Normalized Ground States for the Sobolev Critical Fractional Kirchhoff Equation with at Least Mass Critical Growth

Abstract

In this paper, we delve into the following nonlinear fractional Kirchhoff-type problem ${(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2{)(-\Delta )}^{s}u+\lambda u=g\left(u\right)+{\left|u\right|}^{2_s^{*}-2}u$ in ${\mathbb{R}}^3$ with prescribed mass ${\int }_{{\mathbb{R}}^3}{\left|u\right|}^{2}dx=\rho >0,$ where $s\in ({\displaystyle \frac{3}{4}},1),\lambda \in \mathbb{R},2_s^{*}={\displaystyle \frac{6}{3-2s}}$. Under some general growth assumptions imposed on g, we employ minimization of the energy functional on the linear combination of Nehari and Poho$\breve{z}$aev constraints intersected with the closed ball in the $L^2\left({\mathbb{R}}^3\right)$ of radius $\rho$ to prove the existence of normalized ground state solutions to the equation. Moreover, we provide a detailed description for the asymptotic behavior of the ground state energy map.

1. Introduction and Main Results

In this paper, we consider the existence of normalized solutions for the following fractional Kirchhoff equation

$$\begin{array}{c}\hfill {(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2{)(-\Delta )}^{s}u+\lambda u=g\left(u\right)+{\left|u\right|}^{2_s^{*}-2}u\mathrm{in}{\mathbb{R}}^3\end{array}$$

(1)

with prescribed mass

$${\int }_{{\mathbb{R}}^3}{\left|u\right|}^{2}dx=\rho >0,$$

where $s\in ({\displaystyle \frac{3}{4}},1),\lambda \in \mathbb{R},{\overline{2}}_s=2+{\displaystyle \frac{8s}{3}}$ and $2_s^{*}={\displaystyle \frac{6}{3-2s}}$. The operator ${(-\Delta )}^s$ is the fractional Laplacian defined by

$${(-\Delta )}^{s}u\left(x\right):=C_{s}P.V.{\int }_{{\mathbb{R}}^3}{\displaystyle \frac{u\left(x\right)-u\left(y\right)}{{|x-y|}^{3+2s}}}dy,x\in {\mathbb{R}}^3,$$

and

$${\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^2={\int }_{{\mathbb{R}}^6}{\displaystyle \frac{{|u\left(x\right)-u\left(y\right)|}^2}{{|x-y|}^{3+2s}}}dxdy,$$

where $P.V.$ is the Cauchy principal value on the integral and $C_s$ is some positive normalization constant [1].

Equation (1) has its origin in the theory of nonlinear vibration. In the case of $s=1$ and $h(x,u)=g\left(u\right)+{\left|u\right|}^{2_s^{*}-2}u$, problem (1) reduces to the following Kirchhoff equation

$$\begin{array}{c}\hfill {(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2{)(-\Delta )}^{s}u+\lambda u=g\left(u\right)+{\left|u\right|}^{2_s^{*}-2}u\mathrm{in}{\mathbb{R}}^3.\end{array}$$

(2)

which is related to the stationary case of the following equation:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{u}_{tt}-(a+b{\int }_{\Omega }{|\nabla u|}^2{dx)\Delta u=g\left(u\right)+|u|}^{2_s^{*}-2}u,\mathrm{in}\Omega \times (0,+\infty ),\hfill \\ u(x,t)=0,\mathrm{on}\partial \Omega \times [0,+\infty ).\hfill \end{array}\right.\end{array}$$

(3)

where a is the initial tension, b is related to the intrinsic properties of the string, $\Omega \subset {\mathbb{R}}^3$ is a smooth domain, u stands for displacement, and $f\in C(\Omega \times \mathbb{R},\mathbb{R})$ is the external force. Equation (3) comes from the following model for the modified d’Alembert wave equation:

$$\begin{array}{c}\hfill \rho {\displaystyle \frac{{\partial }^{2}u}{\partial t^2}}-\left({\displaystyle \frac{{\rho }_0}{h}}+{\displaystyle \frac{E}{2L}}{\int }_0^L{\left|{\displaystyle \frac{\partial u}{\partial x}}\right|}^{2}dx\right){\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}=f(x,u),\end{array}$$

(4)

for free vibrations of elastic strings, which was proposed by G.Kirchhoff [2] in 1883 and considered theoretically or experimentally by several physicists [3,4]. Arosio and Panizzi [5] Introduced the research background and significance of this issue, and listed existing research results on this issue. Moreover, they considered the Cauchy problem for the quasilinear hyperbolic integro-differential equation.

In 2013, Fiscella and Valdinoci [6] proved the existence of non-negative solutions for a Kirchhoff type problem driven by a nonlocal integrodifferential operator:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-{M\left(\right|\left|u\right||}_Z^2){\mathcal{L}}_{K}u=\lambda f(x,u)+{\left|u\right|}^{2_s^{*}-2}u\mathrm{in}\Omega ,\hfill \\ u=0\mathrm{in}{\mathbb{R}}^N\setminus \Omega ,\hfill \end{array}\right.\end{array}$$

(5)

where ${\mathcal{L}}_K$ is a nonlocal operator, defined as follows:

$$\begin{array}{c}\hfill {\mathcal{L}}_K={\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}(u(x+y)+u(x-y)-2u\left(x\right))K\left(y\right)dy,\end{array}$$

(6)

for all $x\in {\mathbb{R}}^N$ and $K:{\mathbb{R}}^N\setminus \left\{0\right\}\to (0,+\infty )$ is a measurable function with some properties. When $K\left(x\right)={\left|x\right|}^{-N+2s}$, the problem (5) becomes

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-M\left({\int }_{{\mathbb{R}}^{2N}}{\displaystyle \frac{{|u\left(x\right)-u\left(y\right)|}^2}{{|x-y|}^{N+2s}}}dxdy\right){\mathcal{L}}_{K}u=\lambda f(x,u)+{\left|u\right|}^{2_s^{*}-2}u\mathrm{in}\Omega ,\hfill \\ u=0\mathrm{in}{\mathbb{R}}^N\setminus \Omega ,\hfill \end{array}\right.\end{array}$$

(7)

which is the fractional Kirchhoff-type problem. In 2014, Autuori, Fiscella, and Pucci [7] dealt with the existence and asymptotic behavior of non-negative solutions of (7) when M is zero at zero; that is, the problem is degenerate. In view of variational methods, Pucci and Saldi [8] studied the existence and multiplicity of nontrivial non-negative weak solutions of (7). Figueiredo, Ikoma and J$\acute{u}$nior [9] studied the existence and concentration results of a class of Kirchhoff-type equations with general nonlinear terms. Pucci and R$\check{a}$dulescu [10] mainly discussed the progress of nonlinear Kirchhoff problems and introduced the research background and application fields of the Kirchhoff problem, especially for the study of closed Riemann manifolds. In addition, one can refer to [11,12] to know more about the mathematical and physical background of (3).

In recent years, stationary Kirchhoff problems have been widely studied. However, there are few papers on the fractional Kirchhoff-type problem. Under the condition $g:\mathbb{R}\to \mathbb{R}$ is an odd function satisfying Berestycki–Lions-type assumptions, with the aid of minimax arguments, Ambrosio and Isernia [13] established a multiplicity result for

$$\begin{array}{c}\hfill \left(p+q(1+s){\int }_{{\mathbb{R}}^N}|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u|}^{2}dx\right){(-\Delta )}^{s}u=g\left(u\right)\mathrm{in}{\mathbb{R}}^N,\end{array}$$

(8)

provided that q is sufficiently small. Without the $\left(AR\right)$-condition, Mingqi Xiang, Binlin Zhang and Miaomiao Yang [14] used variational methods combined with a cut-off function technique to investigate the existence of radial solutions for a fractional Kirchhoff-type problem:

$$\begin{array}{c}\hfill \left(a+b{\left({\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u|}^{2}dx+{\int }_{{\mathbb{R}}^N}{|u|}^{2}dx\right)}^{\theta -1}\right)({(-\Delta )}^{s}u+u)=f\left(u\right)\mathrm{in}{\mathbb{R}}^N.\end{array}$$

(9)

Jin and Liu [15] considered the following fractional Kirchhoff equation

$$\begin{array}{c}\hfill \left(a+b{\int }_{{\mathbb{R}}^N}|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u|}^{2}dx\right){(-\Delta )}^{s}u+u=f\left(u\right)\mathrm{in}{\mathbb{R}}^N.\end{array}$$

(10)

and the existence of solutions to (10) without the $AR$-condition is proven when the parameter b is small. Moreover, they studied the asymptotic behavior of solutions as $b\to 0$. Li [16] considered the fractional Kirchhoff equation with critical growth:

$$\begin{array}{c}\hfill \left(a+b{\int }_{{\mathbb{R}}^N}|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u|}^{2}dx\right){(-\Delta )}^{s}u+u=f(x,u)+{\left|u\right|}^{2_s^{*}-2}u\mathrm{in}{\mathbb{R}}^3.\end{array}$$

(11)

and proved the existence of ground state solutions. Gu, Ming, and Yang [17] made use of variational methods and compactness analysis to investigate the existence of ground state solutions for a non-autonomous fractional-order Kirchhoff equation.

Motivated by the works above, we are concerned with (1). Let $G\left(u\right):={\int }_0^{u}g\left(s\right)ds$, $H\left(u\right):=g\left(u\right)u-2G\left(u\right)$ and $h=H^{\prime }$. Now, we make the following conditions:

(${\mathrm{A}}_0$) $g,h\in C\left(\mathbb{R}\right)$ and there exists a constant $C>0$ such that

$$\left|h\left(t\right)\right|\le C\left(\left|t\right|+{\left|t\right|}^{2_s^{*}-1}\right)\mathrm{for}t\in \mathbb{R}.$$

(${\mathrm{A}}_1$) $\eta :=\underset{\left|t\right|\to 0}{lim\; sup}{\displaystyle \frac{G\left(t\right)}{{\left|t\right|}^{{\overline{2}}_s}}}<\infty$.

(${\mathrm{A}}_2$) $\underset{\left|t\right|\to \infty }{lim}{\displaystyle \frac{G\left(t\right)}{{\left|t\right|}^{{\overline{2}}_s}}}=\infty$.

(${\mathrm{A}}_3$) $\underset{\left|t\right|\to \infty }{lim}{\displaystyle \frac{G\left(t\right)}{{\left|t\right|}^{2_s^{*}}}}=0$.

(${\mathrm{A}}_4$) $\left(2+{\displaystyle \frac{8s}{3}}\right)H\left(t\right)\le h\left(t\right)t$ for $t\in \mathbb{R}$.

(${\mathrm{A}}_5$) $\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(t\right)\le g\left(t\right)t\le 2_s^{*}G\left(t\right)$ for any $t\in \mathbb{R}$.

(${\mathrm{A}}_6$) $H\left({\zeta }_0\right)>0$ for some ${\zeta }_0\ne 0$.

Before we present our main results, for any $s\in (0,1)$, we recall the definition of the fractional Sobolev space

$$H^s\left({\mathbb{R}}^3\right)=\left\{u\in L^2\left({\mathbb{R}}^3\right):\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^2<+\infty \right\}$$

endowed with the norm

$${\left|\right|u\left|\right|}_{H^s\left({\mathbb{R}}^3\right)}^2={\int }_{{\mathbb{R}}^3}{\left(\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{|}^2+u^2)dx.$$

Now, we set

$$\mathcal{S}:=\{u\in H^s\left({\mathbb{R}}^3\right)|\left|\right|u|{|}_2^2=\rho \}.,$$

and

$$\mathcal{D}:=\{u\in H^s\left({\mathbb{R}}^3\right):|\left|u\right|{|}_2^2\le \rho \}.$$

The solutions to (1) are critical points of the energy functional

$$\begin{array}{cc}\hfill E\left(u\right)=& {\displaystyle \frac{a}{2}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+{\displaystyle \frac{b}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^4-{\int }_{{\mathbb{R}}^N}G\left(u\right)dx-{\displaystyle \frac{1}{2_s^{*}}}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}.\hfill \end{array}$$

(12)

where $G\left(u\right)={\int }_0^{u}g\left(s\right)ds$. From [18], we obtain that the following Poho$\breve{z}$aev identity

$$\begin{array}{c}\hfill {\displaystyle \frac{3-2s}{2}}\left(a+{b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^2\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2=3{\int }_{{\mathbb{R}}^3}G\left(u\right)dx+{\displaystyle \frac{3}{2_s^{*}}}{\int }_{{\mathbb{R}}^3}{\left|u\right|}^{2_s^{*}}dx.\end{array}$$

(13)

Furthermore, let $H\left(u\right)=g\left(u\right)u-2G\left(u\right)$ and $h=H^{\prime }$, we define the Poho$\breve{z}$aev manifold

$$\mathcal{P}=\{u\in H^s\left({\mathbb{R}}^3\right)\setminus \left\{0\right\}:P\left(u\right)=0\},$$

where

$$\begin{array}{c}\hfill P\left(u\right)=s\left(a+{b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^2\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2-{\displaystyle \frac{3}{2}}{\int }_{{\mathbb{R}}^3}H\left(u\right)dx-{s\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}.\end{array}$$

(14)

The constraint manifold $\mathcal{P}$ constitutes the complete solution space of nontrivial solutions to (1) and exhibits parameter invariance with respect to $\lambda$. All nontrivial solutions to (1) belongs to $\mathcal{S}\cap \mathcal{P}\subset \mathcal{D}\cap \mathcal{P}$. Therefore, any solution u that simultaneously satisfies (1) and achieves the energy minimization condition $E\left(u\right)={inf}_{\mathcal{S}\cap \mathcal{P}}E$ can be characterized as a normalized ground state solution to the system (1).

Remark 1.

Bieganowski and Mederski [19] showed the existence of ground state solutions to the problem $-\Delta u+\lambda u=g\left(u\right)$ with the $L^2$-norm constrained manifold. Moreover, Li, R$\check{a}$dulescu and Zhang [20] extended the results in [19] from the Sobolev subcritical setting to the Sobolev critical framework. Taking inspiration from the above paper, can we extend some of the results in the article to the fractional Kirchhoff Equation (1)? However, the emergence of a fractional operator greatly limits the range of dimensions studied and increases the difficulty of calculation. Furthermore, nonlocal terms greatly increase the difficulty of energy estimation. In our opinion, the range estimation of the minimization of energy functional cannot be directly applied to the analysis in the above article. Of course, we believe that with the continuous deepening of research, better methods will be found in the future to solve this problem.

Theorem 1.

Let (${\mathit{A}}_0$)–(${\mathit{A}}_6$) hold and

$$\begin{array}{c}\hfill 2_s^{*}\eta C_{s,{\overline{2}}_s,p}^{{\overline{2}}_s}{\rho }^{{\displaystyle \frac{4s-3}{3}}}<b,\end{array}$$

(15)

$$\begin{array}{c}\hfill a^{4s-3}{S}^{-3}<b^{2s}-b^{3-2s}\end{array}$$

(16)

hold. Then, there is $u\in \mathcal{D}\cap \mathcal{P}$, such that

$$\begin{array}{c}\hfill E\left(u\right)=\underset{\mathcal{D}\cap \mathcal{P}}{inf}E>0,\end{array}$$

(17)

and if, in addition, g is odd, then $\underset{\mathcal{D}\cap \mathcal{P}}{inf}E=\underset{\mathcal{S}\cap \mathcal{P}}{inf}E$ and $u\in \mathcal{D}\cap \mathcal{P}$ is a positive and radially symmetric normalized ground state solution to (1).

For any $\rho >0$, we define

$${\mathcal{D}}_{\rho }:=\{u\in H^s\left({\mathbb{R}}^3\right){:\left|\right|u\left|\right|}_2^2\le \rho \}$$

and

$${\mathcal{S}}_{\rho }:=\{u\in H^s\left({\mathbb{R}}^3\right){:\left|\right|u\left|\right|}_2^2=\rho \}.$$

Then, we obtain the asymptotic behavior of the ground state energy map $\rho \mapsto {inf}_{{\mathcal{S}}_{\rho }\cap \mathcal{P}}E$ as follows:

Theorem 2.

Under the assumptions of Theorem 1, then

• (a) the ground state energy map $\rho \mapsto \underset{{\mathcal{S}}_{\rho }\cap \mathcal{P}}{inf}E$ is strictly decreasing;
• (b) if $\eta \ne 0$,
  •   (1) the map $\rho \mapsto \underset{{\mathcal{S}}_{\rho }\cap \mathcal{P}}{inf}E$ is continuous;
  •   (2) $\underset{{\mathcal{S}}_{\rho }\cap \mathcal{P}}{inf}E\to 0^{+}$ as $\rho \to \infty$.

2. Preliminaries

First, we establish some estimates. Under assumptions (${\mathrm{A}}_0$), (${\mathrm{A}}_1$), (${\mathrm{A}}_3$), and (${\mathrm{A}}_5$), for each $\epsilon >0$, a constant $C_{\epsilon }>0$ can be found, which satisfying

$$\begin{array}{c}\hfill H\left(t\right)\le (2_s^{*}-2)G\left(u\right)\le (2_s^{*}-2)\left({\epsilon \left|t\right|}^{2_s^{*}}+{(\epsilon +\eta )\left|t\right|}^{{\overline{2}}_s}+C_{\epsilon }{\left|t\right|}^q\right)\end{array}$$

(18)

for any $t\in \mathbb{R}$, where $2\le q\le 2_s^{*}$. Alternatively, in view of (${\mathrm{A}}_5$), one obtains that

$$\begin{array}{c}\hfill G\left(t\right),H\left(t\right)\ge 0,\forall t\in \mathbb{R}.\end{array}$$

(19)

According to [21], we know

$$\begin{array}{c}\hfill S=\underset{D^{s,2}\left({\mathbb{R}}^3\right)\setminus \left\{0\right\}}{inf}{\displaystyle \frac{{\left|\right|(-\Delta )}^{\frac{s}{2}}u{\left|\right|}_2^2}{{\left|\right|u\left|\right|}_{2_s^{*}}^2}}\end{array}$$

(20)

where S is the best Sobolev constant and $D^{s,2}\left({\mathbb{R}}^3\right):=\{u\in L^{2_s^{*}}\left({\mathbb{R}}^3\right):\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2<\infty \}$.

It is known that [22] S is attained by the functions $\tilde{u}\left(x\right)=\kappa ({\mu }^2+|x-x_0{{|}^2)}^{-{\displaystyle \frac{3-2s}{2}}}$, $x\in {\mathbb{R}}^3$, where $\kappa \in \mathbb{R}\setminus \left\{0\right\},\kappa >0$ and $x_0\in {\mathbb{R}}^3$ are fixed constants. By [23], it is clear that $u^{*}$ is a solution to the problem

$$\begin{array}{c}\hfill {(-\Delta )}^{s}u={\left|u\right|}^{2_s^{*}-2}u\mathrm{in}{\mathbb{R}}^3\end{array}$$

(21)

and $\left|\right|u^{*}{\left|\right|}_{2_s^{*}}^{2_s^{*}}=S^{{\displaystyle \frac{3}{2s}}}$. For any $\epsilon >0$, set $U_{\epsilon }\left(x\right)={\epsilon }^{-{\displaystyle \frac{3-2s}{2}}}{u}^{*}\left({\displaystyle \frac{x}{\epsilon }}\right)$. Fix $\phi \in C_0^{\infty }({\mathbb{R}}^3,[0,1])$, such that $\mathrm{ssup}\phi \subset B_{2r}\left(0\right)$ and $\phi \left(x\right)=1$ for $x\in B_r\left(0\right)$. Let $v_{\epsilon }\left(x\right)=\phi \left(x\right)U_{\epsilon }\left(x\right)$. From [24] and [25], it follows that

$$\begin{array}{c}\hfill {\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_{\epsilon }\left(x\right){\left|\right|}_2^2\le S^{{\displaystyle \frac{3}{2s}}}+O\left({\epsilon }^{3-2s}\right),\end{array}$$

(22)

$$\begin{array}{c}\hfill \left|\right|v_{\epsilon }\left(x\right){\left|\right|}_{2_s^{*}}^{2_s^{*}}=S^{{\displaystyle \frac{3}{2s}}}+O\left({\epsilon }^3\right),\end{array}$$

(23)

$$\begin{array}{cc}\hfill \left|\right|v_{\epsilon }(x){\left|\right|}_p^p& =\left\{\begin{array}{c}O({\epsilon }^{{\displaystyle \frac{(3-2s)p}{2}}}),p<{\displaystyle \frac{3}{3-2s}},\hfill \\ O({\epsilon }^{{\displaystyle \frac{3(2-p)+2sp}{2}}}|log\epsilon |),p={\displaystyle \frac{3}{3-2s}},\hfill \\ O({\epsilon }^{{\displaystyle \frac{3(2-p)+2sp}{2}}}),{\displaystyle \frac{3}{3-2s}}<p<{\displaystyle \frac{6}{3-2s}}.\hfill \end{array}\right.\hfill \end{array}$$

(24)

Proposition 1

(Fractional Gagliardo–Nirenberg inequality [26]). Let $s\in ({\displaystyle \frac{3}{4}},1)$, $q\in (2,2_s^{*})$, then, for all $u\in H^s\left({\mathbb{R}}^3\right)$, there exists a constant $B_{q,s}>0$ depending on $p,s$ such that

$$\begin{array}{c}{\left|\right|u\left|\right|}_q^q\le B_{s,q}^q{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^{{\displaystyle \frac{N(q-2)}{2s}}}{\left|\right|u\left|\right|}_2^{q-{\displaystyle \frac{N(q-2)}{2s}}}.\hfill \end{array}$$

(25)

Proposition 2

([27]). Let $s\in (0,1)$. For any $u\in H^s\left({\mathbb{R}}^3\right)$, the following equation holds

$$\begin{array}{c}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2\le {\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}^{*}{\left|\right|}_2^2.\hfill \end{array}$$

(26)

3. Proof of Theorem 1

Lemma 1.

Assume that (${\mathit{A}}_0$), (${\mathit{A}}_1$), (${\mathit{A}}_3$), (${\mathit{A}}_5$), (${\mathit{A}}_6$) and (15) hold. For any $u\in \mathcal{D}\cap \mathcal{P}$, there exists $\delta >0$, such that ${\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^2\ge \delta .$

Proof. 

Taking any ${\overline{p}}_s<q<p_s^{*}$. For any $u\in \mathcal{D}\cap \mathcal{P}\subset \mathcal{D}$, from (14) and (18), we can obtain that

$$\begin{array}{cc}\hfill & s\left(a+{b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^2\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^2\hfill \\ & ={\displaystyle \frac{3}{2}}{\int }_{{\mathbb{R}}^3}[g\left(u\right)u-2G\left(u\right)]dx+{s\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & ={\displaystyle \frac{3}{2}}{\int }_{{\mathbb{R}}^3}H\left(u\right)dx+{s\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & \le {\displaystyle \frac{3}{2}}(2_s^{*}-2){\int }_{{\mathbb{R}}^3}\left({\epsilon \left|u\right|}^{2_s^{*}}+{(\epsilon +\eta )\left|u\right|}^{{\overline{2}}_s}+C_{\epsilon }{\left|u\right|}^q\right)dx+{s\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & =s(2_s^{*}\epsilon +1){\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}+2_s^{*}s(\epsilon +\eta ){\left|\right|u\left|\right|}_{{\overline{2}}_s}^{{\overline{2}}_s}+s{2}_s^{*}{C}_{\epsilon }{\left|\right|u\left|\right|}_q^q.\hfill \end{array}$$

In view of (20) and (25), we have

$$\begin{array}{cc}\hfill & {a\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+{b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^4\hfill \\ & \le (2_s^{*}\epsilon +1){\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}+2_s^{*}(\epsilon +\eta ){\left|\right|u\left|\right|}_{{\overline{2}}_s}^{{\overline{2}}_s}+2_s^{*}{C}_{\epsilon }{\left|\right|u\left|\right|}_q^q\hfill \\ & \le (2_s^{*}\epsilon +1)S^{-{\displaystyle \frac{2_s^{*}}{2}}}\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^{2_s^{*}}+2_s^{*}(\epsilon +\eta )B_{s,{\overline{2}}_s}^{{\overline{2}}_s}{\rho }^{{\displaystyle \frac{4s-3}{3}}}\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^4\hfill \\ & +2_s^{*}{C}_{\epsilon }{B}_{s,q}^q{\rho }^{q(1-{\gamma }_{q,s})}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^{q{\gamma }_{q,s}},\hfill \end{array}$$

that is,

$$\begin{array}{cc}\hfill & {a\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+(b-2_s^{*}(\epsilon +\eta )B_{s,{\overline{2}}_s}^{{\overline{2}}_s}{\rho }^{{\displaystyle \frac{4s-3}{3}}}){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^4\hfill \\ & \le (2_s^{*}\epsilon +1)S^{-{\displaystyle \frac{2_s^{*}}{2}}}\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^{2_s^{*}}+2_s^{*}{C}_{\epsilon }{B}_{s,q}^q{\rho }^{q(1-{\gamma }_{q,s})}\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^{q{\gamma }_{q,s}},\hfill \end{array}$$

As $q{\gamma }_{q,s}>4$ and $2_s^{*}\eta B_{s,{\overline{2}}_s}^{{\overline{2}}_s}{\rho }^{{\displaystyle \frac{4s-3}{3}}}<b$, taking $\epsilon <{\displaystyle \frac{b-2_s^{*}\eta B_{s,{\overline{2}}_s}^{{\overline{2}}_s}{\rho }^{\frac{4s-3}{N}}}{p_s^{*}\eta B_{s,{\overline{2}}_s}^{{\overline{2}}_s}{\rho }^{\frac{4s-3}{3}}}}$, we complete the proof. □

Let $u\in H^s\left({\mathbb{R}}^3\right)\setminus \left\{0\right\}$ satisfy

$$\begin{array}{c}\hfill 4\eta B_{s,{\overline{p}}_s}^{{\overline{p}}_s}{\left({\int }_{{\mathbb{R}}^3}{u}^{2}dx\right)}^{{\displaystyle \frac{4s-3}{3}}}<b\end{array}$$

(27)

For each $\lambda >0$, we define the scaling functional $\varphi \left(\lambda \right):=E({\lambda }^{{\displaystyle \frac{3}{2}}}u(\lambda ·))$. The following Lemma holds:

Lemma 2.

Suppose that $u\in H^s\left({\mathbb{R}}^N\right)\setminus \left\{0\right\}$ satisfies (27) and (${\mathrm{A}}_0$), (${\mathrm{A}}_1$), (${\mathrm{A}}_3$)–(${\mathrm{A}}_6$) hold. Then, there is an interval ${\lambda }_0\subset (0,+\infty )$, such that

• (i) $\varphi \left({\lambda }_0\right)\ge \varphi \left(\lambda \right)$ for any $\lambda \in (0,+\infty )$ and the strict inequality holds for $\lambda \in (0,+\infty )\setminus \left\{{\lambda }_0\right\}$.
• (ii)

  $$\begin{array}{c}\hfill P({\lambda }^{{\displaystyle \frac{3}{2}}}u(\lambda ·))\left\{\begin{array}{c}>0if\lambda \in (0,{\lambda }_0),\hfill \\ =0if\lambda ={\lambda }_0,\hfill \\ <0if\lambda \in ({\lambda }_0,+\infty ).\hfill \end{array}\right.\end{array}$$

  (28)

Proof. 

Let $u\in H^s\left({\mathbb{R}}^N\right)\setminus \left\{0\right\}$ satisfy (27). From the (${\mathrm{A}}_1$) and (25), one obtain

$$\begin{array}{cc}\hfill & \varphi \left(\lambda \right)=E\left({\lambda }^{{\displaystyle \frac{3}{2}}}u\left(\lambda x\right)\right)\hfill \\ & ={\displaystyle \frac{a{\lambda }^{2s}}{2}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+{\displaystyle \frac{b{\lambda }^{4s}}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^4-{\int }_{{\mathbb{R}}^3}{\displaystyle \frac{G\left({\lambda }^{\frac{3}{2}}u\left(x\right)\right)}{{\lambda }^3}}dx-{\displaystyle \frac{{\lambda }^{2_s^{*}s}}{2_s^{*}}}{\left|\right|u\left(x\right)\left|\right|}_{2_s^{*}}^{2_s^{*}}.\hfill \end{array}$$

Obviously, $\varphi \left(\lambda \right)\to 0$ as $\lambda \to 0^{+}$. Moreover, $\varphi \left(\lambda \right)\to -\infty$ as $\lambda \to \infty$ because of (19). Set $R:={\left|\right|u\left|\right|}_2^2=\left|\right|{\lambda }^{{\displaystyle \frac{3}{2}}}u(\lambda ·){\left|\right|}_2^2>0$. From (18), (20), (25), it is not difficult to find that for any $ϵ>0$, there exists $C_{ϵ}>0$, such that

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^N}G\left(u\right)dx& \le {(\epsilon +\eta )\left|\right|u\left|\right|}_{{\overline{p}}_s}^{{\overline{p}}_s}+C_{\epsilon }{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & \le (\epsilon +\eta )B_{s,{\overline{2}}_s}^{{\overline{2}}_s}{R}^{{\displaystyle \frac{4s-3}{3}}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^4+C_{\epsilon }{S}^{-{\displaystyle \frac{2_s^{*}}{2}}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^{2_s^{*}}.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill {\displaystyle \frac{\varphi \left(\lambda \right)}{{\lambda }^{4s}}}& ={\displaystyle \frac{a}{2{\lambda }^{2s}}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+{\displaystyle \frac{b}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^4-{\displaystyle \frac{1}{{\lambda }^{4s}}}{\int }_{{\mathbb{R}}^3}G\left({\lambda }^{{\displaystyle \frac{3}{2}}}u\left(\lambda x\right)\right)dx-{\displaystyle \frac{{\lambda }^{(2_s^{*}-4)s}}{2_s^{*}}}{\left|\right|u\left(x\right)\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & \ge {\displaystyle \frac{a}{2{\lambda }^{2s}}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+\left({\displaystyle \frac{b}{4}}-(\epsilon +\eta )B_{s,{\overline{2}}_s}^{{\overline{2}}_s}{R}^{{\displaystyle \frac{4s-3}{3}}}\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^4\hfill \\ & -{\lambda }^{(2_s^{*}-4)s}{C}_{\epsilon }{S}^{-{\displaystyle \frac{2_s^{*}}{2}}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^{2_s^{*}}-{\displaystyle \frac{{\lambda }^{(2_s^{*}-4)s}}{2_s^{*}}}{\left|\right|u\left(x\right)\left|\right|}_{2_s^{*}}^{2_s^{*}}.\hfill \end{array}$$

By (27), one can obtain $\varphi \left(\lambda \right)>0$ for small enough $\epsilon >0$ and $\lambda >0$. Hence, there exists some ${\lambda }_0>0$, such that ${\varphi }_{max}=\varphi \left({\lambda }_0\right)$. In particular, ${\varphi }^{\prime }\left({\lambda }_0\right)=0$. Since

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\varphi }^{\prime }\left(\lambda \right)}{{\lambda }^{4s-1}}}=& as{\lambda }^{-2s}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+{bs\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^4-{\displaystyle \frac{3}{2}}{\lambda }^{-3-4s}{\int }_{{\mathbb{R}}^3}H({\lambda }^{{\displaystyle \frac{3}{2}}}u)dx-s{\lambda }^{2_s^{*}s-4s}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \end{array}$$

and combined with the (14), we can easily see that $P({\lambda }_0^{{\displaystyle \frac{N}{p}}}u\left({\lambda }_{0}x\right))={\lambda }_0{\varphi }^{\prime }\left({\lambda }_0\right)=0$. Set

$$\begin{array}{cc}\hfill & m_1\left(\lambda \right):=as{\lambda }^{-2s}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+{bs\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^4,\hfill \\ & m_2\left(\lambda \right):={\displaystyle \frac{3}{2}}{\lambda }^{-3-4s}{\int }_{{\mathbb{R}}^3}H({\lambda }^{{\displaystyle \frac{N}{p}}}u)dx+s{\lambda }^{2_s^{*}s-4s}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}.\hfill \end{array}$$

It is easy to know that $m_1\left(\lambda \right)$ is strictly decreasing on $\lambda \in (0,+\infty )$. By virtue of (${\mathrm{A}}_4$), we have the function $m_2\left(\lambda \right)$, which is strictly increasing on $\lambda \in (0,+\infty )$ and tends to $+\infty$ as $n\to \infty$. This means that ${\varphi }^{\prime }\left(\lambda \right)>0$ if $\lambda \in (0,{\lambda }_0)$ and ${\varphi }^{\prime }\left(\lambda \right)<0$ if $\lambda \in ({\lambda }_0,+\infty )$. □

Lemma 3.

Assume that (${\mathit{A}}_0$), (${\mathit{A}}_1$), (${\mathit{A}}_3$)–(${\mathit{A}}_6$), and (15) hold. Then, E is coercive on $\mathcal{D}\cap \mathcal{P}$.

Proof. 

For any $u\in \mathcal{D}\cap \mathcal{P}$, according to Lemma 1, (1) and (${\mathrm{A}}_5$), one has

$$\begin{array}{cc}\hfill E\left(u\right)& =E\left(u\right)-{\displaystyle \frac{1}{4s}}P\left(u\right)\hfill \\ & ={\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left(u\right)u-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(u\right)\right)dx+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}}){\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}>0\hfill \end{array}$$

(29)

Then, E is bounded from below on $\mathcal{D}\cap \mathcal{P}$. Now, we suppose that there exits a sequence $\left\{u_n\right\}\subset \mathcal{D}\cap \mathcal{P}$, such that $\left|\right|u_n\left|\right|\to \infty$ and $E\left(u_n\right)$ is bounded from above. Since $u_n\in \mathcal{D}$, we see that ${\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2\to \infty$ as $n\to \infty$. Set ${\lambda }_n^s:={\displaystyle \frac{1}{{\left|\right|(-\Delta )}^{\frac{s}{2}}{u}_n{\left|\right|}_2}}>0$ and $v_n:={\lambda }_n^{{\displaystyle \frac{3}{2}}}{u}_n({\lambda }_n·)$. Obviously, ${\lambda }_n\to 0^{+}$ as $n\to \infty$. Hence, we obtain

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^N}|v_n{|}^{2}dx={\lambda }_n^3{\int }_{{\mathbb{R}}^3}|u_n\left({\lambda }_{n}x\right){|}^{2}dx={\int }_{{\mathbb{R}}^N}{\left|u_n\right|}^{2}dx\le \rho \end{array}$$

and

$${\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2={\lambda }_n^{2s}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2={\displaystyle \frac{1}{{\left|\right|(-\Delta )}^{\frac{s}{2}}{u}_n{\left|\right|}_2^2}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2=1,$$

which implies that $\left\{v_n\right\}\in \mathcal{D}$ is bounded in $H^s\left({\mathbb{R}}^N\right)$.

Set $l\left(z\right):={\int }_{B_1\left(z\right)}{\left|v_n\right|}^{2_s^{*}}dx$. From the integral absolute continuity and $v_n\ne 0$, we have that $l\left(z\right)$ is continuous on ${\mathbb{R}}^N$ and ${\int }_{B_1\left(z_0\right)}{\left|v_n\right|}^{2_s^{*}}dx:=\xi >0$ for some $z_0\in {\mathbb{R}}^3$. Then, we can find a large $R>0$, such that ${\int }_{{\mathbb{R}}^3\setminus B_R\left(0\right)}{\left|v_n\right|}^{2_s^{*}}dx<\xi$, which means $l\left(z\right)={\int }_{B_1\left(z\right)}{\left|v_n\right|}^{2_s^{*}}dx<\xi$ for any $z\in {\mathbb{R}}^3\setminus B_{R+1}\left(0\right)$ and ${sup}_{z\in {\mathbb{R}}^3}l\left(z\right)={sup}_{z\in {\overline{B}}_{R+1}\left(0\right)}l\left(z\right)$.

In view of the compactness of ${\overline{B}}_{R+1}\left(0\right)$, we deduce that there exits $y_n\in {\overline{B}}_{R+1}\left(0\right)$, such that $l\left(y_n\right)={sup}_{z\in {\overline{B}}_{R+1}\left(0\right)}l\left(z\right)$, that is,

$${\int }_{B_1\left(y_n\right)}|v_n{|}^{2_s^{*}}dx=\underset{z\in {\mathbb{R}}^3}{sup}{\int }_{B_1\left(z\right)}{\left|v_n\right|}^{2_s^{*}}dx.$$

In the following, we claim $\underset{n\to \infty }{lim}{\int }_{B_1\left(y_n\right)}{\left|v_n\right|}^{2_s^{*}}dx=0$, which we prove through contradiction with $\underset{n\to \infty }{lim\; sup}{\int }_{B_1\left(y_n\right)}{\left|v_n\right|}^{2_s^{*}}dx>0$. Then, $v\ne 0$, by virtue of (29) and (19), we have

$$\begin{array}{cc}\hfill 0<{\displaystyle \frac{E\left(u_n\right)}{{\left|\right|(-\Delta )}^{\frac{s}{2}}{u}_n{\left|\right|}_2^{p_s^{*}}}}& \le {\displaystyle \frac{a}{2}}{\lambda }_n^{(2_s^{*}-2)s}+{\displaystyle \frac{b}{4}}{\lambda }_n^{(2·2_s^{*}-2)s}-{\displaystyle \frac{1}{2_s^{*}}}\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & \le {\displaystyle \frac{a}{2}}{\lambda }_n^{(2_s^{*}-2)s}+{\displaystyle \frac{b}{4}}{\lambda }_n^{(2·2_s^{*}-2)s}-{\displaystyle \frac{1}{2_s^{*}}}{\int }_{B_1\left(y_n\right)}{\left|v_n\right|}^{2_s^{*}}dx<0\hfill \end{array}$$

if n is large enough. Obviously, it is impossible. Similar to [28], we can obtain $\underset{n\to \infty }{lim}{\int }_{B_1\left(y_n\right)}{\left|v_n\right|}^{2_s^{*}}dx=0$. It is not difficult to obtain $v_n\to 0$ in $L^{2+{\displaystyle \frac{8s}{3}}}$. Combining (18) and (19), for any $\lambda >0$, we can deduce that

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^N}G\left({\lambda }^{{\displaystyle \frac{3}{2}}}{v}_n\left(\lambda x\right)\right)dx\to 0\end{array}$$

as $n\to \infty$. It is obvious to know that $u_n\in \mathcal{D}\cap \mathcal{P}$ satisfies (27) when (15) holds. By Lemma 2, we have

$$\begin{array}{cc}\hfill E\left(u_n\right)& \ge E\left({\lambda }^{{\displaystyle \frac{3}{2}}}{v}_n\left(\lambda x\right)\right)={\displaystyle \frac{a{\lambda }^{2s}}{2}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2+{\displaystyle \frac{b{\lambda }^{4s}}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^4+o\left(1\right),\hfill \end{array}$$

which is a contradiction if $\lambda$ is large enough. □

Lemma 4.

Assume that $u\in H^s\left({\mathbb{R}}^3\right)\setminus \left\{0\right\}$ satisfies (27) and (${\mathit{A}}_0$)–(${\mathit{A}}_6$) hold. Then,

$$c:=\underset{\mathcal{D}\cap \mathcal{P}}{inf}E\left(u\right)<c^{*}:={\displaystyle \frac{a{S}^{\frac{3}{2s}}}{2}}{T}^{2s}+{\displaystyle \frac{b{S}^{\frac{3}{s}}}{4}}{T}^{4s}-{\displaystyle \frac{S^{\frac{3}{2s}}}{2_s^{*}}}{T}^{2_s^{*}s},$$

where $T>0$ is the unique maximum point of the function $\mathcal{A}\left(t\right)={\displaystyle \frac{a{S}^{\frac{3}{2s}}}{2}}{t}^{2s}+{\displaystyle \frac{b{S}^{\frac{3}{s}}}{4}}{t}^{4s}-{\displaystyle \frac{S^{\frac{3}{2s}}}{2_s^{*}}}{t}^{2_s^{*}s},t>0$.

Proof. 

From (22), (23), and (24), we set

$$\begin{array}{c}\hfill u_{\epsilon }\left(x\right)=({\rho }^{-{\displaystyle \frac{1}{2s}}}\left|\right|v_{\epsilon }{\left|\right|}_2^{{\displaystyle \frac{1}{s}}}{)}^{{\displaystyle \frac{3-2s}{2}}}{v}_{\epsilon }({\rho }^{-{\displaystyle \frac{1}{2s}}}\left|\right|v_{\epsilon }{\left|\right|}_2^{{\displaystyle \frac{1}{s}}}x).\end{array}$$

Hence, it is not difficult to find that $\left|\right|u_{\epsilon }{\left|\right|}_2^2=\rho$, that is, $u_{\epsilon }\in \mathcal{S}$. Moreover, one obtains

$$\begin{array}{cc}\hfill & {\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_{\epsilon }{\left|\right|}_2^2={\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_{\epsilon }{\left|\right|}_2^2\le S^{{\displaystyle \frac{3}{2s}}}+O\left({\epsilon }^{3-2s}\right),\hfill \\ & \left|\right|u_{\epsilon }{\left|\right|}_{2_s^{*}}^{2_s^{*}}=\left|\right|v_{\epsilon }{\left|\right|}_{2_s^{*}}^{2_s^{*}}=S^{{\displaystyle \frac{3}{2s}}}+O\left({\epsilon }^3\right)\hfill \end{array}$$

and

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^3}|u_{\epsilon }{|}^{{\overline{2}}_s}dx=({\rho }^{-{\displaystyle \frac{1}{2s}}}\left|\right|v_{\epsilon }{\left|\right|}_2^{{\displaystyle \frac{1}{s}}}{)}^{{\displaystyle \frac{2s(3-4s)}{3}}}{\int }_{{\mathbb{R}}^3}{\left|v_{\epsilon }\right|}^{{\overline{2}}_s}dx.\end{array}$$

From Lemma 2, it is clear that there exists ${\lambda }_{\epsilon }>0$, such that ${\lambda }_{\epsilon }^{{\displaystyle \frac{3}{2}}}{u}_{\epsilon }({\lambda }_{\epsilon }·)\in \mathcal{P}$. Due to $u_{\epsilon }\in \mathcal{S}$, we have ${\lambda }_{\epsilon }^{{\displaystyle \frac{3}{2}}}{u}_{\epsilon }({\lambda }_{\epsilon }·)\in \mathcal{S}\cap \mathcal{P}\subset \mathcal{D}\cap \mathcal{P}$. In view of (29) and (19), we have

$$\begin{array}{cc}\hfill 0<c& =\underset{\mathcal{D}\cap \mathcal{P}}{inf}E\hfill \\ & \le E\left({\lambda }_{\epsilon }^{{\displaystyle \frac{3}{2}}}{u}_{\epsilon }({\lambda }_{\epsilon }·)\right)\hfill \\ & ={\displaystyle \frac{a{\lambda }_{\epsilon }^{2s}}{2}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_{\epsilon }{\left|\right|}_2^2+{\displaystyle \frac{b{\lambda }_{\epsilon }^{4s}}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_{\epsilon }{\left|\right|}_2^4-{\int }_{{\mathbb{R}}^3}{\displaystyle \frac{G({\lambda }_{\epsilon }^{\frac{3}{2}}u\left(x\right))}{{\lambda }_{\epsilon }^3}}dx-{\displaystyle \frac{{\lambda }_{\epsilon }^{2_s^{*}s}}{2_s^{*}}}\left|\right|u_{\epsilon }{\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & \le {\displaystyle \frac{a{\lambda }_{\epsilon }^{2s}}{2}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_{\epsilon }{\left|\right|}_2^2+{\displaystyle \frac{b{\lambda }_{\epsilon }^{4s}}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_{\epsilon }{\left|\right|}_2^4-{\displaystyle \frac{{\lambda }_{\epsilon }^{2_s^{*}s}}{2_s^{*}}}\left|\right|u_{\epsilon }{\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \end{array}$$

Let

$$l\left({\lambda }_{\epsilon }\right)={\displaystyle \frac{a{\lambda }_{\epsilon }^{2s}}{2}}\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_{\epsilon }{\left|\right|}_2^2+{\displaystyle \frac{b{\lambda }_{\epsilon }^{4s}}{4}}\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_{\epsilon }{\left|\right|}_2^4-{\displaystyle \frac{{\lambda }_{\epsilon }^{2_s^{*}s}}{2_s^{*}}}\left|\right|u_{\epsilon }{\left|\right|}_{2_s^{*}}^{2_s^{*}}.$$

Obviously, $l\left({\lambda }_{\epsilon }\right)\to -\infty$ as ${\lambda }_{\epsilon }\to +\infty$ and $l\left({\lambda }_{\epsilon }\right)>0$ for ${\lambda }_{\epsilon }$ is small enough. Moreover, we can easily find that there exits two constants ${\lambda }_{{\epsilon }_1},{\lambda }_{{\epsilon }_2}\in {\mathbb{R}}^{+}$, independent of $\epsilon$, such that ${sup}_{{\lambda }_{\epsilon }>0}l\left({\lambda }_{\epsilon }\right)={sup}_{{\lambda }_{\epsilon }\in [{\lambda }_{{\epsilon }_1},{\lambda }_{{\epsilon }_2}]}l\left({\lambda }_{\epsilon }\right)$.

Making use of (${\mathrm{A}}_2$) and (19), for any $M>0$, there exists $C_M>$, such that $G\left(t\right)\ge {M\left|t\right|}^{{\overline{2}}_s}-C_M{\left|t\right|}^2$, $\forall t\in \mathbb{R}$. Therefore,

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^3}G\left({\lambda }_{\epsilon }^{{\displaystyle \frac{3}{2}}}{u}_{\epsilon }\left({\lambda }_{\epsilon }x\right)\right)& \ge M{\int }_{{\mathbb{R}}^3}|{\lambda }_{\epsilon }^{{\displaystyle \frac{3}{2}}}{u}_{\epsilon }\left({\lambda }_{\epsilon }x\right){|}^{{\overline{2}}_s}dx-C_M{\int }_{{\mathbb{R}}^3}{|{\lambda }_{\epsilon }^{{\displaystyle \frac{3}{2}}}{u}_{\epsilon }\left({\lambda }_{\epsilon }x\right)|}^{2}dx\hfill \\ & =M{\lambda }_{\epsilon }^{4s}{\int }_{{\mathbb{R}}^3}|u_{\epsilon }{|}^{{\overline{2}}_s}dx-C_M{\int }_{{\mathbb{R}}^3}{\left|u_{\epsilon }\right|}^{2}dx\hfill \\ & =M{\lambda }_{\epsilon }^{4s}({\rho }^{-{\displaystyle \frac{1}{2s}}}\left|\right|v_{\epsilon }{\left|\right|}_2^{{\displaystyle \frac{1}{s}}}{)}^{{\displaystyle \frac{2s(3-4s)}{3}}}{\int }_{{\mathbb{R}}^3}{\left|v_{\epsilon }\right|}^{{\overline{2}}_s}dx-C_M\rho \hfill \\ & \ge M{\lambda }_{\epsilon }^{4s}{\rho }^{{\displaystyle \frac{3-2s}{3}}}{K}_{1,s}{\epsilon }^{{\displaystyle \frac{{(4s-3)}^2}{3}}}-C_M\rho ,\hfill \end{array}$$

where $K_{1,s}>0$ is a constant. Combining (22) and (23), one obtains

$$\begin{array}{cc}\hfill l\left({\lambda }_{\epsilon }\right)& \le {\displaystyle \frac{a{\lambda }_{\epsilon }^{2s}}{2}}\left(S^{{\displaystyle \frac{3}{2s}}}+O\left({\epsilon }^{3-2s}\right)\right)+{\displaystyle \frac{b{\lambda }_{\epsilon }^{4s}}{4}}\left(S^{{\displaystyle \frac{3}{s}}}+O\left({\epsilon }^{3-2s}\right)\right)-{\displaystyle \frac{{\lambda }_{\epsilon }^{2_s^{*}s}}{2_s^{*}}}\left(S^{{\displaystyle \frac{3}{2s}}}+O\left({\epsilon }^3\right)\right)\hfill \end{array}$$

Since M can be arbitrarily large, we can obtain $c\le {sup}_{t\ge 0}\mathcal{A}\left(t\right),$ where

$$\begin{array}{c}\hfill \mathcal{A}\left(t\right)={\displaystyle \frac{a{S}^{\frac{3}{2s}}}{2}}{t}^{2s}+{\displaystyle \frac{b{S}^{\frac{3}{s}}}{4}}{t}^{4s}-{\displaystyle \frac{S^{\frac{3}{2s}}}{2_s^{*}}}{t}^{2_s^{*}s}.\end{array}$$

Then

$$\begin{array}{c}\hfill {\mathcal{A}}^{\prime }\left(t\right)=s{S}^{{\displaystyle \frac{3}{2s}}}{t}^{2s-1}\left(a+b{S}^{{\displaystyle \frac{3}{2s}}}{t}^{2s}-t^{(2_s^{*}-2)s}\right).\end{array}$$

Let ${\mathcal{A}}^{\prime }\left(t\right)=s{S}^{{\displaystyle \frac{3}{2s}}}{t}^{2s-1}\tilde{\mathcal{A}}\left(t\right)$, where

$$\begin{array}{cc}\hfill & \tilde{\mathcal{A}}\left(t\right)=a+b{S}^{{\displaystyle \frac{3}{2s}}}{t}^{2s}-t^{(2_s^{*}-2)s},\hfill \\ & {\tilde{\mathcal{A}}}^{\prime }\left(t\right)=s{t}^{2s-1}\left(2b{S}^{{\displaystyle \frac{3}{2s}}}-(2_s^{*}-2)t^{(2_s^{*}-4)s}\right).\hfill \end{array}$$

Since $s\in ({\displaystyle \frac{3}{4}},1)$, we can find a unique T, such that $\tilde{\mathcal{A}}\left(t\right)>0$ for $t\in (0,T)$ and $\tilde{\mathcal{A}}\left(t\right)<0$ for $t>T$. Then, ${\mathcal{A}}^{\prime }\left(T\right)=0$ and T is the unique maximum point of $\mathcal{A}\left(t\right)$. □

Due to

$$\underset{t\to 0}{lim}{\displaystyle \frac{H\left(t\right)}{{\left|t\right|}^2}}=\underset{\left|t\right|\to \infty }{lim}{\displaystyle \frac{H\left(t\right)}{{\left|t\right|}^{2_s^{*}}}}=0,$$

we have the following profile decomposition.

Theorem 3

(Profile decomposition). Assume that $\left\{u_n\right\}$ is bounded in $H^s\left({\mathbb{R}}^3\right)$. Then, there exists sequence ${\left\{{\tilde{u}}_i\right\}}_{i=0}^{\infty }\subset H^s\left({\mathbb{R}}^3\right)$, ${\left\{y_n^i\right\}}_{i=0}^{\infty }\subset {\mathbb{R}}^3$ for any $n\ge 1$, such that $y_0^n=0$, $|y_n^i-y_n^j|\to +\infty$ as $n\to \infty$ for $i\ne j$, and up to a sequence, the following conditions hold for any $i\ge 0$:

$$\begin{array}{c}\hfill u_n(·+y_n^i)\rightharpoonup {\tilde{u}}_i\mathrm{in}{H}^s\left({\mathbb{R}}^3\right)\mathrm{as}n\to \infty ,\end{array}$$

(30)

$$\begin{array}{c}\hfill A^2:=\underset{n\to \infty }{lim}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2=\sum _{j=0}^i{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2+\underset{n\to \infty }{lim}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n^i{\left|\right|}_2^2,\end{array}$$

(31)

where $v_n^i(·):=u_n-{\sum }_{j=0}^i{\tilde{u}}_j(·-y_n^j)$ and

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}{\int }_{{\mathbb{R}}^N}H\left(u_n\right)dx=\sum _{j=0}^{\infty }H\left({\tilde{u}}_j\right)dx,\end{array}$$

(32)

For more details, please refer to [19,20,29,30,31,32,33].

Lemma 5.

Suppose that (${\mathit{A}}_0$)–(${\mathit{A}}_6$) and (15) hold. Then, $c:=\underset{\mathcal{D}\cap \mathcal{P}}{inf}E$ is attained. If, in addition, g is odd, then ρ is attained by a nonnegative and radially symmetric function in $\mathcal{D}\cap \mathcal{P}$.

Proof. 

From Lemma 3, for any sequence $\left\{u_n\right\}\subset \mathcal{D}\cap \mathcal{P}$, such that $E\left(u_n\right)\to c$, we know that $\left\{u_n\right\}$ is bounded in $H^s\left({\mathbb{R}}^3\right)$. Set $\mathcal{M}:=\{i\ge 0:{\tilde{u}}_i\ne 0\}$. Then, we claim that $\mathcal{M}\ne 0$. We prove this through contradiction with $\mathcal{M}=0$.

In view of (19) and (32), we have $\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^N}H\left(u_n\right)dx=0$. Therefore, it is easy to know

$$\left(a+{b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2-\left|\right|u_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}=o\left(1\right).$$

By Lemma 1, up to a subsequence, denote by $l\ge 0$, such that

$$\begin{array}{c}\hfill l:=\underset{n\to \infty }{lim}\left|\right|u_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}=\underset{n\to \infty }{lim}\left(a+{b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2\end{array}$$

and

$$\begin{array}{c}\hfill A^2:=\underset{n\to \infty }{lim}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2>0.\end{array}$$

From (20), we have

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{S}^{-{\displaystyle \frac{2_s^{*}}{2}}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^{2_s^{*}}\ge \underset{n\to \infty }{lim}\left(a+{b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2.\end{array}$$

(33)

Then

$$\begin{array}{cc}\hfill S^{-{\displaystyle \frac{2_s^{*}}{2}}}{A}^{2_s^{*}}& \ge a{A}^2+b{A}^4>b{A}^4.\hfill \end{array}$$

(34)

that is,

$$\begin{array}{c}\hfill A^{2_s^{*}-4}>b{S}^{{\displaystyle \frac{2_s^{*}}{2}}}\end{array}$$

(35)

Making use of Lemma 4, one obtains

$$\begin{array}{c}\hfill \left\{\begin{array}{c}a+b{S}^{{\displaystyle \frac{3}{2s}}}{T}^{2s}-T^{(2_s^{*}-2)s}=0,\hfill \\ c^{*}={\displaystyle \frac{a{S}^{\frac{3}{2s}}}{2}}{T}^{2s}+{\displaystyle \frac{b{S}^{\frac{3}{s}}}{4}}{T}^{4s}-{\displaystyle \frac{S^{\frac{3}{2s}}}{2_s^{*}}}{T}^{2_s^{*}s}.\hfill \end{array}\right.\end{array}$$

Then

$$\begin{array}{cc}\hfill c^{*}& ={\displaystyle \frac{a{S}^{\frac{3}{2s}}}{2}}{T}^{2s}+{\displaystyle \frac{b{S}^{\frac{3}{s}}}{4}}{T}^{4s}-{\displaystyle \frac{S^{\frac{3}{2s}}}{2_s^{*}}}{T}^{2_s^{*}s}\hfill \\ & =S^{{\displaystyle \frac{3}{2s}}}{T}^{2s}\left({\displaystyle \frac{a}{2}}+{\displaystyle \frac{b}{4}}{S}^{{\displaystyle \frac{3}{2s}}}{T}^{2s}-{\displaystyle \frac{1}{2_s^{*}}}{T}^{(2_s^{*}-2)s}\right)\hfill \\ & =S^{{\displaystyle \frac{3}{2s}}}{T}^{2s}\left(\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2_s^{*}}}\right)a+\left({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}}\right)b{S}^{{\displaystyle \frac{3}{2s}}}{T}^{2s}\right).\hfill \end{array}$$

Consequently,

$$\begin{array}{cc}\hfill E\left(u_n\right)& =\left({\displaystyle \frac{a}{2}}+{\displaystyle \frac{b}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2-{\displaystyle \frac{1}{2_s^{*}}}\left(a+{b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2+o\left(1\right)\hfill \\ & =\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2_s^{*}}}\right){a\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2+\left({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}}\right){b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^4+o\left(1\right).\hfill \\ & <\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2_s^{*}}}\right)a{S}^{{\displaystyle \frac{3}{2s}}}{T}^{2s}+\left({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}}\right)b{S}^{{\displaystyle \frac{3}{s}}}{T}^{4s}.\hfill \end{array}$$

Therefore,

$$\begin{array}{c}\hfill A^2=\underset{n\to \infty }{lim}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2\le S^{{\displaystyle \frac{3}{2s}}}{T}^{2s}.\end{array}$$

(36)

Combining (35), we have

$$\begin{array}{c}\hfill S^{{\displaystyle \frac{3}{2s}}}{T}^{2s}>{(b{S}^{{\displaystyle \frac{2_s^{*}}{2}}})}^{{\displaystyle \frac{2}{2_s^{*}-4}}},\end{array}$$

(37)

that is,

$$\begin{array}{c}\hfill T>b^{{\displaystyle \frac{3-2s}{2s(4s-3)}}}{S}^{{\displaystyle \frac{3(3-2s)}{4s^2(4s-3)}}}.\end{array}$$

(38)

It follows from (16) and (38) that

$$\begin{array}{c}\hfill \tilde{\mathcal{A}}(b^{{\displaystyle \frac{3-2s}{2s(4s-3)}}}{S}^{{\displaystyle \frac{3(3-2s)}{4s^2(4s-3)}}})=a+b^{{\displaystyle \frac{3-2s}{4s-3}}}{S}^{{\displaystyle \frac{3}{4s-3}}}-b^{{\displaystyle \frac{2s}{4s-3}}}{S}^{{\displaystyle \frac{3}{4s-3}}}<0,\end{array}$$

(39)

which is a contradiction to Lemma 4. Thus, $\mathcal{M}\ne 0$.

In the following, we claim that for every $i\ge 0$, it holds that

$$\begin{array}{c}\hfill u_n(·+y_n^i)\to {\tilde{u}}_i,\end{array}$$

(40)

in $D^{s,2}\left({\mathbb{R}}^3\right)$. Assume that there exists $i\ge 0$, such that

$$\begin{array}{c}\hfill \mathcal{B}:=\underset{n\to \infty }{lim}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2=\underset{n\to \infty }{lim}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}(u_n(·+y_n^i)-{\tilde{u}}_i){\left|\right|}_2^2>0\end{array}$$

and

$$\begin{array}{c}\hfill {(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2>{\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left({\tilde{u}}_i\right)dx+\left|\right|{\tilde{u}}_i{\left|\right|}_{2_s^{*}}^{2_s^{*}}\end{array}$$

(41)

where $v_n(·):=u_n(·+y_n^i)-{\tilde{u}}_i(·)$. Set $\mathcal{K}(·)={\displaystyle \frac{3}{2s}}H(·)+{|·|}^{2_s^{*}}$, in view of Vitali’s convergence theorem, one obtains

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^3}(\mathcal{K}\left(u_n\right)-\mathcal{K}\left(v_n\right))dx& ={\int }_{{\mathbb{R}}^3}{\int }_0^1-{\displaystyle \frac{d}{ds}}\mathcal{K}(u_n-t{\tilde{u}}_i)dtdx\hfill \\ & ={\int }_{{\mathbb{R}}^3}{\int }_0^{1}k(u_n-t{\tilde{u}}_i){\tilde{u}}_{i}dtdx\hfill \\ & \to {\int }_0^1{\int }_{{\mathbb{R}}^3}k({\tilde{u}}_i-t{\tilde{u}}_i){\tilde{u}}_{i}dtdx\hfill \\ & ={\int }_0^1{\int }_{{\mathbb{R}}^3}-{\displaystyle \frac{d}{ds}}\mathcal{K}(u_n-t{\tilde{u}}_i)dtdx={\int }_{{\mathbb{R}}^3}\mathcal{K}\left({\tilde{u}}_i\right)dx\hfill \end{array}$$

as $n\to \infty$. There holds that

$$\begin{array}{cc}\hfill & {(a+b(\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2+{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2{\left)\right)\left(\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2+{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2)\hfill \\ & ={\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left(v_n\right)dx+\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}+{\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left({\tilde{u}}_i\right)dx+\left|\right|{\tilde{u}}_i{\left|\right|}_{2_s^{*}}^{2_s^{*}}+o\left(1\right).\hfill \end{array}$$

Combining with (41), it is clear to know that

$$\begin{array}{c}\hfill {(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2\le {\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left(v_n\right)dx+\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}+o\left(1\right)\end{array}$$

(42)

In view of (${\mathrm{A}}_6$) and arguing as in [32], for any $R>0$, one can find a radial function $u\in H_0^s\left(B(0,R)\right)\cap L^{\infty }\left(B(0,R)\right)$, such that ${\int }_{{\mathbb{R}}^N}H\left(u\right)dx>0$. We define $r(·)$ as

$$\begin{array}{c}\hfill r\left(u\right):={\left({\displaystyle \frac{-a+\sqrt{a^2+4b(\frac{3}{2s}{\int }_{{\mathbb{R}}^3}H\left(u\right)dx+{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}})}{{2b\left|\right|(-\Delta )}^{\frac{s}{2}}u{\left|\right|}_2^2}}\right)}^{{\displaystyle \frac{1}{2s}}}\end{array}$$

(43)

and $R_n=r\left(v_n\right)$. Obviously, we can see that $u\left(r\right(u)·)\in \mathcal{P}$, so $\mathcal{P}$ is nonempty.

Now, we claim that $R_n\to 1$ as $n\to \infty$. If

$$\begin{array}{c}\hfill {(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2>{\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left(v_n\right)dx+\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}\end{array}$$

holds for a.e. on n, by (42), it is easy to know $R_n\to 1$ as $n\to \infty$.

On the other hand,

$$\begin{array}{c}\hfill {(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2\le {\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left(v_n\right)dx+\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}\end{array}$$

implies that $R_n\ge 1$. Since $\underset{n\to \infty }{lim}\left(\right||u_n{\left|\right|}_2^2-\left|\right|v_n{\left|\right|}_2^2)=||{\tilde{u}}_i{\left|\right|}_2^2>0$, we have $v_n\in \mathcal{D}$ and $v_n(R_n·)\in \mathcal{D}\cap \mathcal{P}$ for a.e. on n.

Making use of the Br$\acute{e}$zis-Lieb Lemma, one obtains

$$\begin{array}{cc}\hfill c& =\underset{\mathcal{D}\cap \mathcal{P}}{inf}E\le E\left(v_n(R_n·)\right)\hfill \\ & =E\left(v_n(R_n·)\right)-{\displaystyle \frac{1}{4s}}P\left(v_n(R_n·)\right)\hfill \\ & ={\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n(R_n·){\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|v_n(R_n·){\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left(v_n(R_n·)\right)v_n(R_n·)-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(v_n(R_n·)\right)\right)dx\hfill \\ & ={\displaystyle \frac{a}{4}}{R}_n^{-3+2s}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})R_n^{-3}\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}+{\displaystyle \frac{3}{8s}}{R}_n^{-3}{\int }_{{\mathbb{R}}^3}\left(g\left(v_n\right)v_n-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(v_n\right)\right)dx\hfill \\ & \le {\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}+{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left(v_n\right)v_n-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(v_n\right)\right)dx\hfill \\ & ={\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2-{\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|u_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}-({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|{\tilde{u}}_i{\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left(u_n\right)u_n-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(u_n\right)\right)dx-{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left({\tilde{u}}_i\right){\tilde{u}}_i-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left({\tilde{u}}_i\right)\right)dx+o\left(1\right)\hfill \\ & \le {\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|u_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}+{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left(u_n\right)u_n-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(u_n\right)\right)dx+o\left(1\right)\hfill \\ & =E\left(u_n\right)-{\displaystyle \frac{1}{4s}}P\left(u_n\right)+o\left(1\right)=E\left(u_n\right)+o\left(1\right)=c+o\left(1\right)\hfill \end{array}$$

which implies that $R_n\to 1$ as $n\to \infty$.

From Theorem 3, it is clear that

$$\begin{array}{c}\hfill {(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2={\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left(v_n\right)dx+\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}+o\left(1\right)=\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}+o\left(1\right).\end{array}$$

Similar to the above analysis, we can easily know that $\underset{n\to \infty }{lim}E\left(v_n\right)\ge c^{*}$. Moreover,

$$\begin{array}{cc}\hfill E\left({\tilde{u}}_i\right)& =E\left({\tilde{u}}_i\right)-{\displaystyle \frac{1}{4s}}P\left({\tilde{u}}_i\right)\hfill \\ & ={\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|{\tilde{u}}_i{\left|\right|}_{2_s^{*}}^{2_s^{*}}+{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left({\tilde{u}}_i\right){\tilde{u}}_i-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left({\tilde{u}}_i\right)\right)dx\ge 0,\hfill \end{array}$$

since $E\left(u_n\right)=E\left(v_n\right)+E\left({\tilde{u}}_i\right)+o\left(1\right)$ and (${\mathrm{A}}_0$) hold. Hence, we have

$$\begin{array}{c}\hfill c=\underset{n\to \infty }{lim}E\left(u_n\right)=\underset{n\to \infty }{lim}\left(E\left(v_n\right)+E\left({\tilde{u}}_i\right)\right)=\underset{n\to \infty }{lim}E\left(v_n\right)+E\left({\tilde{u}}_i\right)\ge c^{*},\end{array}$$

which is a contradiction to Lemma 4. Hence, we obtain

$$\begin{array}{c}\hfill {0<(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2\le {\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left({\tilde{u}}_i\right)dx+\left|\right|{\tilde{u}}_i{\left|\right|}_{2_s^{*}}^{2_s^{*}},\end{array}$$

which yields that $r\left({\tilde{u}}_i\right)\ge 1$ and ${\tilde{u}}_i(r\left({\tilde{u}}_i\right)·)\in \mathcal{P}\cap \mathcal{D}$.

If $r\left({\tilde{u}}_i\right)>1$, then passing to a subsequence $u_n(·+y_n^i)\to {\tilde{u}}_i\left(x\right)$ for a.e. $x\in {\mathbb{R}}^3$. Using the Fatou Lemma, it is not difficult to deduce that

$$\begin{array}{cc}\hfill c=\underset{\mathcal{D}\cap \mathcal{P}}{inf}E& \le E\left({\tilde{u}}_i\left(r\left({\tilde{u}}_i\right)x\right)\right)=E\left({\tilde{u}}_i\left(r\left({\tilde{u}}_i\right)x\right)\right)-{\displaystyle \frac{1}{4s}}P\left({\tilde{u}}_i\left(r\left({\tilde{u}}_i\right)x\right)\right)\hfill \\ & ={\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i\left(r\left({\tilde{u}}_i\right)x\right){\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|{\tilde{u}}_i\left(r\left({\tilde{u}}_i\right)x\right){\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left({\tilde{u}}_i\left(r\left({\tilde{u}}_i\right)x\right)\right){\tilde{u}}_i\left(r\left({\tilde{u}}_i\right)x\right)-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left({\tilde{u}}_i\left(r\left({\tilde{u}}_i\right)x\right)\right)\right)dx\hfill \\ & ={\displaystyle \frac{a}{4}}r{\left({\tilde{u}}_i\right)}^{-3+2s}\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i\left(x\right){\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})r{\left({\tilde{u}}_i\right)}^{-3}\left|\right|{\tilde{u}}_i\left(x\right){\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}r{\left({\tilde{u}}_i\right)}^{-3}{\int }_{{\mathbb{R}}^3}\left(g\left({\tilde{u}}_i\left(x\right)\right){\tilde{u}}_i\left(x\right)-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left({\tilde{u}}_i\left(x\right)\right)\right)dx\hfill \\ & <{\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left(x\right)\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|{\tilde{u}}_i\left(x\right){\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left({\tilde{u}}_i\left(x\right)\right){\tilde{u}}_i\left(x\right)-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left({\tilde{u}}_i\left(x\right)\right)\right)dx\hfill \\ & \le \underset{n\to \infty }{lim\; inf}{\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n(·+y_n^i){\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|u_n(·+y_n^i){\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left(u_n(·+y_n^i)\right)u_n(·+y_n^i)-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(u_n(·+y_n^i)\right)\right)dx\hfill \\ & \le \underset{n\to \infty }{lim}E\left(u_n\right)=c,\hfill \end{array}$$

which is a contradiction. Thus, $r\left({\tilde{u}}_i\right)=1$, ${\tilde{u}}_i\in \mathcal{D}\cap \mathcal{P}$, and $u_n(·+y_n^i)\to {\tilde{u}}_i\left(x\right)$ in $D^{s,2}\left({\mathbb{R}}^3\right)$. Moreover, by (19), we have

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}H\left(u_n(·+y_n^i)\right)dx={\int }_{{\mathbb{R}}^3}H\left({\tilde{u}}_i\right)dx\end{array}$$

and

$$\underset{n\to \infty }{lim}E\left(u_n\right)=E\left({\tilde{u}}_i\right)=c.$$

In addition, we suppose that g is odd. Then, G and H are even, so that $G\left(\right|u\left|\right)=G\left(u\right)$ and $H\left(\right|u\left|\right)=H\left(u\right)$ for all $u\in H^s\left({\mathbb{R}}^3\right)$. Define ${\tilde{v}}_i:={\left|{\tilde{u}}_i\right|}^{*}$ as the Schwarz symmetrization of $|{\tilde{u}}_i|$. Then $\left|\right|{\tilde{v}}_i{\left|\right|}_2=\left|\right|{\tilde{u}}_i{\left|\right|}_2$, that is, ${\tilde{v}}_i\in \mathcal{D}$. Furthermore, due to

$$\begin{array}{cc}\hfill {\displaystyle \frac{a}{2}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{v}}_i{\left|\right|}_2^2+{\displaystyle \frac{b}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{v}}_i{\left|\right|}_2^4& \le {\displaystyle \frac{a}{2}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^2+{\displaystyle \frac{b}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{\tilde{u}}_i{\left|\right|}_2^4\hfill \\ & ={\int }_{{\mathbb{R}}^N}G\left({\tilde{u}}_i\right)dx+{\displaystyle \frac{1}{2_s^{*}}}\left|\right|{\tilde{u}}_i{\left|\right|}_{2_s^{*}}^{2_s^{*}}={\int }_{{\mathbb{R}}^N}G\left({\tilde{v}}_i\right)dx+{\displaystyle \frac{1}{2_s^{*}}}\left|\right|{\tilde{v}}_i{\left|\right|}_{2_s^{*}}^{2_s^{*}},\hfill \end{array}$$

we obtain that $r\left({\tilde{v}}_i\right)\ge 1$ and ${\tilde{v}}_i\left(r\left({\tilde{v}}_i\right)\right)\in \mathcal{P}$, where $r(·)$ is defined as (43). Using a similar argument as before, we can obtain that $r\left({\tilde{v}}_i\right)=1$, so ${\tilde{v}}_i\in \mathcal{P}$. It is clear that $E\left({\tilde{v}}_i\right)=c$ and ${\tilde{v}}_i$ is radially symmetric. □

Lemma 6.

Suppose that (${\mathit{A}}_0$)–(${\mathit{A}}_6$), and (15) hold. If g is odd, $\underset{\mathcal{S}\cap \mathcal{P}}{inf}E<E\left(u\right)$ for any $u\in (\mathcal{D}\setminus \mathcal{S})\cap \mathcal{P}$.

Proof. 

Suppose, by contradiction, that there exits $\tilde{u}\in (\mathcal{D}\setminus \mathcal{S})\cap \mathcal{P}$, such that

$$\begin{array}{c}\hfill \underset{\mathcal{D}\cap \mathcal{P}}{inf}E=c=E\left(\tilde{u}\right)\le \underset{\mathcal{S}\cap \mathcal{P}}{inf}E.\end{array}$$

which implies that $\tilde{u}$ is a local minimizer for E on $\mathcal{D}\cap \mathcal{P}$. Since $\mathcal{D}\cap \mathcal{P}$ is an open set in $\mathcal{P}$, we know that $\tilde{u}$ is a local minimizer for E on $\mathcal{P}$. This resules in a Lagrange multiplier $\mu \in \mathbb{R}$, such that

$$\begin{array}{c}\hfill E^{\prime }\left(\tilde{u}\right)v+\mu P^{\prime }\left(\tilde{u}\right)v=0,\end{array}$$

for any $v\in H^s\left({\mathbb{R}}^3\right)$. Namely, $\tilde{u}$ is a weak solution to

$$\begin{array}{c}\hfill {(a(1+\mu )+b(1+2\mu )||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2{)(-\Delta )}^s\tilde{u}=g\left(\tilde{u}\right)+{\displaystyle \frac{3}{4s}}\mu h\left(\tilde{u}\right)+\left(1+{\displaystyle \frac{2_s^{*}}{2}}\mu \right){\left|\tilde{u}\right|}^{2_s^{*}-2}\tilde{u}.\end{array}$$

In particular, $\tilde{u}$ satisfies the following Nehari-type identity

$$\begin{array}{cc}\hfill & {(a(1+\mu )+b(1+2\mu )||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2\hfill \\ & ={\int }_{{\mathbb{R}}^3}\left(g\left(\tilde{u}\right)\tilde{u}+{\displaystyle \frac{3}{4s}}\mu h\left(\tilde{u}\right)\tilde{u}+\left(1+{\displaystyle \frac{2_s^{*}}{2}}\mu \right){\left|\tilde{u}\right|}^{2_s^{*}}\right)dx.\hfill \end{array}$$

(44)

Moreover, since $\tilde{u}\in \mathcal{P}$, then $P\left(\tilde{u}\right)=0$, that is,

$$\begin{array}{c}\hfill \left(a+{b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2\right){\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2={\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left(\tilde{u}\right)dx+\left|\right|\tilde{u}{\left|\right|}_{2_s^{*}}^{2_s^{*}}.\end{array}$$

(45)

On the other hand, $\tilde{u}$ satisfies Pohozaev and Nehari identities. Therefore,

$$\begin{array}{cc}\hfill & {(a(1+\mu )+b(1+2\mu )||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2\hfill \\ & ={\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}\left(H\left(\tilde{u}\right)+{\displaystyle \frac{3}{4s}}\mu (h\left(\tilde{u}\right)\tilde{u}-2H\left(\tilde{u}\right))+\left(1-{\displaystyle \frac{2}{2_s^{*}}}\right)\left(1+{\displaystyle \frac{2_s^{*}}{2}}\mu \right){\left|\tilde{u}\right|}^{2_s^{*}}\right)dx\hfill \end{array}$$

Combining these two identities, it is clear to know

$$\begin{array}{cc}\hfill & \mu \left({a\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2+{2b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^4-{\displaystyle \frac{9}{8s^2}}{\int }_{{\mathbb{R}}^3}\left(h\left(\tilde{u}\right)\tilde{u}-2H\left(\tilde{u}\right)\right)dx-{\displaystyle \frac{3}{3-2s}}\left|\right|\tilde{u}{\left|\right|}_{2_s^{*}}^{2_s^{*}}\right)=0\hfill \end{array}$$

By (${\mathrm{A}}_6$) and (45), we have

$$\begin{array}{cc}\hfill & \left({a\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2+{2b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^4-{\displaystyle \frac{9}{8s^2}}{\int }_{{\mathbb{R}}^3}\left(h\left(\tilde{u}\right)\tilde{u}-2H\left(\tilde{u}\right)\right)dx-{\displaystyle \frac{3}{3-2s}}\left|\right|\tilde{u}{\left|\right|}_{2_s^{*}}^{2_s^{*}}\right)\hfill \\ & \le \left(-{a\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2+{\displaystyle \frac{3-4s}{3-2s}}\left|\right|\tilde{u}{\left|\right|}_{2_s^{*}}^{2_s^{*}}\right)<0.\hfill \end{array}$$

which indicates $\mu =0$. Hence, $\tilde{u}$ is a weak solution to

$$\begin{array}{c}\hfill {(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2{)(-\Delta )}^s\tilde{u}=g\left(\tilde{u}\right)+{\left|\tilde{u}\right|}^{2_s^{*}-2}\tilde{u}.\end{array}$$

(46)

and

$$\begin{array}{c}\hfill {(a+b||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2={\int }_{{\mathbb{R}}^3}g\left(\tilde{u}\right)\tilde{u}dx+\left|\right|\tilde{u}{\left|\right|}_{2_s^{*}}^{2_s^{*}}.\end{array}$$

By (45), we have

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^3}g\left(\tilde{u}\right)\tilde{u}dx={\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}H\left(\tilde{u}\right)dx,\end{array}$$

i.e.,

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^3}\left(g\left(\tilde{u}\right)\tilde{u}-2_s^{*}G\left(\tilde{u}\right)\right)dx=0.\end{array}$$

From the elliptic regularity theory and (${\mathrm{A}}_6$), $\tilde{u}$ is continuous in ${\mathbb{R}}^3$ and $g\left(\tilde{u}\right)\tilde{u}=2_s^{*}G\left(\tilde{u}\right)$ for $x\in {\mathbb{R}}^3$. Due to $\tilde{u}\in H^s\left({\mathbb{R}}^3\right)$, there is an open interval $\Omega \in \mathbb{R}$ such that $0\in \overline{\Omega }$ and $2G\left(u\right)=g\left(u\right)u$ for $u\in \overline{\Omega }$. If g is odd, in the light of Lemma 5, we may assume that $\tilde{u}$ is nonnegative and radially symmetric. However, it is impossible to show that $\tilde{u}$ is a solution to

$$\begin{array}{c}\hfill (a+b{A}^2){(-\Delta )}^s\tilde{u}=(2_s^{*}C+1){\left|\tilde{u}\right|}^{2_s^{*}-2}\tilde{u},\end{array}$$

(47)

so that $\tilde{u}$ is not a solution to (46), where $A^2={\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2$. Because the nonnegative and radial solution of (47) is a Aubin–Talenti instanton, up to a scaling and a translation, which is not $L^2$-integrable. □

Proof of Theorem 1.

Considering Lemmas 5 and 6, we know that $c=\underset{\mathcal{S}\cap \mathcal{P}}{inf}E$ is attained. Furthermore, we can infer that there exists Lagrange multipliers $\lambda ,\mu \in \mathbb{R}$, such that $\tilde{u}\in \mathcal{S}\cap \mathcal{P}$ is a solution to

$$\begin{array}{cc}\hfill & {(a(1+\mu )+b(1+2\mu )||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2{)(-\Delta )}^s\tilde{u}+\lambda \tilde{u}\hfill \\ & =g\left(\tilde{u}\right)+{\displaystyle \frac{3}{4s}}\mu h\left(\tilde{u}\right)+\left(1+{\displaystyle \frac{2_s^{*}}{2}}\mu \right){\left|\tilde{u}\right|}^{2_s^{*}-2}\tilde{u}.\hfill \end{array}$$

(48)

In view of the Nehari and Pohozaev identities for (48), one deduces that

$$\begin{array}{cc}\hfill & {(a(1+\mu )+b(1+2\mu )||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2{\left)\right||(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2\hfill \\ & ={\displaystyle \frac{3}{2s}}{\int }_{{\mathbb{R}}^3}\left(H\left(\tilde{u}\right)+{\displaystyle \frac{3}{4s}}\mu (h\left(\tilde{u}\right)\tilde{u}-2H\left(\tilde{u}\right))+\left(1-{\displaystyle \frac{2}{2_s^{*}}}\right)\left(1+{\displaystyle \frac{2_s^{*}}{2}}\mu \right){\left|\tilde{u}\right|}^{2_s^{*}}\right)dx\hfill \end{array}$$

Due to $\tilde{u}\in \mathcal{S}\cap \mathcal{P}\subset \mathcal{P}$, we have $P\left(\tilde{u}\right)=0$, so

$$\begin{array}{cc}\hfill & \mu \left({a\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^2+{2b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left|\right|}_2^4-{\displaystyle \frac{9}{8s^2}}{\int }_{{\mathbb{R}}^3}\left(h\left(\tilde{u}\right)\tilde{u}-2H\left(\tilde{u}\right)\right)dx-{\displaystyle \frac{3}{3-2s}}\left|\right|\tilde{u}{\left|\right|}_{2_s^{*}}^{2_s^{*}}\right)=0.\hfill \end{array}$$

Similar to the proof of Lemma 6, it is clear that $\mu =0$. Combining Lemma 5, under the condition g is odd, we can deduce that $\tilde{u}\in \mathcal{S}\cap \mathcal{P}$ is a positive and radially symmetric normalized ground state solution to (1). □

4. Proof of Theorem 2

Proof of Theorem 2 (a).

Taking into account Lemmas 5 and 6, there exists $u_{{\rho }_1}\in {\mathcal{S}}_{{\rho }_1}\cap \mathcal{P},u_{{\rho }_2}\in {\mathcal{S}}_{{\rho }_2}\cap \mathcal{P}$ such that ${\vartheta }_1=E\left(u_{{\rho }_1}\right),{\vartheta }_2=E\left(u_{{\rho }_2}\right)$ for ${\rho }_1>{\rho }_2>0$. Thus, it is easy to obtain that $u_{{\rho }_1}\in ({\mathcal{D}}_{{\rho }_1}\setminus {\mathcal{S}}_{{\rho }_1})\cap \mathcal{P}$, so that ${\vartheta }_{{\rho }_1}=\underset{{\mathcal{S}}_{{\rho }_1}\cap \mathcal{P}}{inf}E<E\left(u_{{\rho }_2}\right)={\vartheta }_{{\rho }_2}$. □

Proof of Theorem 2 (b).

Assume that $\eta \ne 0$.

(1) Now, we suppose that ${\rho }_n\to {\rho }^{+}$ as $n\to \infty$. It follows from Lemma 5 that there exists $u_n\in {\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}$, such that $E\left(u_n\right)=\underset{{\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}}{inf}E$. Similar to the proof of Lemma 5, up to a translation and up to a subsequence, it is not difficult to know $u_n\rightharpoonup \tilde{u}$ in $H^s\left({\mathbb{R}}^3\right)$ and $r\left(\tilde{u}\right)\ge 1$. Furthermore, $\tilde{u}(r\left(\tilde{u}\right)·)\in {\mathcal{D}}_{\rho }\cap \mathcal{P}\subset {\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}$.If $r\left(\tilde{u}\right)>0$, from Fatou lemma and (${\mathrm{A}}_5$), we have

$$\begin{array}{cc}\hfill c=\underset{{\mathcal{D}}_{\rho }\cap \mathcal{P}}{inf}E& \le E\left(\tilde{u}\left(r\left(\tilde{u}\right)x\right)\right)=E\left(\tilde{u}\left(r\left(\tilde{u}\right)x\right)\right)-{\displaystyle \frac{1}{4s}}P\left(\tilde{u}\left(r\left(\tilde{u}\right)x\right)\right)\hfill \\ & ={\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}\left(r\left(\tilde{u}\right)x\right){\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|\tilde{u}\left(r\left(\tilde{u}\right)x\right){\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left(\tilde{u}\left(r\left(\tilde{u}\right)x\right)\right)\tilde{u}\left(r\left(\tilde{u}\right)x\right)-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left({\tilde{u}}_i\left(r\left({\tilde{u}}_i\right)x\right)\right)\right)dx\hfill \\ & ={\displaystyle \frac{a}{4}}r{\left(\tilde{u}\right)}^{-3+2s}\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}\left(x\right){\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})r{\left(\tilde{u}\right)}^{-3}\left|\right|\tilde{u}\left(x\right){\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}r{\left(\tilde{u}\right)}^{-3}{\int }_{{\mathbb{R}}^3}\left(g\left(\tilde{u}\left(x\right)\right)\tilde{u}\left(x\right)-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(\tilde{u}\left(x\right)\right)\right)dx\hfill \\ & <{\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}\tilde{u}{\left(x\right)\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|\tilde{u}\left(x\right){\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left(\tilde{u}\left(x\right)\right)\tilde{u}\left(x\right)-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(\tilde{u}\left(x\right)\right)\right)dx\hfill \\ & \le \underset{n\to \infty }{lim\; inf}{\displaystyle \frac{a}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u}_n{\left|\right|}_2^2+({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2_s^{*}}})\left|\right|u_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}\hfill \\ & +{\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}\left(g\left(u_n\right)u_n-\left(2+{\displaystyle \frac{8s}{3}}\right)G\left(u_n\right)\right)dx\hfill \\ & \le \underset{n\to \infty }{lim}E\left(u_n\right)\le \underset{n\to \infty }{lim}\underset{{\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}}{inf}E\le \underset{n\to \infty }{lim}\underset{{\mathcal{D}}_{\rho }\cap \mathcal{P}}{inf}E=\underset{{\mathcal{D}}_{\rho }\cap \mathcal{P}}{inf}E,\hfill \end{array}$$

which implies that $r\left(\tilde{u}\right)=1$. Similar to the proof of Lemma 5, we can easily deduce that

$$\begin{array}{c}\hfill E\left(\tilde{u}\right)=\underset{{\mathcal{D}}_{\rho }\cap \mathcal{P}}{inf}E=\underset{n\to \infty }{lim}\underset{{\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}}{inf}E.\end{array}$$

On the other hand, we suppose that ${\rho }_n\to {\rho }^{-}$ as $n\to \infty$. Choosing $u\in {\mathcal{D}}_{\rho }\cap \mathcal{P}$ so that $E\left(u\right)=\underset{{\mathcal{D}}_{\rho }\cap \mathcal{P}}{inf}E$. Similar to [33], we set ${\theta }_n:={\displaystyle \frac{{\rho }_n}{\rho }}\to 1^{-}$ as $n\to \infty$, $v_n={\theta }_{n}u$. Taking into account Lemma 2, it is easy to find ${\lambda }_n>0$, such that ${\lambda }_n^{{\displaystyle \frac{3}{2}}}{v}_n({\lambda }_n·)\in \mathcal{P}$.

Now, we claim that there exists $\lambda >0$, such that ${\lambda }_n\to \lambda$ as $n\to \infty$. We prove this through indirection. Suppose ${\lambda }_n\to \infty$ as $n\to \infty$ passing to a subsequence. By (${\mathrm{A}}_5$), we have that

$$\begin{array}{c}\hfill a{\lambda }_n^{2s}{\theta }_n^2{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+b{\lambda }_n^{4s}{\theta }_n^4{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^4\ge {\lambda }_n^{2_s^{*}s}{\theta }_n^{2_s^{*}}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}},\end{array}$$

that is,

$$\begin{array}{c}\hfill {b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^4\ge {\lambda }_n^{2_s^{*}s-4s}{\theta }_n^{2_s^{*}-4}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}-{\displaystyle \frac{{a\left|\right|(-\Delta )}^{\frac{s}{2}}u{\left|\right|}_2^2}{{\lambda }_n^{2s}{\theta }_n^2}}\to \infty ,\end{array}$$

as $n\to \infty$, which is a contradiction. In addition, we need to exclude ${\lambda }_n\to 0$ as $n\to \infty$ passing to a subsequence. If $\underset{n\to \infty }{lim}{\lambda }_n\to 0$, it is clear to know that

$$\begin{array}{c}\hfill {b\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^4={\displaystyle \frac{3}{2s}}{\theta }^{{\displaystyle \frac{8s}{3}}}{\int }_{{\mathbb{R}}^3}{\displaystyle \frac{H\left({\lambda }_n^{\frac{3}{2}}{\theta }_{n}u\left(x\right)\right)}{|{\lambda }_n^{\frac{3}{2}}{\theta }_{n}u\left(x\right){\left)\right|}^{2+\frac{8s}{3}}}}{\left|u\right|}^{2+{\displaystyle \frac{8s}{3}}}dx+{\lambda }_n^{2_s^{*}s-4s}{\theta }_n^{2_s^{*}-4}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}-{\displaystyle \frac{{a\left|\right|(-\Delta )}^{\frac{s}{2}}u{\left|\right|}_2^2}{{\lambda }_n^{2s}{\theta }_n^2}}\to -\infty ,\end{array}$$

which is impossible. Thus, up to a subsequence, ${\lambda }_n\to \lambda$ as $n\to \infty$ and ${\lambda }^{{\displaystyle \frac{3}{2}}}u(\lambda ·)\in \mathcal{P}$. In view of Lemma 2, we have

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}\underset{{\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}}{inf}E\le \underset{n\to \infty }{lim}E({\lambda }_n^{{\displaystyle \frac{3}{2}}}{v}_n(\lambda ·))=E({\lambda }^{{\displaystyle \frac{3}{2}}}u(\lambda ·))=E(u)=\underset{{\mathcal{D}}_{\rho }\cap \mathcal{P}}{inf}E\end{array}$$

Furthermore, we can infer that

$$\begin{array}{c}\hfill \underset{{\mathcal{D}}_{\rho }\cap \mathcal{P}}{inf}E\le \underset{n\to \infty }{lim\; sup}\underset{{\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}}{inf}E\le \underset{n\to \infty }{lim\; sup}\underset{{\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}}{inf}E\le \underset{{\mathcal{D}}_{\rho }\cap \mathcal{P}}{inf}E,\end{array}$$

that is,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\underset{{\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}}{inf}E=\underset{{\mathcal{D}}_{\rho }\cap \mathcal{P}}{inf}E,\end{array}$$

since ${\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}\subset {\mathcal{D}}_{\rho }\cap \mathcal{P}$. Hence, the continuity of the ground state energy map is proven.

(2) Set $\underset{n\to \infty }{lim}{\rho }_n=+\infty$ and $u\in H^s\left({\mathbb{R}}^3\right)$ is a ground state solution to the Equation (1) with $\rho =1$. Moreover, $E\left(u\right)=\underset{{\mathcal{D}}_1\cap \mathcal{P}}{inf}E=\underset{{\mathcal{S}}_1\cap \mathcal{P}}{inf}E$, Without a loss of generality, we may suppose that ${\rho }_n>1$, and define $u_n={\rho }_{n}u$. Making use of Lemma 2, due to $u_n\in {\mathcal{S}}_{{\rho }_n}\subset {\mathcal{D}}_{{\rho }_n}$, there exists ${\lambda }_n>0$ such that $v_n:={\lambda }_n^{{\displaystyle \frac{3}{2}}}{u}_n({\lambda }_n·)\in \mathcal{P}$. Since $\left|\right|v_n{\left|\right|}_2^2=\left|\right|u_n{\left|\right|}_2^2={\rho }_n$, we have $v_n\in {\mathcal{D}}_{{\rho }_n}\cap \mathcal{P}$. It results that

$$\begin{array}{c}\hfill 0<\underset{{\mathcal{D}}_n\cap \mathcal{P}}{inf}E\le E\left(v_n\right)\le {\displaystyle \frac{a}{2}}{\lambda }_n^{2s}{\rho }_n^2\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+{\displaystyle \frac{b}{4}}{\lambda }_n^{4s}{\rho }_n^4\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^4\end{array}$$

(49)

As $v_n\in \mathcal{P}$, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{a}{2}}{\lambda }_n^{2s}{\rho }_n^2{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{u\left|\right|}_2^2+{\displaystyle \frac{b}{4}}{\lambda }_n^{4s}{\rho }_n^4{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^4\hfill \\ & ={\displaystyle \frac{a}{2}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2+{\displaystyle \frac{b}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^4\hfill \\ & ={\displaystyle \frac{3}{8s}}{\int }_{{\mathbb{R}}^3}H\left(v_n\right)dx+{\displaystyle \frac{1}{4}}\left|\right|v_n{\left|\right|}_{2_s^{*}}^{2_s^{*}}+{\displaystyle \frac{a}{4}}\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}{v}_n{\left|\right|}_2^2\hfill \\ & ={\displaystyle \frac{3}{2s}}{\lambda }_n^{-3}{\int }_{{\mathbb{R}}^3}H({\lambda }_n^{{\displaystyle \frac{3}{2}}}{\rho }_{n}u)dx+{\lambda }_n^{2_s^{*}s}{\rho }^{2_s^{*}}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}+{\displaystyle \frac{a}{4}}{\lambda }_n^{2s}{\rho }_n^2\left|\right|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^2,\hfill \end{array}$$

that is,

$$\begin{array}{cc}\hfill {\displaystyle \frac{b}{4}}{\left|\right|(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{\left|\right|}_2^4& ={\displaystyle \frac{3}{2s}}{\rho }_n^{-4}{\lambda }_n^{-3-4s}{\int }_{{\mathbb{R}}^3}H({\lambda }_n^{{\displaystyle \frac{3}{2}}}{\rho }_{n}u)dx+{\lambda }_n^{2_s^{*}s-4s}{\rho }_n^{2_s^{*}-4}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}-{\displaystyle \frac{a}{4}}{\displaystyle \frac{{\left|\right|(-\Delta )}^{\frac{s}{2}}u{\left|\right|}_2^2}{{\lambda }_n^{2s}{\rho }_n^2}}\hfill \\ & ={\displaystyle \frac{3}{2s}}{\rho }_n^{{\displaystyle \frac{8s}{3}}}{\int }_{{\mathbb{R}}^3}{\displaystyle \frac{H\left({\lambda }_n^{\frac{3}{2}}{\rho }_{n}u\left(x\right)\right)}{|{\lambda }_n^{\frac{3}{2}}{\rho }_{n}u\left(x\right){\left)\right|}^{2+\frac{8s}{3}}}}{\left|u\right|}^{2+{\displaystyle \frac{8s}{3}}}dx+{\lambda }_n^{2_s^{*}s-4s}{\rho }_n^{2_s^{*}-4}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}-{\displaystyle \frac{a}{4}}{\displaystyle \frac{{\left|\right|(-\Delta )}^{\frac{s}{2}}u{\left|\right|}_2^2}{{\lambda }_n^{2s}{\rho }_n^2}}.\hfill \end{array}$$

(50)

Considering $\underset{n\to \infty }{lim}{\rho }_n=\infty$, if ${\lambda }_n^s{\rho }_n\to l>0$ as $n\to \infty$, we have ${\lambda }_n^{{\displaystyle \frac{3}{2}}}{\rho }_n\to 0$ as $n\to \infty$. As $\eta \ne 0$, we can obtain

$${\displaystyle \frac{3}{2s}}{\rho }_n^{{\displaystyle \frac{8s}{3}}}{\int }_{{\mathbb{R}}^3}{\displaystyle \frac{H\left({\lambda }_n^{\frac{3}{2}}{\rho }_{n}u\left(x\right)\right)}{|{\lambda }_n^{\frac{3}{2}}{\rho }_{n}u\left(x\right){\left)\right|}^{2+\frac{8s}{3}}}}{\left|u\right|}^{2+{\displaystyle \frac{8s}{3}}}dx+{\lambda }_n^{2_s^{*}s-4s}{\rho }_n^{2_s^{*}-4}{\left|\right|u\left|\right|}_{2_s^{*}}^{2_s^{*}}-{\displaystyle \frac{a}{4}}{\displaystyle \frac{{\left|\right|(-\Delta )}^{\frac{s}{2}}u{\left|\right|}_2^2}{{\lambda }_n^{2s}{\rho }_n^2}}\to \infty ,$$

as $n\to \infty$, which is a contradiction. Obviously, it is impossible that (50) holds when ${\lambda }_n^s{\rho }_n\to \infty$ as $n\to \infty$. Thus, we can obtain ${\lambda }_n^s{\rho }_n\to 0$ and ${\lambda }_n^{{\displaystyle \frac{3}{2}}}{\rho }_n\to 0$ as $n\to \infty$. It follows from (49) that $\underset{n\to \infty }{lim}\underset{{\mathcal{D}}_n\cap \mathcal{P}}{inf}E=0$, that is, $\underset{{\mathcal{D}}_n\cap \mathcal{P}}{inf}E\to 0^{+}$ as $\rho \to \infty$. □

5. Conclusions

This article mainly studies a class of nonlinear fractional Kirchhoff equation with at least critical mass growth. We focus on the existence and properties of solutions to Equation (1) and establish a series of assumptions ($A_0$)–($A_6$) to ensure that the solutions in question satisfy specific conditions. By establishing a novel variational framework that combines Nehari and Pohozaev constraint methods within an $L^2$-ball of prescribed radius $\rho$, we show the existence of solutions. Under the condition where g is odd, the solution is a positive and radially symmetric normalized ground state solution to Equation (1). In addition, we give a detailed description for the asymptotic behavior of the ground state energy map. This study has significant physical implications and provides important theoretical basis for in-depth research on nonlinear optics and Bose Einstein condensates