for any $0< b< a< a^{\ast }$, we have

\[ e(b)\leq J_{b}(u_{a})=e(a)+\frac{a-b}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x. \]

Then, lemma 4.1 shows that

(4.11)\begin{equation} \frac{1}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x\geq\frac{e(b)-e(a)}{a-b}\geq\frac{C_{1}(a^{{\ast}}-b)^{\frac{2}{2+s}}-C_{2}(a^{{\ast}}-a)^{\frac{1}{1+s}}}{a-b}. \end{equation}

Taking $b=a-C_{3}(a^{\ast }-a)^{\frac {2+s}{2+2s}}$ in (4.11), where $C_{3}>0$ is large enough such that $C_{1}C_{3}^{\frac {2}{2+s}}>2C_{2}$. Then, we can see that

\[ \int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x\geq C(a^{{\ast}}-a)^{-\frac{s}{2+2s}}, \]

which implies that the lower bounded of (4.10).

Proof. $(i)$ By lemma 2.1 and lemma 4.1, we deduce that

(4.17)\begin{equation} \int_{\mathbb{R}^{2}}V(x)|u_{a}|^{2}\,{\rm d}x\leq e(a)\leq C_{2}(a^{{\ast}}-a)^{\frac{1}{1+s}}, \text{as }a\nearrow a^{{\ast}}, \end{equation}

and

(4.18)\begin{align} 0 & \leq\int_{\mathbb{R}^{2}}|(-\Delta)^{\frac{s}{2}}u_{a}|^{2}\,{\rm d}x-\frac{a}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x\nonumber\\ & =\epsilon_{a}^{{-}2s}-\frac{a}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x\leq e(a)\stackrel{a\nearrow a^{{\ast}}}{\longrightarrow}0. \end{align}

Lemma 4.2 implies that

(4.19)\begin{equation} \int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x\rightarrow+\infty, \text{ as }a\nearrow a^{{\ast}}. \end{equation}

By (4.18)–(4.19), we have

\[ 0\leq\frac{\epsilon_{a}^{{-}2s}}{\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x}-\frac{a}{1+s}\leq\frac{e(a)}{\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x}\rightarrow0, \text{ as }a\nearrow a^{{\ast}}, \]

that is,

\[ \frac{\epsilon_{a}^{{-}2s}}{\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x}\rightarrow\frac{a^{{\ast}}}{1+s}, \quad \text{as }a\nearrow a^{{\ast}}, \]

which implies that

(4.20)\begin{equation} 0<\frac{1}{m}\epsilon_{a}^{{-}2s}\leq\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x\leq m\epsilon_{a}^{{-}2s}, \quad \text{as }a\nearrow a^{{\ast}}, \end{equation}

where $m=\max \{\frac {2}{a^{\ast }},\, a^{\ast }\}$. Thus, from (4.20) and lemma 4.2, there exist $C_{3},\, C_{4}>0$ such that

(4.21)\begin{equation} C_{3}(a^{{\ast}}-a)^{-\frac{s}{2+2s}}\leq\epsilon_{a}^{{-}2s}\leq C_{4}(a^{{\ast}}-a)^{-\frac{s}{1+s}}, \quad \text{as }a\nearrow a^{{\ast}}, \end{equation}

which implies that $\epsilon _{a}\rightarrow 0$ as $a\nearrow a^{\ast }$.

$(ii)$ Set

(4.22)\begin{equation} \tilde{w}_{a}(x):=\epsilon_{a}u_{a}(\epsilon_{a}x). \end{equation}

By (4.12) and (4.20), we have

(4.23)\begin{equation} \int_{\mathbb{R}^{2}}|(-\Delta)^{\frac{s}{2}}\tilde{w}_{a}|^{2}\,{\rm d}x=\int_{\mathbb{R}^{2}}|\tilde{w}_{a}|^{2}\,{\rm d}x=1,\quad \frac{1}{m}\leq\int_{\mathbb{R}^{2}}|\tilde{w}_{a}|^{2+2s}\,{\rm d}x\leq m. \end{equation}

Next, we show that there exist a sequence $\{y_{\epsilon _{a}}\}\subset \mathbb {R}^{2}$ and $R_{0},\, \eta >0$ such that

(4.24)\begin{equation} \liminf_{\epsilon_{a}\rightarrow0}\int_{B_{R_{0}}(y_{\epsilon_{a}})}|\tilde{w}_{a}|^{2}\,{\rm d}x\geq\eta>0. \end{equation}

Suppose by contradiction, for any $R>0$, there exists a sequence $\{\tilde {w}_{a_{k}}\}$ with $a\nearrow a^{\ast }$ such that

\[ \lim_{k\rightarrow\infty}\sup_{y\in\mathbb{R}^{2}}\int_{B_{R}(y)}|\tilde{w}_{a_{k}}|^{2}\,{\rm d}x=0. \]

From lemma 2.4, we get that $\tilde {w}_{a_{k}}\stackrel {k}{\longrightarrow }0$ in $L^{r}(\mathbb {R}^{2})$ for $2< r<2_{s}^{\ast }$. Hence, $\tilde {w}_{a_{k}}\stackrel {k}{\longrightarrow }0$ in $L^{2+2s}(\mathbb {R}^{2})$, which contradicts with (4.23). Therefore, from (4.22) and (4.24), we have

\[ \liminf_{a\nearrow a^{{\ast}}}\int_{B_{R_{0}}(0)}|w_{a}|^{2}\,{\rm d}x\geq\eta>0. \]

$(iii)$ From (4.13) and (4.17), it follows that

(4.25)\begin{equation} \int_{\mathbb{R}^{2}}(|x|-M)^{2}|u_{a}|^{2}\,{\rm d}x=\int_{\mathbb{R}^{2}}(|\epsilon_{a}x+\epsilon_{a}y_{\epsilon_{a}}|-M)^{2}|w_{a}|^{2}\,{\rm d}x\rightarrow0,\quad \text{as }a\nearrow a^{{\ast}}. \end{equation}

Now, we prove that

(4.26)\begin{equation} \lim_{\epsilon_{a}\rightarrow0}|\epsilon_{a}y_{\epsilon_{a}}|=M. \end{equation}

Indeed, assume by contradiction that there exist a constant $\alpha >0$ and a subsequence $\{a_{n}\}$ with $a_{n}\nearrow a^{\ast }$ as $n\rightarrow \infty$ such that

\[ \epsilon_{n}:=\epsilon_{a_{n}}\rightarrow0, \quad \hbox{and}\quad ||\epsilon_{n}y_{\epsilon_{n}}|-M|\geq\alpha>0,\quad \text{as }n\rightarrow\infty. \]

By (4.14) and Fatou's lemma, we can see that

\[ \lim_{n\rightarrow\infty}\int_{\mathbb{R}^{2}}(|\epsilon_{n}x+\epsilon_{n}y_{\epsilon_{n}}|-M)^{2}|w_{a_{n}}|^{2}\,{\rm d}x\geq\frac{\alpha^{2}}{2}\eta>0, \]

which gives a contradiction by (4.25). Thus, (4.26) shows that $\{\epsilon _{a}y_{\epsilon _{a}}\}$ is bounded uniformly as $\epsilon _{a}\rightarrow 0$ and (4.15) holds true.

Next, we prove that (4.16) holds. Since $u_{a}$ is a nonnegative minimizer of (1.3), we have

(4.27)\begin{equation} (-\Delta)^{s}u_{a}+(|x|-M)^{2}u_{a}=\lambda_{a}u_{a}+au^{2s+1}_{a},\quad \text{in }\mathbb{R}^2, \end{equation}

where $\lambda _{a}\in \mathbb {R}$ is a Lagrange multiplier. Moreover, we also have

(4.28)\begin{equation} \lambda_{a}=e(a)-\frac{sa}{1+s}\int_{\mathbb{R}^{2}}|u_{a}|^{2+2s}\,{\rm d}x. \end{equation}

From (4.20), (4.28) and lemma 4.1, we can see that there exist $C_{5},\, C_{6}>0$, independent of $a,\, s$, such that

\[{-}C_{5}<\epsilon_{a}^{2s}\lambda_{a}<{-}C_{6}<0, \quad \text{as }a\nearrow a^{{\ast}}. \]

By (4.13) and (4.27), we deduce that

(4.29)\begin{equation} (-\Delta)^{s}w_{a}+\epsilon_{a}^{2s}(|\epsilon_{a}x+\epsilon_{a}y_{\epsilon_{a}}|-M)^{2}w_{a}=\epsilon_{a}^{2s}\lambda_{a}w_{a}+aw^{2s+1}_{a},\quad \text{in }\mathbb{R}^2. \end{equation}

Passing if necessary to a subsequence of $\{a_{k}\}$, still denoted by $\{a_{k}\}$, we may assume that

\[ \left\{ \begin{array}{@{}ll@{}} \epsilon_{k}^{2s}\lambda_{a_{k}}\stackrel{k}{\longrightarrow}-\beta_{1}^{2}<0, & \text{for some }\beta_{1}>0,\\ w_{a_{k}}\stackrel{k}{\rightharpoonup}w_{0}\geq0, & \text{in }H^{s}(\mathbb{R}^{2}). \end{array} \right. \]

From the boundedness of $\{\epsilon _{a}y_{\epsilon _{a}}\}$, by passing to the weak limit of (4.29), we get

(4.30)\begin{equation} (-\Delta)^{s}w_{0}={-}\beta_{1}^{2}w_{0}+a^{{\ast}}w^{2s+1}_{0},\quad \text{in }\mathbb{R}^2. \end{equation}

Clearly, (4.14) implies that $w_{0}\not \equiv 0$. Similar argument to the proof of proposition 4.4 in [Reference Teng and Agarwal21], we know that $w_{0}\in C^{1, \alpha }$ for some $\alpha \in (0,\, 1)$. Then, by lemma 3.2 in [Reference Di Nezza, Palatucci and Valdinoci7], we have

\[ (-\Delta)^{s}w_{0}(x)={-}\frac{1}{2}C(2, s)\int_{\mathbb{R}^{2}}\frac{w_{0}(x+y)+w_{0}(x-y)-2w_{0}(x)}{|x-y|^{2+2s}}{\rm d}x{\rm d}y, \quad \forall x\in\mathbb{R}^{2}. \]

Next, we show that $w_{0}>0$. Assume by contradiction that there exists $x_{0}\in \mathbb {R}^{2}$ such that $w_{0}(x_{0})=0$, then we can see that

\[ (-\Delta)^{s}w_{0}(x_{0})={-}\frac{1}{2}C(2, s)\int_{\mathbb{R}^{2}}\frac{w_{0}(x_{0}+y)+w_{0}(x_{0}-y)}{|x_{0}-y|^{2+2s}}{\rm d}x{\rm d}y<0, \]

since $w_{0}\geq 0$ and $w_{0}\not \equiv 0$. However, it is easy to see that

\[ (-\Delta)^{s}w_{0}(x_{0})={-}\beta_{1}^{2}w_{0}(x_{0})+a^{{\ast}}w^{2s+1}_{0}(x_{0})=0, \]

which gives a contradiction. Hence $w_{0}>0$ for all $x\in \mathbb {R}^{2}$. Now, by (4.30) and $Q$ is the unique positive radial solution of (1.7), we can deduce that

(4.31)\begin{equation} w_{0}(x)=\frac{\beta_{1}}{s^{\frac{1}{2s}}\|Q\|_{2}}Q\left(\frac{\beta_{1}}{s^{\frac{1}{2s}}}|x-\bar{y_{0}}|\right), \text{for some }\bar{y_{0}}\in\mathbb{R}^2. \end{equation}

By simple computation, we know that $\|w_{0}\|_{2}^{2}=1$. From the norm preservation, we get that $w_{a_{k}}\stackrel {k}{\longrightarrow }w_{0}$ in $L^{2}(\mathbb {R}^{2})$. Hence, by the boundedness of $\{w_{a_{k}}\}$ in $H^{s}(\mathbb {R}^{2})$, we have

\[ w_{a_{k}}\stackrel{k}{\longrightarrow}w_{0}, \text{ in }L^{p}(\mathbb{R}^{2})\text{ for }p\in[2, 2_{s}^{{\ast}}). \]

Therefore, in view of (4.29) and (4.30), we know that $w_{a_{k}}\stackrel {k}{\longrightarrow }w_{0}$ in $H^{s}(\mathbb {R}^{2})$, and thus (4.16) holds.

Proof. The proof is parallel to lemma 2.4 in [Reference Guo, Zeng and Zhou11], for the reader's convenience, we give a brief proof. From lemma 4.3, we know that $\epsilon _{a_{k}}y_{\epsilon _{a_{k}}}\stackrel {k}{\longrightarrow }x_{0}$ with $|x_{0}|=M>0$. Hence. we get

(4.32)\begin{equation} y_{\epsilon_{a_{k}}}\stackrel{k}{\longrightarrow}\infty, \end{equation}

which implies that

(4.33)\begin{equation} \frac{2x\cdot y_{\epsilon_{a_{k}}}}{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}\stackrel{k}{\longrightarrow}0, \quad \hbox{uniformly for} x\in B_{R}(0). \end{equation}

Without loss of generality, we may assume that $x_{0}=(M,\, 0)$. Then, arg $y_{\epsilon _{a_{k}}}\stackrel {k}{\longrightarrow }0$. By setting $0<\theta <\frac {\pi }{16}$ small enough, we get that

(4.34)\begin{equation} -\theta< arg y_{\epsilon_{a_{k}}}<\theta, \text{as }\epsilon_{a_{k}}\rightarrow0. \end{equation}

Let

(4.35)\begin{align} \Delta^{1}_{\epsilon_{a_{k}}}: & =\left\{x\in B_{R}(0): \sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}\leq\frac{M}{\epsilon_{a_{k}}}\right\}\nonumber\\ & =\left\{x\in B_{R}(0): |x|^{2}\leq\left(\frac{M}{\epsilon_{a_{k}}}\right)^{2}-|y_{\epsilon_{a_{k}}}|^{2}\right\}, \end{align}

and

(4.36)\begin{align} \Delta^{2}_{\epsilon_{a_{k}}}: & =\left\{x\in B_{R}(0): \sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}>\frac{M}{\epsilon_{a_{k}}}\right\}\nonumber\\ & =\left\{x\in B_{R}(0): \left(\frac{M}{\epsilon_{a_{k}}}\right)^{2}-|y_{\epsilon_{a_{k}}}|^{2}<|x|^{2}< R^{2}\right\}. \end{align}

Clearly, $B_{R}(0)=\Delta ^{1}_{\epsilon _{a_{k}}}\cup \Delta ^{2}_{\epsilon _{a_{k}}}$ and $\Delta ^{1}_{\epsilon _{a_{k}}}\cap \Delta ^{2}_{\epsilon _{a_{k}}}=\emptyset$. Next, we consider the following two cases.

Case 1: $|\Delta ^{1}_{\epsilon _{a_{k}}}|\geq \frac {\pi R^{2}}{2}$. It is easy to check that $B_{\frac {R}{\sqrt {2}}}(0)\subset \Delta ^{1}_{\epsilon _{a_{k}}}$. Let

\[ \Delta^{1}:=\left(B_{\frac{R}{\sqrt{2}}}(0)\backslash B_{\frac{R}{2}}(0)\right)\cap\left\{x: \frac{\pi}{2}+2\theta< arg x<\frac{3\pi}{2}-2\theta\right\}\subset\Delta^{1}_{\epsilon_{a_{k}}}. \]

By simple computation, we get

(4.37)\begin{equation} |\Delta^{1}|=\frac{(\pi-4\theta)R^{2}}{8}. \end{equation}

By (4.34), we have

(4.38)\begin{equation} x\cdot y_{\epsilon_{a_{k}}}=|x||y_{\epsilon_{a_{k}}}|\cos\langle x, y_{\epsilon_{a_{k}}}\rangle<0,\text{ for }x\in\Delta^{1}, \end{equation}

and

(4.39)\begin{equation} |\cos\langle x, y_{\epsilon_{a_{k}}}\rangle|>{-}\cos(\frac{\pi}{2}+\theta),\quad \text{ for }x\in\Delta^{1}. \end{equation}

Moreover, by (4.33) and the Taylor expansion, we have

(4.40)\begin{align} & \frac{1}{\epsilon_{a_{k}}^{2}}(|\epsilon_{a_{k}}x+\epsilon_{a_{k}}y_{\epsilon_{a_{k}}}|-M)^{2}=\left(|x+y_{\epsilon_{a_{k}}}|-\frac{M}{\epsilon_{a_{k}}}\right)^{2}\nonumber\\ & \quad=\left|\sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}\sqrt{1+\frac{2x\cdot y_{\epsilon_{a_{k}}}}{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}}-\frac{M}{\epsilon_{a_{k}}}\right|^{2}\nonumber\\ & \quad=\left|\sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}\left(1+\frac{x\cdot y_{\epsilon_{a_{k}}}}{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}+O\left(\frac{1}{|y_{\epsilon_{a_{k}}}|^{2}}\right)\right)-\frac{M}{\epsilon_{a_{k}}}\right|^{2}\nonumber\\ & \quad=\left|\sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}+\frac{x\cdot y_{\epsilon_{a_{k}}}}{\sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}}-\frac{M}{\epsilon_{a_{k}}}+O\left(\frac{1}{|y_{\epsilon_{a_{k}}}|}\right)\right|^{2}. \end{align}

From (4.35), (4.38), (4.39) and (4.40), it follows that

(4.41)\begin{align} & \sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}+\frac{x\cdot y_{\epsilon_{a_{k}}}}{\sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}}-\frac{M}{\epsilon_{a_{k}}}+O\left(\frac{1}{|y_{\epsilon_{a_{k}}}|}\right)\nonumber\\ & \quad\leq\frac{x\cdot y_{\epsilon_{a_{k}}}}{\sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}}+O\left(\frac{1}{|y_{\epsilon_{a_{k}}}|}\right)\leq\frac{x\cdot y_{\epsilon_{a_{k}}}}{2\sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}}\nonumber\\ & \quad\leq\frac{|x||y_{\epsilon_{a_{k}}}|\cos\left(\frac{\pi}{2}+\theta\right)}{2\sqrt{|x|^{2}+|y_{\epsilon_{a_{k}}}|^{2}}}, \text{ for }x\in\Delta^{1}. \end{align}

Hence, by (4.32), (4.39) and (4.40), we have

\[ \frac{1}{\epsilon_{a_{k}}^{2}}(|\epsilon_{a_{k}}x+\epsilon_{a_{k}}y_{\epsilon_{a_{k}}}|-M)^{2}\geq\frac{\cos^{2}(\frac{\pi}{2}+\theta)|x|^{2}}{8},\text{ for }x\in\Delta^{1}, \]

which implies that

\begin{align*} & \lim_{\epsilon_{a_{k}}\rightarrow0}\frac{1}{\epsilon_{a_{k}}^{2}}\int_{B_{R}(0)}(|\epsilon_{a_{k}}x+\epsilon_{a_{k}}y_{\epsilon_{a_{k}}}|-M)^{2}|w_{a_{k}}|^{2}\,{\rm d}x\\ & \quad\geq\lim_{\epsilon_{a_{k}}\rightarrow0}\frac{1}{\epsilon_{a_{k}}^{2}}\int_{\Delta^{1}}(|\epsilon_{a_{k}}x+\epsilon_{a_{k}}y_{\epsilon_{a_{k}}}|-M)^{2}|w_{a_{k}}|^{2}\,{\rm d}x\\ & \quad\geq\frac{\cos^{2}\frac{11\pi}{20}}{8}\int_{\Delta^{1}}|x|^{2}|w_{0}|^{2}\,{\rm d}x:=C(R)>0, \end{align*}

where $\theta =\frac {\pi }{20}$.

Case 2: $|\Delta ^{2}_{\epsilon _{a_{k}}}|\geq \frac {\pi R^{2}}{2}$. Clearly, $B_{0}(R)\backslash B_{\frac {R}{\sqrt {2}}}(0)\subset \Delta ^{2}_{\epsilon _{a_{k}}}$. Set

\[ \Delta^{2}:=\Bigg(B_{R}(0)\backslash B_{\frac{R}{\sqrt{2}}}(0)\Bigg)\cap\Bigg\{x: -\frac{\pi}{2}+2\theta< arg x<\frac{\pi}{2}-2\theta\Bigg\}\subset\Delta^{2}_{\epsilon_{a_{k}}}. \]

The rest of the proof is very similar to the case 1, we omit it.

Proof. By lemma 4.1, it suffices to prove that there exists a $C>0,$ independent of $a,\, s,$ such that

(4.42)\begin{equation} e(a)\geq C(a^{{\ast}}-a)^{\frac{1}{1+s}},\text{ as }a\nearrow a^{{\ast}}. \end{equation}

From lemma 4.3, we know that for any sequence $\{a_{k}\}$ with $a_{k}\nearrow a^{\ast }$, there exists a convergent subsequence, still denoted by $\{a_{k}\}$, such that $w_{a_{k}}\rightarrow w_{0}>0$ in $L^{2+2s}(\mathbb {R}^{2})$, where $w_{0}$ satisfies (4.31). Thus, there exists $M_{1}>0$, independent of $a_{k},\, s$, such that

(4.43)\begin{equation} \int_{\mathbb{R}^{2}}|u_{a_{k}}|^{2+2s}\,{\rm d}x\geq M_{1},\text{ as }a_{k}\nearrow a^{{\ast}}. \end{equation}

Lemma 4.4 shows that there exists $M_{2}>0$, independent of $a_{k},\, s$, such that

(4.44)\begin{equation} \int_{B_{1}(0)}(|\epsilon_{k}x+\epsilon_{k}y_{\epsilon_{k}}|-M)^{2}|w_{a_{k}}|^{2}\,{\rm d}x\geq M_{2}\epsilon_{k}^{2},\text{ as }a_{k}\nearrow a^{{\ast}}. \end{equation}

In view of (4.43)–(4.44), we deduce that

\begin{align*} e(a_{k})& =J_{a_{k}}(u_{a_{k}})=\frac{1}{\epsilon_{k}^{2s}}\left[\int_{\mathbb{R}^{2}}|(-\Delta)^{\frac{s}{2}}w_{a_{k}}|^{2}\,{\rm d}x-\frac{a^{{\ast}}}{1+s}\int_{\mathbb{R}^{2}}|w_{a_{k}}|^{2+2s}\,{\rm d}x\right]\\ & \quad+\frac{a^{{\ast}}-a_{k}}{(1+s)\epsilon_{k}^{2s}}\int_{\mathbb{R}^{2}}|w_{a_{k}}|^{2+2s}\,{\rm d}x+\int_{\mathbb{R}^{2}}(|\epsilon_{k}x+\epsilon_{k}y_{\epsilon_{k}}|-M)^{2}|w_{a_{k}}|^{2}\,{\rm d}x\\ & \geq\frac{a^{{\ast}}-a_{k}}{(1+s)\epsilon_{k}^{2s}}M_{1}+M_{2}\epsilon_{k}^{2}\\ & \geq\left\{\frac{M_{1}}{1+s}\left[\frac{sM_{1}}{(1+s)M_{2}}\right]^{-\frac{s}{1+s}}+M_{2}\left[\frac{sM_{1}}{(1+s)M_{2}}\right]^{\frac{1}{1+s}}\right\}(a^{{\ast}}-a_{k})^{\frac{1}{1+s}}, \end{align*}

as $a_{k}\nearrow a^{\ast }$ and here the last equality is achieved at

\[ \epsilon_{k}=\left[\frac{sM_{1}(a^{{\ast}}-a_{k})}{(1+s)M_{2}}\right]^{\frac{1}{2+2s}}. \]

Thus, (4.42) holds for the subsequence $\{a_{k}\}$. Actually, the above argument can be carried out for any subsequence $\{a_{k}\}$ satisfying $a_{k}\nearrow a^{\ast }$, which then implies that (4.42) holds for all $a\nearrow a^{\ast }$.

Proof. We divide the proof into four steps. Step 1. By (4.27) and (5.4), we get

(5.8)\begin{equation} (-\Delta)^{s}w_{k}+\varepsilon_{k}^{2s}(|\varepsilon_{k}x+\varepsilon_{k}y_{\varepsilon_{k}}|-M)^{2}w_{k}=\varepsilon_{k}^{2s}\lambda_{k}w_{k}+a_{k}w^{2s+1}_{k},\text{ in }\mathbb{R}^2, \end{equation}

where $\lambda _{k}\in \mathbb {R}^{2}$ is a Lagrange multiplier. Similar to the proof of lemma 4.3-$(iii)$, we know that (5.6) holds.

Step 2. For all $k$, we assume that $v_{k}\geq 0$ satisfies

(5.9)\begin{equation} (-\Delta)^{s}v_{k}-\varepsilon_{k}^{2s}\lambda_{k}v_{k}=a_{k}w^{2s+1}_{k},\text{ in }\mathbb{R}^2. \end{equation}

Next, we prove that $\|v_{k}\|_{\infty }\leq C$, for all $k$. In fact, (5.8) shows that

(5.10)\begin{equation} (-\Delta)^{s}w_{k}-\varepsilon_{k}^{2s}\lambda_{k}w_{k}\leq a_{k}w^{2s+1}_{k},\text{ in }\mathbb{R}^2. \end{equation}

From (5.9) and (5.10), it is easy to see that

(5.11)\begin{equation} 0\leq w_{k}\leq v_{k},\text{ a.e. in }\mathbb{R}^2 \text{ and for all}\ k. \end{equation}

For $\beta \geq 1$ and $T>0$, let

\begin{align*} & \varphi(t)=\left\{ \begin{array}{@{}ll@{}} 0, & t\leq0, \\ t^{\beta}, & 0< t< T, \\ \beta T^{\beta-1}(t-T)+T^{\beta}, & t\geq T. \end{array} \right. \end{align*}

Clearly, $\varphi$ is convex and Lipschitz continuous, we get

(5.12)\begin{equation} (-\Delta)^{s}\varphi(v_{k})\leq\varphi'(v_{k})(-\Delta)^{s}v_{k}, \end{equation}

in the weak sense. By using Sobolev inequality, (5.11), (5.12), the fact $\lambda _{k}<0$, $\varphi '(v_{k})\varphi (v_{k})\leq \beta v_{k}^{2\beta -1}$, $v_{k}\varphi '(v_{k})\leq \beta \varphi (v_{k})$ and integrating by parts, we deduce that

(5.13)\begin{align} \|\varphi(v_{k})\|_{2_{s}^{{\ast}}}^{2} & \leq C\int_{\mathbb{R}^{2}}|(-\Delta)^{\frac{s}{2}}\varphi(v_{k})|^{2}\,{\rm d}x=C\int_{\mathbb{R}^{2}}\varphi(v_{k})(-\Delta)^{s}\varphi(v_{k})\,{\rm d}x\nonumber\\ & \leq C\int_{\mathbb{R}^{2}}\varphi(v_{k})\varphi'(v_{k})(-\Delta)^{s}v_{k}\,{\rm d}x\nonumber\\ & =C\int_{\mathbb{R}^{2}}\varphi(v_{k})\varphi'(v_{k})(\varepsilon_{k}^{2s}\lambda_{k}v_{k}+a_{k}w^{2s+1}_{k})\,{\rm d}x\nonumber\\ & \leq C\int_{\mathbb{R}^{2}}\varphi(v_{k})\varphi'(v_{k})(1+v_{k}^{2_{s}^{{\ast}}-1})\,{\rm d}x\nonumber\\ & =C\left(\int_{\mathbb{R}^{2}}\varphi(v_{k})\varphi'(v_{k})\,{\rm d}x+\int_{\mathbb{R}^{2}}\varphi(v_{k})\varphi'(v_{k})v_{k}^{2_{s}^{{\ast}}-1}\,{\rm d}x\right)\nonumber\\ & \leq C\beta\left(\int_{\mathbb{R}^{2}}v_{k}^{2\beta-1}\,{\rm d}x+\int_{\mathbb{R}^{2}}(\varphi(v_{k}))^{2}v_{k}^{2_{s}^{{\ast}}-2}\,{\rm d}x\right), \end{align}

where $C>0$ independent of $k$ and $\beta$.

Note that $\beta \geq 1$ and that $\varphi (v_{k})$ is linear when $v_{k}\geq T$, then we have

\begin{align*} \int_{\mathbb{R}^{2}}(\varphi(v_{k}))^{2}v_{k}^{2_{s}^{{\ast}}-2}\,{\rm d}x& =\int_{\{v_{k}\leq T\}}(\varphi(v_{k}))^{2}v_{k}^{2_{s}^{{\ast}}-2}\,{\rm d}x+\int_{\{v_{k}>T\}}(\varphi(v_{k}))^{2}v_{k}^{2_{s}^{{\ast}}-2}\,{\rm d}x\\ & \leq T^{2\beta-2}\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}}\,{\rm d}x+C\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}}\,{\rm d}x<{+}\infty, \end{align*}

which implies that $\int _{\mathbb {R}^{2}}(\varphi (v_{k}))^{2}v_{k}^{2_{s}^{\ast }-2}\,{\rm d}x$ is well defined for every $T$.

Now, we let $\beta$ in (5.13) such that $2\beta -1=2_{s}^{\ast }$ and define $\beta _{1}=\frac {2_{s}^{\ast }+1}{2}$. Let $R>0$ be fixed later, by Hölder's inequality, we have

(5.14)\begin{align} & \int_{\mathbb{R}^{2}}(\varphi(v_{k}))^{2}v_{k}^{2_{s}^{{\ast}}-2}\,{\rm d}x\nonumber\\ & \quad=\int_{\{v_{k}\leq R\}}(\varphi(v_{k}))^{2}v_{k}^{2_{s}^{{\ast}}-2}\,{\rm d}x+\int_{\{v_{k}>R\}}(\varphi(v_{k}))^{2}v_{k}^{2_{s}^{{\ast}}-2}\,{\rm d}x\nonumber\\ & \quad\leq R^{2_{s}^{{\ast}}-1}\int_{\{v_{k}\leq R\}}\frac{(\varphi(v_{k}))^{2}}{v_{k}}\,{\rm d}x\nonumber\\ & \qquad+\left(\int_{\{v_{k}>R\}}v_{k}^{2_{s}^{{\ast}}}\,{\rm d}x\right)^{\frac{2_{s}^{{\ast}}-2}{2_{s}^{{\ast}}}}\left(\int_{\mathbb{R}^{2}}(\varphi(v_{k}))^{2_{s}^{{\ast}}}\,{\rm d}x\right)^{\frac{2}{2_{s}^{{\ast}}}}. \end{align}

Similar to the proof of lemma 4.3-$(iii)$, we know that $\{v_{k}\}$ converges strongly in $H^{s}(\mathbb {R}^{2})$, then $\{v_{k}\}$ converges strongly in $L^{2_{s}^{\ast }}(\mathbb {R}^{2})$, so we can choose $R$ sufficiently large such that

(5.15)\begin{equation} \left(\int_{\{v_{k}>R\}}v_{k}^{2_{s}^{{\ast}}}\,{\rm d}x\right)^{\frac{2_{s}^{{\ast}}-2}{2_{s}^{{\ast}}}}\leq\frac{1}{2C\beta_{1}}. \end{equation}

From (5.13)–(5.15), it follows that

(5.16)\begin{equation} \left(\int_{\mathbb{R}^{2}}(\varphi(v_{k}))^{2_{s}^{{\ast}}}\,{\rm d}x\right)^{\frac{2}{2_{s}^{{\ast}}}}\leq2C\beta_{1}\left(\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}}\,{\rm d}x+R^{2_{s}^{{\ast}}-1}\int_{\mathbb{R}^{2}}\frac{(\varphi(v_{k}))^{2}}{v_{k}}\,{\rm d}x\right). \end{equation}

Thus, by applying $\varphi (v_{k})\leq v_{k}^{\beta _{1}}$ and letting $T\rightarrow \infty$, we have

\[ \left(\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}\beta_{1}}\,{\rm d}x\right)^{\frac{2}{2_{s}^{{\ast}}}}\leq2C\beta_{1}\left(\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}}\,{\rm d}x+R^{2_{s}^{{\ast}}-1}\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}}\,{\rm d}x\right)<\infty, \]

which implies that

(5.17)\begin{equation} v_{k}\in L^{2_{s}^{{\ast}}\beta_{1}}(\mathbb{R}^{2}). \end{equation}

Assume that $\beta >\beta _{1}$. By taking $T\rightarrow \infty$ in (5.13), we can see that

(5.18)\begin{equation} \left(\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}\beta}\,{\rm d}x\right)^{\frac{2}{2_{s}^{{\ast}}}}\leq C\beta\left(\int_{\mathbb{R}^{2}}v_{k}^{2\beta-1}\,{\rm d}x+\int_{\mathbb{R}^{2}}v_{k}^{2\beta+2_{s}^{{\ast}}-2}\,{\rm d}x\right). \end{equation}

Let

\[ v_{k}^{2\beta-1}=v_{k}^{l}v_{k}^{m}, \]

where $l=\frac {2_{s}^{\ast }(2_{s}^{\ast }-1)}{2(\beta -1)}$ and $m=2\beta -1-l$. Moreover, $\beta >\beta _{1}$ implies that $0< l,\, m<2_{s}^{\ast }$, by using Young's inequality, we get

(5.19)\begin{align} \int_{\mathbb{R}^{2}}v_{k}^{2\beta-1}\,{\rm d}x & \leq\frac{l}{2_{s}^{{\ast}}}\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}}\,{\rm d}x+\frac{2_{s}^{{\ast}}-l}{2_{s}^{{\ast}}}\int_{\mathbb{R}^{2}}v_{k}^{\frac{2_{s}^{{\ast}}m}{2_{s}^{{\ast}}-l}}\,{\rm d}x\nonumber\\ & \leq\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}}\,{\rm d}x+\int_{\mathbb{R}^{2}}v_{k}^{2\beta+2_{s}^{{\ast}}-2}\,{\rm d}x\nonumber\\ & \leq C\left(1+\int_{\mathbb{R}^{2}}v_{k}^{2\beta+2_{s}^{{\ast}}-2}\,{\rm d}x\right). \end{align}

In view of (5.18) and (5.19), we have

(5.20)\begin{equation} \left(\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}\beta}\,{\rm d}x\right)^{\frac{2}{2_{s}^{{\ast}}}}\leq C\beta\left(1+\int_{\mathbb{R}^{2}}v_{k}^{2\beta+2_{s}^{{\ast}}-2}\,{\rm d}x\right), \end{equation}

which shows that

(5.21)\begin{equation} \left(1+\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}\beta}\,{\rm d}x\right)^{\frac{1}{2_{s}^{{\ast}}(\beta-1)}}\leq (C\beta)^{\frac{1}{2(\beta-1)}}\left(1+\int_{\mathbb{R}^{2}}v_{k}^{2\beta+2_{s}^{{\ast}}-2}\,{\rm d}x\right)^{\frac{1}{2(\beta-1)}}. \end{equation}

Iterating this argument, we obtain

(5.22)\begin{align} & \left(1+\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}\beta_{i+1}}\,{\rm d}x\right)^{\frac{1}{2_{s}^{{\ast}}(\beta_{i+1}-1)}}\nonumber\\ & \quad\leq (C\beta_{i+1})^{\frac{1}{2(\beta_{i+1}-1)}}\left(1+\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}\beta_{i}}\,{\rm d}x\right)^{\frac{1}{2(\beta_{i}-1)}}, \end{align}

where

\[ 2\beta_{i+1}+2_{s}^{{\ast}}-2=2_{s}^{{\ast}}\beta_{i} \quad \hbox{and} \quad \beta_{i+1}=(\frac{2_{s}^{{\ast}}}{2})^{i}(\beta_{i}-1)+1. \]

Setting $C_{i+1}=C\beta _{i+1}$ and

\[ K_{i}=\left(1+\int_{\mathbb{R}^{2}}v_{k}^{2_{s}^{{\ast}}\beta_{i}}\,{\rm d}x\right)^{\frac{1}{2_{s}^{{\ast}}(\beta_{i}-1)}}. \]

We can see that there exists a constant $C>0$ independent of $i$ , such that

\[ K_{i+1}\leq\Pi_{i=2}^{i+1}C_{i}^{\frac{1}{2(\beta_{i}-1)}}K_{1}\leq CK_{1}. \]

Hence, we have

\[ \|v_{k}\|_{\infty}\leq C, \quad \hbox{for all}\ k. \]

Step 3. We prove that $w_{k}(x)\rightarrow 0$ as $|x|\rightarrow \infty$ uniformly in $k$.

In fact, we rewrite problem (5.9) as follows

\[ (-\Delta)^{s}v_{k}+v_{k}=h_{k}(x),\quad x\in\mathbb{R}^2, \]

where $h_{k}(x)=v_{k}+\varepsilon _{k}^{2s}\lambda _{k}v_{k}+a_{k}w^{2s+1}_{k}$. Thus, step 2 shows that $h_{k}\in L^{\infty }(\mathbb {R}^{2})$ and is uniformly bounded. From interpolation inequality and $\{v_{k}\}$ converges strongly in $H^{s}(\mathbb {R}^{2})$, we know that $h_{k}\rightarrow h$ in $L^{q}(\mathbb {R}^{2})$ for $q\in [2,\, +\infty )$. Thus, by [Reference Felmer, Quaas and Tan8], we deduce that

\[ v_{k}=\int_{\mathbb{R}^{2}}\mathcal{K}(x-y)h_{k}(y){\rm d}y, \]

where $\mathcal {K}$ is a Bessel potential and it satisfies

$(\mathcal {K}_{1})$ $\mathcal {K}$ is positive, radially symmetric and smooth in $\mathbb {R}^{2}\backslash \{0\}$.

$(\mathcal {K}_{2})$ There exists a $C>0$ such that $\mathcal {K}(x)\leq \frac {C}{|x|^{2+2s}}$ for $x\in \mathbb {R}^{2}\backslash \{0\}$.

$(\mathcal {K}_{3})$ $\mathcal {K}\in L^{r}(\mathbb {R}^{2})$ for $r\in [1,\, \frac {1}{1-s})$.

Now, for any $\zeta >0$, we have

\begin{align*} 0\leq v_{k}& \leq\int_{\mathbb{R}^{2}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y\\ & =\int_{\{|x-y|\geq\frac{1}{\zeta}\}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y+\int_{\{|x-y|<\frac{1}{\zeta}\}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y. \end{align*}

By step 1 and $(\mathcal {K}_{2})$, we can see that

(5.23)\begin{equation} \int_{\{|x-y|\geq\frac{1}{\zeta}\}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y\leq C\zeta\int_{\{|x-y|\geq\frac{1}{\zeta}\}}\frac{1}{|x-y|^{2+2s}}{\rm d}y=C\zeta^{2s}. \end{equation}

Moreover, by using Hölder's inequality and $(\mathcal {K}_{3})$, we deduce that

\begin{align*} & \int_{\{|x-y|<\frac{1}{\zeta}\}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y\\ & \quad\leq\int_{\{|x-y|<\frac{1}{\zeta}\}}\mathcal{K}(x-y)|h_{k}-h|{\rm d}y+\int_{\{|x-y|<\frac{1}{\zeta}\}}\mathcal{K}(x-y)|h|{\rm d}y\\ & \quad\leq\Bigg(\!\int_{\mathbb{R}^{2}}|\mathcal{K}|^{2}{\rm d}y\!\Bigg)^{\frac{1}{2}}\Bigg(\!\int_{\mathbb{R}^{2}}|h_{k}-h|^{2}{\rm d}y\!\Bigg)^{\frac{1}{2}}+\Bigg(\!\int_{\mathbb{R}^{2}}|\mathcal{K}|^{2}{\rm d}y\!\Bigg)^{\frac{1}{2}}\Bigg(\int_{\{|x-y|<\frac{1}{\zeta}\}}|h|^{2}{\rm d}y\Bigg)^{\frac{1}{2}}, \end{align*}

which implies that there exist $K_{0}\in \mathbb {N}$ and $R_{0}>0$ independent of $\zeta >0$ such that

(5.24)\begin{equation} \int_{\{|x-y|<\frac{1}{\zeta}\}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y\leq\zeta, \forall k\geq K_{0}\text{ and }|x|\geq R_{0}, \end{equation}

where we have used the fact $s>\frac {1}{2}$ so that $2<\frac {1}{1-s}$ and $(\int _{\{|x-y|<\frac {1}{\zeta }\}}|h|^{2}{\rm d}y)^{\frac {1}{2}}\rightarrow 0$ as $|x|\rightarrow \infty$. Thus, by (5.23) and (5.24), we know that

\[ \int_{\mathbb{R}^{2}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y\leq C\zeta^{2s}+\zeta, \quad \forall k\geq K_{0}\ \hbox{and}\ |x|\geq R_{0}. \]

On the other hand, for all $k\in \{1,\, 2,\,\cdot \cdot \cdot,\, K_{0}-1\}$, there exists $R_{k}>0$ such that

\[ \left(\int_{\{|x-y|<\frac{1}{\zeta}\}}|h_{k}|^{2}{\rm d}y\right)^{\frac{1}{2}}\leq\zeta, \text{ as }|x|\geq R_{k}, \]

which implies that

\begin{align*} \int_{\mathbb{R}^{2}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y& \leq C\zeta^{2s}+\int_{\{|x-y|<\frac{1}{\zeta}\}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y\\ & \quad\leq C\zeta^{2s}+\|\mathcal{K}\|_{2}\Bigg(\int_{\{|x-y|<\frac{1}{\zeta}\}}|h_{k}|^{2}{\rm d}y\Bigg)^{\frac{1}{2}}\\ & \quad\leq C(\zeta^{2s}+\zeta). \end{align*}

Thus, setting $R=\max \{R_{0},\, R_{1},\,\cdot \cdot \cdot R_{K_{0}-1}\}$, we conclude that

\[ 0\leq v_{k}\leq\int_{\mathbb{R}^{2}}\mathcal{K}(x-y)|h_{k}(y)|{\rm d}y\leq C(\zeta^{2s}+\zeta),\quad \hbox{for all}\ |x|\geq R, \]

which implies

(5.25)\begin{equation} \lim_{|x|\rightarrow\infty}v_{k}(x)=0, \text{ uniformly in }k. \end{equation}

From (5.11) and (5.25), it follows that

(5.26)\begin{equation} \lim_{|x|\rightarrow\infty}w_{k}(x)=0, \text{ uniformly in }k. \end{equation}

Step 4. Combining step 2, step 3 and the proof of theorem 1.1 in [Reference Teng and Cheng22], we can get that

(5.27)\begin{equation} w_{k}(x)\leq\frac{C}{1+|x|^{2+2s}}, \text{ for all }k. \end{equation}

For any $x\in B_{\rho }^{c}(y_{0})$, (5.6) shows that

(5.28)\begin{equation} \frac{|x-z_{k}|}{\varepsilon_{k}}\geq\frac{|x-y_{0}|}{2\varepsilon_{k}}\geq\frac{\rho}{2\varepsilon_{k}}\stackrel{k}{\longrightarrow}+\infty. \end{equation}

From (5.27) and (5.28), it follows that

\begin{align*} u_{k}(x)& =\frac{1}{\varepsilon_{k}}w_{k}\left(\frac{x-z_{k}}{\varepsilon_{k}}\right)\leq\frac{1}{\varepsilon_{k}}\frac{C}{1+|\frac{x-z_{k}}{\varepsilon_{k}}|^{2+2s}}\\ & \leq\frac{1}{\varepsilon_{k}}\frac{C}{1+|\frac{\rho}{2\varepsilon_{k}}|^{2+2s}}\stackrel{k}{\longrightarrow}0,\quad \forall x\in B_{\rho}^{c}(y_{0}). \end{align*}

Proof of theorem 1.2 Set $\varepsilon _{k}:=(a^{\ast }-a_{k})^{\frac {1}{2+2s}}>0$, where $a_{k}\nearrow a^{\ast }$. Define $u_{k}(x):=u_{a_{k}}(x)$ is a nonnegative minimizer of (1.3). Moreover, we set $\bar {z}_{k}$ be any local maximum point of $u_{k}$. Clearly, we have

(5.29)\begin{equation} u_{k}(\bar{z}_{k})\geq\Bigg(\frac{-\lambda_{k}}{a_{k}}\Bigg)^{\frac{1}{2s}}\geq C\varepsilon_{k}^{{-}1}. \end{equation}

Hence, from (5.29) and lemma 5.1, it follows that

(5.30)\begin{equation} \bar{z}_{k}\stackrel{k}{\longrightarrow}y_{0}\in\mathbb{R}^{2}\text{ with }|y_{0}|=M. \end{equation}

Let

(5.31)\begin{equation} \bar{w}_{k}:=\varepsilon_{k}u_{k}(\varepsilon_{k}x+\bar{z}_{k}). \end{equation}

By (5.8), we deduce that

(5.32)\begin{equation} (-\Delta)^{s}\bar{w}_{k}+\varepsilon_{k}^{2s}(|\varepsilon_{k}x+\bar{z}_{k}|-M)^{2}\bar{w}_{k}=\varepsilon_{k}^{2s}\lambda_{k}\bar{w}_{k}+a_{k}\bar{w}^{2s+1}_{k},\text{ in }\mathbb{R}^2. \end{equation}

Next, we prove that $\{\frac {\bar {z}_{k}-z_{k}}{\varepsilon _{k}}\}\subset \mathbb {R}^{2}$ is bounded uniformly in $k$. Assume by contradiction that $|\frac {\bar {z}_{k}-z_{k}}{\varepsilon _{k}}|\rightarrow \infty$ as $k\rightarrow \infty$. (5.27) shows that

\[ u_{k}(\bar{z}_{k})=\frac{1}{\varepsilon_{k}}w_{k}\Bigg(\frac{\bar{z}_{k}-z_{k}}{\varepsilon_{k}}\Bigg)\leq\frac{C}{\varepsilon_{k}}\frac{1}{1+|\frac{\bar{z}_{k}-z_{k}}{\varepsilon_{k}}|^{2+2s}}=o(\varepsilon_{k}^{{-}1}),\text{ as }k\rightarrow\infty, \]

which implies a contradiction by (5.29). Thus, there exists $R_{1}>0$, independent of $k$, such that, $|\frac {\bar {z}_{k}-z_{k}}{\varepsilon _{k}}|<\frac {R_{1}}{2}$. By (5.3), we can see that

(5.33)\begin{equation} \begin{aligned} & \lim_{k\rightarrow\infty}\int_{B_{R_{0}+R_{1}}(0)}|\bar{w}_{k}|^{2}\,{\rm d}x\\ & \quad=\lim_{k\rightarrow\infty}\int_{B_{R_{0}+R_{1}}\left(\frac{\bar{z}_{k}-z_{k}}{\varepsilon_{k}}\right)}|w_{k}|^{2}\,{\rm d}x\geq\int_{B_{R_{0}+\frac{R_{1}}{2}}(0)}|w_{k}|^{2}\,{\rm d}x\geq\eta>0, \end{aligned} \end{equation}

where we have used the fact $\bar {w}_{k}(x)=w_{k}(x+\frac {\bar {z}_{k}-z_{k}}{\varepsilon _{k}})$. Similar to the argument of lemma 4.3-$(iii)$, we know that there exists a subsequence, still denoted by $\{\bar {w}_{k}\}$, of $\{\bar {w}_{k}\}$ such that

(5.34)\begin{equation} \left\{ \begin{array}{@{}ll@{}} \varepsilon_{k}^{2s}\lambda_{k}\stackrel{k}{\longrightarrow}-\beta^{2}<0, & \text{for some }\beta>0, \\ \bar{w}_{k}\stackrel{k}{\longrightarrow}\bar{w}_{0}\geq0, & \text{in }H^{s}(\mathbb{R}^{2}), \end{array} \right. \end{equation}

where $\bar {w}_{0}$ satisfies

(5.35)\begin{equation} (-\Delta)^{s}\bar{w}_{0}={-}\beta^{2}\bar{w}_{0}+a^{{\ast}}\bar{w}^{2s+1}_{0},\quad \text{ in }\mathbb{R}^2. \end{equation}

Note from (5.33) that $\bar {w}_{0}\not \equiv 0$. Thus, similar to the proof of lemma 4.3-$(iii)$, we know that $\bar {w}_{0}>0$ in $\mathbb {R}^{2}$. Since the origin is a critical point of $\bar {w}_{k}$, we get that the origin is also a critical point of $\bar {w}_{0}$. By (5.35) and $Q$ is the unique positive radial solution of (1.7), for the above $\beta >0$, we can deduce that

(5.36)\begin{equation} \bar{w}_{0}(x)=\frac{\beta}{s^{\frac{1}{2s}}\|Q\|_{2}}Q\Bigg(\frac{\beta}{s^{\frac{1}{2s}}}|x|\Bigg). \end{equation}

Clearly, we know that $\bar {w}_{k}\geq (\frac {\beta ^{2}}{2a^{\ast }})^{\frac {1}{2s}}$ at each local maximum point. Hence, lemma 5.1 implies that all the local maximum points of $\bar {w}_{k}$ stay in a finite ball in $\mathbb {R}^{2}$. By (5.26) and the definition of $\bar {w}_{k}$, we can get that

(5.37)\begin{equation} \|\bar{w}_{k}\|_{\infty}\leq C, \quad \text{ uniformly as }k\rightarrow+\infty. \end{equation}

Next, we prove that $\{\bar {w}_{k}\}$ is bounded uniformly in $C^{2, \alpha }_{loc}(\mathbb {R}^{2})$ for some $0<\alpha <1$. In fact, we rewrite (5.32) as follows

(5.38)\begin{equation} (-\Delta)^{s}\bar{w}_{k}(x)=f_{k}(x),\quad \text{in }\mathbb{R}^2, \end{equation}

where $f_{k}(x)=-\varepsilon _{k}^{2s}(|\varepsilon _{k}x+\bar {z}_{k}|-M)^{2}\bar {w}_{k}+\varepsilon _{k}^{2s}\lambda _{k}\bar {w}_{k}+a_{k}\bar {w}^{2s+1}_{k}$. Since $\varepsilon _{k}^{2s}(|\varepsilon _{k}x+\bar {z}_{k}|-M)^{2}$ is locally Lipschitz continuous in $\mathbb {R}^{2}$ and (5.37), we have

(5.39)\begin{equation} \|f_{k}(x)\|_{\infty}\leq C, \text{ uniformly as }k\rightarrow+\infty. \end{equation}

From (5.37), (5.38), (5.39) and lemma 2.3 in [Reference Teng and Wu23], we know that

(5.40)\begin{equation} \bar{w}_{k}\in C^{1, \alpha}(\mathbb{R}^2),\text{ for }\alpha<2s-1, \end{equation}

and

(5.41)\begin{equation} \|\bar{w}_{k}\|_{C^{1, \alpha}(\mathbb{R}^2)}\leq C(\|\bar{w}_{k}\|_{\infty}+\|f_{k}(x)\|_{\infty})\leq C. \end{equation}

From (5.40) and (5.41), it follows that

(5.42)\begin{equation} \|f_{k}(x)\|_{C^{1}_{loc}(\mathbb{R}^{2})}< C,\text{ uniformly as }k\rightarrow+\infty. \end{equation}

Thus, by (5.37), (5.38), (5.42) and lemma 4.4 in [Reference Cabré and Sire4], we know that $\{\bar {w}_{k}\}$ is bounded uniformly in $C^{2, \alpha }_{loc}(\mathbb {R}^{2})$ for some $0<\alpha <1$. Thus, we may assume that there exists $\hat {w}_{0}\in C^{2, \alpha }_{loc}(\mathbb {R}^{2})$ such that $\bar {w}_{k}\rightarrow \hat {w}_{0}$ in $C^{2}_{loc}(\mathbb {R}^{2})$ as $k\rightarrow \infty$. Moreover, (5.34) shows that $\hat {w}_{0}=\bar {w}_{0}$.

Since the origin is the only critical point of $\bar {w}_{0}$, then the content of the appeal discussion shows that all local maximum points of $\{\bar {w}_{k}\}$ must approach the origin and hence stay in a small ball $B_{\varrho }(0)$ as $k\rightarrow \infty$. Letting $\varrho >0$ small enough such that $\bar {w}''_{0}(\tau )<0$ for $0<\tau <\varrho$. By lemma 4.2 in [Reference Ni and Takagi16], we know that $\{\bar {w}_{k}\}$ has no critical points other than the origin. Therefore, we get that there exists a subsequence of $\{u_{k}\}$ concentrating at a unique global minimum point of potential $V(x)=(|x|-M)^{2}$ in $\mathbb {R}^{2}$.

Now, we turn to proving (1.15)–(1.17). In fact, by (5.31), we have

(5.43)\begin{align} e(a_{k}) & =J_{a_{k}}(u_{k})=\frac{1}{\varepsilon_{k}^{2s}}\left[\int_{\mathbb{R}^{2}}|(-\Delta)^{\frac{s}{2}}\bar{w}_{k}|^{2}\,{\rm d}x-\frac{a^{{\ast}}}{1+s}\int_{\mathbb{R}^{2}}|\bar{w}_{k}|^{2+2s}\,{\rm d}x\right]\nonumber\\ & \quad+\frac{\varepsilon_{k}^{2}}{1+s}\int_{\mathbb{R}^{2}}|\bar{w}_{k}|^{2+2s}\,{\rm d}x+\int_{\mathbb{R}^{2}}(|\varepsilon_{k}x+\bar{z}_{k}|-|y_{0}|)^{2}|\bar{w}_{k}|^{2}\,{\rm d}x, \end{align}

where $\bar {z}_{k}$ is the unique global maximum point of $u_{k}$, and $\bar {z}_{k}\rightarrow y_{0}\in \mathbb {R}^{2}$ as $k\rightarrow \infty$ for some $|y_{0}|=M>0$.

Next, we prove that $\{\frac {|\bar {z}_{k}|-|y_{0}|}{\varepsilon _{k}}\}\subset \mathbb {R}$ is bounded uniformly for $k\rightarrow \infty$. Assume by contradiction that there exists a subsequence $\{a_{k}\}$, still denoted by $\{a_{k}\}$ such that $|\frac {|\bar {z}_{k}|-|y_{0}|}{\varepsilon _{k}}|\rightarrow \infty$ as $k\rightarrow \infty$, by (5.33), for any $C>0$, we have

(5.44)\begin{align} & \lim_{k\rightarrow\infty}\varepsilon_{k}^{{-}2}\int_{\mathbb{R}^{2}}(|\varepsilon_{k}x+\bar{z}_{k}|-|y_{0}|)^{2}|\bar{w}_{k}|^{2}\,{\rm d}x\nonumber\\ & \quad=\lim_{k\rightarrow\infty}\int_{\mathbb{R}^{2}}\Bigg(|x+\frac{\bar{z}_{k}}{\varepsilon_{k}}|-\frac{|y_{0}|}{\varepsilon_{k}}\Bigg)^{2}|\bar{w}_{k}|^{2}\,{\rm d}x\geq C. \end{align}

From (5.43)–(5.44), it follows that

\[ e(a_{k})\geq C\varepsilon_{k}^{2}=C(a^{{\ast}}-a_{k})^{\frac{1}{1+s}}, \]

holds for any $C>0$, which implies a contradiction by lemma 4.5. Thus, there exists a subsequence $\{a_{k}\}$, still denoted by $\{a_{k}\}$ such that

(5.45)\begin{equation} \frac{|\bar{z}_{k}|-|y_{0}|}{\varepsilon_{k}}\rightarrow C_{0},\text{ as }k\rightarrow\infty, \end{equation}

for some constant $C_{0}$. Since $Q$ a radially symmetric function and polynomial decay as $|x|\rightarrow \infty$, we then deduce from (5.36) that

(5.46)\begin{align} & \lim_{k\rightarrow\infty}\frac{1}{\varepsilon_{k}^{2}}\int_{\mathbb{R}^{2}}(|\varepsilon_{k}x+\bar{z}_{k}|-|y_{0}|)^{2}|\bar{w}_{k}|^{2}\,{\rm d}x\nonumber\\ & \quad=\lim_{k\rightarrow\infty}\int_{\mathbb{R}^{2}}\Bigg(\frac{|\varepsilon_{k}x+\bar{z}_{k}|-|\bar{z}_{k}|}{\varepsilon_{k}}+\frac{|\bar{z}_{k}|-|y_{0}|}{\varepsilon_{k}}\Bigg)^{2}|\bar{w}_{k}|^{2}\,{\rm d}x\nonumber\\ & \quad=\int_{\mathbb{R}^{2}}\Bigg(\frac{y_{0}\cdot x}{|y_{0}|}+C_{0}\Bigg)^{2}|\bar{w}_{0}|^{2}\,{\rm d}x\geq\int_{\mathbb{R}^{2}}\frac{|y_{0}\cdot x|^{2}}{|M|^{2}}|\bar{w}_{0}|^{2}\,{\rm d}x, \end{align}

where the equality holds if and only if $C_{0}=0$. By (5.43) and (5.46), we have

(5.47)\begin{equation} \begin{aligned} \lim_{k\rightarrow\infty}\frac{e(a_{k})}{(a^{{\ast}}-a_{k})^{\frac{1}{1+s}}} & \geq\frac{1}{1+s}\|\bar{w}_{0}\|_{2+2s}^{2+2s}+\frac{1}{M^{2}}\int_{\mathbb{R}^{2}}|y_{0}\cdot x|^{2}|\bar{w}_{0}|^{2}\,{\rm d}x\\ & =\frac{1}{sa^{{\ast}}}\beta^{2s}+\frac{s^{\frac{1}{s}}}{\|Q\|_{2}^{2}\beta^{2}M^{2}}\int_{\mathbb{R}^{2}}|y_{0}\cdot x|^{2}|Q|^{2}\,{\rm d}x\\ & \geq\frac{1+s}{sa^{{\ast}}}\Bigg(\frac{s^{\frac{1}{s}}}{\|Q\|_{2}^{2-2s}M^{2}}\int_{\mathbb{R}^{2}}|y_{0}\cdot x|^{2}|Q|^{2}\,{\rm d}x\Bigg)^{\frac{s}{1+s}}, \end{aligned} \end{equation}

where the equality is achieved at

\[ \beta=\mu_{0}:=\Bigg(\frac{s^{\frac{1}{s}}}{\|Q\|_{2}^{2-2s}M^{2}}\int_{\mathbb{R}^{2}}|y_{0}\cdot x|^{2}|Q|^{2}\,{\rm d}x\Bigg)^{\frac{1}{2+2s}}. \]

We take

\[ u(x)=\frac{\beta}{s^{\frac{1}{2s}}\varepsilon\|Q\|_{2}}Q\Bigg(\frac{\beta}{s^{\frac{1}{2s}}}\frac{|x-y_{0}|}{\varepsilon}\Bigg) \]

as a trial functional for $J_{a}$, and minimizes over $\beta >0$. (5.47) shows that

(5.48)\begin{equation} \lim_{k\rightarrow\infty}\frac{e(a_{k})}{(a^{{\ast}}-a_{k})^{\frac{1}{1+s}}}=\frac{1+s}{sa^{{\ast}}}\Bigg(\frac{s^{\frac{1}{s}}}{\|Q\|_{2}^{2-2s}M^{2}}\int_{\mathbb{R}^{2}}|y_{0}\cdot x|^{2}|Q|^{2}\,{\rm d}x\Bigg)^{\frac{s}{1+s}}. \end{equation}

Therefore, from (5.48), we get the following several conclusions.

$(I)$ $\beta$ is unique, which is independent of the choice of the subsequence, and takes the value of $\mu _{0}$ as above.

$(II)$ $C_{0}=0$, that is, (1.15) holds.

Finally, by (5.30), (5.34) and (5.36), we have

\[ (a^{{\ast}}-a_{k})^{\frac{1}{2+2s}}u_{a_{k}}\left(x_{k}+(a^{{\ast}}-a_{k})^{\frac{1}{2+2s}}x\right)\stackrel{k}{\longrightarrow}\frac{\mu_{0}Q\Bigg(\frac{\mu_{0}|x|}{s^{\frac{1}{2s}}}\Bigg)}{s^{\frac{1}{2s}}\|Q\|_{2}} \text{ strongly in }H^{s}(\mathbb{R}^{2}), \]

that is, (1.16) holds.

Proof of theorem 1.4 In fact, (5.48) shows that (1.19) holds for the subsequence $\{a_{k}\}$. Moreover, the proof of theorem 1.2 shows that (5.48) is correct for all $\{a_{k}\}$ with $a_{k}\nearrow a^{\ast }$. Therefore, (1.19) holds for all $a\nearrow a^{\ast }$.

Proof of theorem 1.5 It then follows from theorem 1.2 that all nonnegative minimizers of $e(a)$ concentrate at any point on the ring $\{x\in \mathbb {R}^{2}: |x|=M\}$. This further implies that there exists a $a_{\ast }$ satisfying $0< a_{\ast }< a^{\ast }$ such that for any $a\in [a_{\ast },\, a^{\ast })$, $e(a)$ has infinitely many different nonnegative minimizers, each of which concentrates at a specific global minimum point of potential $V(x)=(|x|-M)^{2}$. However, recall from theorem 1.3-(ii) in [Reference L.X. Tian, Wang and Zhang6] that $e(a)$ admits a unique nonnegative minimizer $u_{a}$ for all $a>0$ being small enough $(a< a^{\ast })$, and noting that the trapping potential $V(x)=(|x|-M)^{2}$ $(M>0)$ is radially symmetric. Then similar to the argument of corollary 1.7 in [Reference L.X. Tian, Wang and Zhang6], by rotation $u_{a}$ must be rotational symmetry with respect to the origin.