Ground state solutions for nonlinear Choquard equations with inverse-square potentials 1

Abstract

This paper is concerned with the following Choquard equation $\begin{matrix} \{\begin{matrix} - Δ u + (V (x) - \frac{μ}{| x |^{2}}) u = (\int_{R^{N}} \frac{F (u (y))}{| x - y |^{α}} d y) f (u), \\ u \in H^{1} (R^{N}), \end{matrix} \end{matrix}$ where $N ⩾ 3$ , $0 ⩽ μ < \frac{{(N - 2)}^{2}}{4}$ , $0 < α < N$ , V is 1-periodic in each of $x_{1}, x_{2}, \dots, x_{N}$ and F is the primitive function of f. Under some mild assumptions on V and f, we establish the existence and asymptotical behavior of ground state solutions by variational methods.

Keywords

Choquard equation ground state solutions asymptotical behavior inverse square potential

1. Introduction

In this paper, we study the following Choquard equation $\begin{matrix} (1.1) & \{\begin{matrix} - Δ u + (V (x) - \frac{μ}{| x |^{2}}) u = (\int_{R^{N}} \frac{F (u (y))}{| x - y |^{α}} d y) f (u), \\ u \in H^{1} (R^{N}), \end{matrix} \end{matrix}$ where $N ⩾ 3$ , $0 ⩽ μ < \frac{{(N - 2)}^{2}}{4}$ , $0 < α < N$ , $V : R^{N} \to R$ and $f : R \to R$ satisfy the following assumptions:

$V \in C (R^{N}, R)$ is 1-periodic in each of $x_{1}, x_{2}, \dots, x_{N}$ and ${inf}_{x \in R^{N}} V (x) > 0$ ;

$f \in C (R, R)$ , $| f (t) | ⩽ C_{0} (| t |^{p_{1} - 1} + | t |^{p_{2} - 1})$ , for some $C_{0} > 0$ and $\frac{2 N - α}{N} ⩽ p_{1} ⩽ p_{2} < \frac{2 N - α}{N - 2}$ ;

$f (t)$ is non-decreasing on $R$ ;

${lim}_{| t | \to 0} \frac{f (t)}{| t |^{\frac{N - α}{N}}} = 0$ and ${lim}_{| t | \to \infty} | f (t) | = + \infty$ ;

there exists a constant $θ \in (0, 1)$ such that $F (t) ⩽ θ f (t) t$ , $\forall t \in R$ , where $F (t) = \int_{0}^{t} f (s) d s$ .

The stationary Choquard equation $\begin{matrix} (1.2) & - Δ u + V (x) u = (\int_{R^{N}} \frac{| u (y) |^{p}}{| x - y |^{α}} d y) | u |^{p - 2} u, u \in H^{1} (R^{N}), \end{matrix}$ where $N ⩾ 3$ , $0 < α < N$ , arises in many interesting physical situations and especially plays an important role in the theory of Bose–Einstein condensation where it accounts for the finite-range many-body interactions. The case $N = 3$ , $α = 1$ , $V = 1$ and $p = 2$ goes back to the description of a polaron at rest in the quantum theory by Pekar [20] in 1954, and it can act as an approximation to Hartree–Fock theory of one-component plasma, see [11]. This equation was also proposed by Penrose in [14] as a model for self-gravitating particles and in that context it is known as the Schrödinger–Newton equation. In this case the existence of solutions has been proved through variational methods by Lieb, Lions and Menzala [12–14] and also by ordinary differential equations techniques [5,14].

In [2], Alves and Yang considered the following Choquard equation $\begin{matrix} (1.3) & - ε^{2} Δ u + V (x) u = ε^{μ - N} (\int_{R^{N}} \frac{Q (y) F (u (y))}{| x - y |^{μ}} d y) Q (x) f (u), u \in H^{1} (R^{N}) . \end{matrix}$ Under some suitable assumptions, they established a new concentration behavior of solutions for (1.3) by variational methods. In [15], Moroz and Schaftingen proved the existence of solution to the nonlinear Choquard equation $\begin{matrix} (1.4) & - Δ u + u = (I_{α} * F (u)) F^{'} (u), in R^{N}, \end{matrix}$ where $I_{α}$ is a Riesz potential, under almost necessary conditions on the nonlinearity F in the spirit of Berestycki and Lions.

Recently, a great attention has been devoted to the existence and concentration of solutions for the Schrödinger equation with inverse square potential (see [1,3,4,6–10,22,24]): $\begin{matrix} (1.5) & - Δ u + (V (x) - \frac{μ}{| x |^{2}}) u = f (x, u), x \in R^{N} . \end{matrix}$ If $μ \neq 0$ , the singular potential $- \frac{μ}{| x |^{2}}$ does not belong to the Kato’s class [21] and cannot be regarded as a lower order perturbation term of $- Δ + V$ . Besides, the additional singular potential breaks the compactness of embedding $H^{1} (R^{N}) ↪ L^{2} (R^{N}, | x |^{- 2} d x)$ , which leads some new difficulties. For more results related to unbounded and indefinite potential, we refer to Papageorgiou and Rădulescu [16–18], Szulkin and Weth [25].

In [10], Guo and Mederski proved that (1.5) has a ground state solution for $μ ⩾ 0$ sufficiently small. Firstly, a global compactness lemma for bounded Palais–Smale sequences of strongly indefinite functionals involving sum of periodic and inverse square potentials was proved, which obviously works for Cerami sequences. Then they used this lemma to overcome the main obstacle that the embedding $H^{1} (R^{N}) ↪ L^{2} (B_{R} (0), | x |^{- 2} d x)$ is the non-compactness for any $R > 0$ , where $B_{R} (0) = {x \in R^{N} : | x | < R}$ .

Motivated by [10], in the present paper, we consider the Choquard equation (1.1) with inverse square potential, where the subcritical term f satisfies a weak monotonicity condition (F2). To state our main results, we give some notations.

Let $E = H^{1} (R^{N})$ be the usual Sobolev space equipped with norm $\begin{matrix} (1.6) & ‖ u ‖ : = {(\int_{R^{N}} (| \nabla u |^{2} + | u |^{2}) d x)}^{\frac{1}{2}} . \end{matrix}$ Define an inner product $\begin{matrix} (1.7) & {(u, v)}_{μ} : = \int_{R^{N}} (\nabla u \cdot \nabla v + (V (x) - \frac{μ}{| x |^{2}}) u v) d x, u, v \in E, \end{matrix}$ and the corresponding norm $\begin{matrix} (1.8) & ‖ u ‖_{μ} : = {(\int_{R^{N}} (| \nabla u |^{2} + (V (x) - \frac{μ}{| x |^{2}}) | u |^{2}) d x)}^{\frac{1}{2}}, u \in E . \end{matrix}$ By (V) and Hardy inequality, $E = H^{1} (R^{N})$ with equivalent norms. The energy functional associated to Problem (1.1) is defined by $\begin{matrix} (1.9) & I_{μ} (u) = \frac{1}{2} \int_{R^{N}} (| \nabla u |^{2} + (V (x) - \frac{μ}{| x |^{2}}) u^{2}) d x - \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) F (u (x))}{| x - y |^{α}} d x d y . \end{matrix}$ From the growth assumptions on f and the Hardy–Littlewood–Sobolev inequality, we get that $I_{μ}$ is well defined on E and belongs to $C^{1}$ with its derivative given by $\begin{array}{rcl} ⟨ I_{μ}^{'} (u), v ⟩ & = & \int_{R^{N}} (\nabla u \cdot \nabla v + (V (x) - \frac{μ}{| x |^{2}}) u v) d x \\ (1.10) & - \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) f (u (x)) v (x)}{| x - y |^{α}} d x d y . \end{array}$ Therefore, the solutions of Problem (1.1) correspond to critical points of the functional $I_{μ}$ . We denote by $N_{μ}$ the Nehari manifold associated to $I_{μ}$ defined by ${u \in E ∖ {0} : ⟨ I_{μ}^{'} (u), u ⟩ = 0}$ , namely $\begin{array}{rcl} N_{μ} & = & {u \in E ∖ {0} : \int_{R^{N}} (| \nabla u |^{2} + (V (x) - \frac{μ}{| x |^{2}}) u^{2}) d x \\ (1.11) & = & \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) f (u (x)) u (x)}{| x - y |^{α}} d x d y} . \end{array}$ Set $m_{μ} : = {inf}_{N_{μ}} I_{μ}$ . If there exists $u_{0} \in N_{μ}$ such that $I_{μ} (u_{0}) = m_{μ}$ , we call it a ground state solution for (1.1).

The main results of this paper is stated as follows:

Theorem 1.1.
Assume that (V), (F1), (F2) and (F3) hold. If $0 ⩽ μ < \frac{{(N - 2)}^{2}}{4}$ , $0 < α < N$ , then Problem ( 1.1 ) possesses a solution $u_{0} \in H^{1} (R^{N})$ such that $I_{μ} (u_{0}) = {inf}_{N_{μ}} I_{μ} > 0$ .
Theorem 1.2.
Assume that (V), (F1), (F2), (F3) and (F4) hold. If $0 < α < N$ , $μ = μ_{n} \to 0^{+}$ , then there exists a sequence ${x_{n}} \subset R^{N}$ such that $u_{n} (\cdot + x_{n})$ converges in $H^{1} (R^{N})$ , to a ground state solution $\bar{u}$ of $I_{0}$ , where $u_{n}$ is a solution obtained in Theorem 1.1 .

By using variational method, we establish the existence and asymptotic behavior of Nehari-type ground state solution for Problem (1.1). To prove Theorem 1.1, we first construct a Cerami sequence ${u_{n}}$ of $I_{μ}$ (see Lemma 2.8), unlike the Nehari manifold method, our approach lies on finding a minimizing Cerami sequence for energy functional outside the manifold by using diagonal method (see [26–30]), then we prove the boundedness of the Cerami sequence. To overcome the difficulty caused by the non-compactness of the embedding $H^{1} (R^{N}) ↪ L^{2} (B_{R} (0), | x |^{- 2} d x)$ , we prove that the level of periodic functional $I_{0}$ is actually below that of the functional $I_{μ}$ related to (1.1), see (3.3) in Section 3. Moreover, we find a new method to deal with the weak monotonicity condition (F2), see Lemma 2.5. To prove Theorem 1.2, we first take advantage of the relation of the functionals and derivatives between the equation and its limit equation to find a ground state solution of $I_{0}$ , and then deduce the asymptotic behavior of ground state solution by the Ambrosetti–Rabinowitz type condition (F4).

The paper is organized as follows. In Section 2, we give some preliminary lemmas. The proofs of Theorems 1.1 and 1.2 are given in Section 3.

Throughout this paper, we denote the norm of $L^{s} (R^{N})$ ( $1 ⩽ s < \infty$ ) by $| u |_{s} = {(\int_{R^{N}} | u |^{s} d x)}^{1 / s}$ , and positive constants possibly different in different places, by $C_{1}, C_{2}, \dots$ .
2. Preliminary lemmas

In this section, we present some useful lemmas. Firstly, let us recall the following Cerami-type version of the Mountain pass Theorem.

Lemma 2.1 ([23], [19, Theorem 5.4.6]).

Let E be a real Banach space, $φ \in C^{1} (E, R)$ , $e \in E$ and $r > 0$ be such that $‖ e ‖ > r$ and $\begin{matrix} inf_{‖ u ‖ = r} φ (u) > φ (0) ⩾ φ (e) . \end{matrix}$ Then there exists a sequence ${u_{n}} \subset E$ satisfying $\begin{matrix} φ (u_{n}) \to c, ‖ φ^{'} (u_{n}) ‖ (1 + ‖ u_{n} ‖) \to 0, \end{matrix}$ where $\begin{matrix} c : = sup_{O \in O} inf_{u \in \partial O} φ (u) \end{matrix}$ and $O$ be the set of all bounded open sets $O \subset E$ such that $0 \in O$ but $e \notin \bar{O}$ .

Lemma 2.2.
If $0 ⩽ μ_{n} ⩽ μ_{0} < \frac{{(N - 2)}^{2}}{4}$ , then there exist two constants $θ_{1}$ , $θ_{2} > 0$ such that $\begin{matrix} θ_{1} ‖ u ‖_{μ_{n}} ⩽ ‖ u ‖ ⩽ θ_{2} ‖ u ‖_{μ_{n}}, \forall n \in N_{+}, \forall u \in H^{1} (R^{N}) . \end{matrix}$
Proof.
By (V) and $0 ⩽ μ_{n}$ , there exists a constant $C > 0$ such that $\begin{matrix} ‖ u ‖_{μ_{n}}^{2} ⩽ ‖ u ‖_{0}^{2} ⩽ C ‖ u ‖^{2} . \end{matrix}$ By Hardy inequality, we have $\begin{matrix} ‖ u ‖_{μ_{n}}^{2} & = ε \int_{R^{N}} (| \nabla u |^{2} + V (x) u^{2}) d x + (1 - ε) \int_{R^{N}} (| \nabla u |^{2} + V (x) u^{2}) d x - μ_{n} \int_{R^{N}} \frac{u^{2}}{| x |^{2}} d x \\ ⩾ ε ‖ u ‖_{0}^{2} + (1 - ε) \frac{{(N - 2)}^{2}}{4} \int_{R^{N}} \frac{u^{2}}{| x |^{2}} d x - μ_{n} \int_{R^{N}} \frac{u^{2}}{| x |^{2}} d x \\ = ε ‖ u ‖_{0}^{2} + ((1 - ε) \frac{{(N - 2)}^{2}}{4} - μ_{n}) \int_{R^{N}} \frac{u^{2}}{| x |^{2}} d x . \end{matrix}$ Since $0 ⩽ μ_{n} ⩽ μ_{0} < \frac{{(N - 2)}^{2}}{4}$ , there exist $ε_{0} > 0$ and $ε_{1} > 0$ such that $(1 - ε_{0}) \frac{{(N - 2)}^{2}}{4} - μ_{n} ⩾ (1 - ε_{0}) \frac{{(N - 2)}^{2}}{4} - μ_{0} ⩾ 0$ , then $‖ u ‖_{μ_{n}}^{2} ⩾ ε_{0} ‖ u ‖_{0}^{2} ⩾ ε_{1} ‖ u ‖^{2}$ . □

We set $\begin{matrix} ψ (u) = \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (u (x)) F (u (y))}{| x - y |^{α}} d x d y, \forall u \in H^{1} (R^{N}) . \end{matrix}$ Lemma 2.3.
Assume that (F1), (F2) and (F3) hold. Then ψ is nonnegative, weakly sequentially lower semi-continuous, and $ψ^{'}$ is weakly sequentially continuous.
Proof.
It is clear that $\begin{matrix} ψ (u) ⩾ 0, \forall u \in H^{1} (R^{N}) . \end{matrix}$ Let $u_{n} ⇀ u$ in $H^{1} (R^{N})$ , then $u_{n} \to u$ , a.e. on $R^{N}$ . By Fatou’s lemma, we obtain $\begin{array}{rcl} ψ (u) & = & \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) F (u (x))}{| x - y |^{α}} d x d y \\ ⩽ & {lim_{_}}_{n \to \infty} \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (y)) F (u_{n} (x))}{| x - y |^{α}} d x d y \\ = & {lim_{_}}_{n \to \infty} ψ (u_{n}) . \end{array}$ So ψ is nonnegative and weakly sequentially lower semi-continuous.

Next, we show that $ψ^{'}$ is weakly sequentially continuous. Assume that $u_{n} ⇀ u$ in $H^{1} (R^{N})$ . For any $v \in H^{1} (R^{N})$ , we have $\begin{array}{l} ⟨ ψ^{'} (u_{n}), v ⟩ = \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (y)) f (u_{n} (x)) v (x)}{| x - y |^{α}} d x d y, \\ ⟨ ψ^{'} (u), v ⟩ = \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) f (u (x)) v (x)}{| x - y |^{α}} d x d y . \end{array}$ Then $\begin{matrix} (2.1) & \begin{matrix} ⟨ ψ^{'} (u_{n}), v ⟩ - ⟨ ψ^{'} (u), v ⟩ = & \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (y)) [f (u_{n} (x)) - f (u (x))] v (x)}{| x - y |^{α}} d x d y \\ + \int_{R^{N}} \int_{R^{N}} \frac{[F (u_{n} (y)) - F (u (y))] f (u (x)) v (x)}{| x - y |^{α}} d x d y \\ : = & I_{1} + I_{2} . \end{matrix} \end{matrix}$

Since $u_{n} ⇀ u$ in $H^{1} (R^{N})$ . By (F1), (F2) and (F3) and Sobolev embedding theorem, we have $F (u_{n})$ is bounded in $L^{\frac{2 N}{2 N - α}} (R^{N})$ , and $F (u_{n} (x)) \to F (u (x))$ , a.e. $x \in R^{N}$ . Passing to a subsequence, we have $F (u_{n}) ⇀ F (u)$ in $L^{\frac{2 N}{2 N - α}} (R^{N})$ . Using Hardy–Littlewood–Sobolev inequality we get $\begin{matrix} \int_{R^{N}} \frac{F (u_{n} (y))}{| x - y |^{α}} d y ⇀ \int_{R^{N}} \frac{F (u (y))}{| x - y |^{α}} d y in L^{\frac{2 N}{α}} (R^{N}) . \end{matrix}$ Since $f (u) v \in L^{\frac{2 N}{2 N - α}} (R^{N}) = {(L^{\frac{2 N}{α}} (R^{N}))}^{}$ , then $\begin{matrix} (2.2) & I_{2} = o (1) . \end{matrix}$ Since $u_{n} ⇀ u$ in $H^{1} (R^{N})$ , passing to a subsequence, we have $u_{n} \to u$ a.e. on $R^{N}$ and $u_{n} \to u$ in $L_{loc}^{s} (R^{N})$ , $2 ⩽ s < 2^{}$ . For any $v \in C_{0}^{\infty} (R^{N})$ with the support Ω, by the Hardy–Littlewood–Sobolev inequality, one can get $\begin{matrix} (2.3) & | I_{1} | ⩽ C {| F (u_{n}) |}_{\frac{2 N}{2 N - α}} {| (f (u_{n}) - f (u)) v |}_{\frac{2 N}{2 N - α}} . \end{matrix}$ We obtain, by the Lebesgue’s dominated convergence theorem, that $\begin{matrix} (2.4) & \begin{matrix} \int_{R^{N}} {| (f (u_{n}) - f (u)) v |}^{\frac{2 N}{2 N - α}} d x & = \int_{Ω} {| (f (u_{n}) - f (u)) v |}^{\frac{2 N}{2 N - α}} d x \\ = o (1) . \end{matrix} \end{matrix}$ By (2.1)–(2.4), we get $\begin{matrix} (2.5) & ⟨ ψ^{'} (u_{n}), v ⟩ \to ⟨ ψ^{'} (u), v ⟩, \forall v \in C_{0}^{\infty} (R^{N}) . \end{matrix}$ Since ${ψ^{'} (u_{n})}$ is bounded in ${(H^{1} (R^{N}))}^{}$ and $C_{0}^{\infty} (R^{N})$ is dense in $H^{1} (R^{N})$ , we conclude that (2.5) holds for any $v \in H^{1} (R^{N})$ . So $ψ^{'}$ is weakly sequentially continuous. □
Lemma 2.4.
Assume that (F1), (F2), (F3) and (V) hold. Then* $\begin{matrix} (2.6) & I_{μ} (u) ⩾ I_{μ} (t u) + \frac{1 - t^{2}}{2} ⟨ I_{μ}^{'} (u), u ⟩, \forall t ⩾ 0, u \in H^{1} (R^{N}) . \end{matrix}$ In particular, if $u \in N_{μ}$ , then $\begin{matrix} (2.7) & I_{μ} (u) = max_{t ⩾ 0} I_{μ} (t u) . \end{matrix}$
Proof.
We compute directly $\begin{matrix} I_{μ} (u) - I_{μ} (t u) - \frac{1 - t^{2}}{2} ⟨ I_{μ}^{'} (u), u ⟩ \\ = \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (t u (y)) F (t u (x))}{| x - y |^{α}} d x d y + \frac{1 - t^{2}}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) f (u (x)) u (x)}{| x - y |^{α}} d x d y \\ - \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) F (u (x))}{| x - y |^{α}} d x d y = : h (t) . \end{matrix}$ Then $\begin{matrix} (2.8) & h^{'} (t) = \int_{R^{N}} \int_{R^{N}} \frac{[F (t u (y)) f (t u (x)) - t F (u (y)) f (u (x))] u (x)}{| x - y |^{α}} d x d y . \end{matrix}$ By (F1), (F2) and (F3), we can prove $h (1) = 0$ , $h^{'} (1) = 0$ ; $h^{'} (1) ⩾ 0$ for all $t > 1$ ; $h^{'} (1) ⩽ 0$ for all $0 < t < 1$ . So that $h (t) ⩾ 0$ , $\forall t ⩾ 0$ . □
Lemma 2.5.
Assume that (F1), (F2), (F3) and (V) hold. Then for all $u \neq 0$ , there exists a unique $t_{u} > 0$ such that $t_{u} u \in N_{μ}$ and $I_{μ} (t_{u} u) = {max}_{t ⩾ 0} I_{μ} (t u) > 0$ .
Proof.
Let $g (t) : = I_{μ} (t u) = \frac{t^{2}}{2} ‖ u ‖_{μ}^{2} - \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (t u (x)) F (t u (y))}{| x - y |^{α}} d x d y$ . By (F2), (F2) and (F3), it is easy to see that $g (t) > 0$ when t is small, and $g (t) < 0$ when t is large. Then there exists $t_{u} > 0$ such that $g^{'} (t_{u}) = 0$ , moreover, $t_{u} u \in N_{μ}$ . Let $t_{2} > t_{1} > 0$ and $g^{'} (t_{1}) = g^{'} (t_{2}) = 0$ , then $\begin{matrix} \int_{R^{N}} \int_{R^{N}} \frac{F (t_{1} u (y)) f (t_{1} u (x)) u (x)}{t_{1} | x - y |^{α}} d x d y = \int_{R^{N}} \int_{R^{N}} \frac{F (t_{2} u (y)) f (t_{2} u (x)) u (x)}{t_{2} | x - y |^{α}} d x d y . \end{matrix}$ Since $\begin{matrix} \frac{F (t_{1} u (y)) f (t_{1} u (x)) u (x)}{t_{1}} ⩽ \frac{F (t_{2} u (y)) f (t_{2} u (x)) u (x)}{t_{2}} . \end{matrix}$ So we have $\begin{matrix} \frac{F (t_{1} u (y)) f (t_{1} u (x)) u (x)}{t_{1}} = \frac{F (t_{2} u (y)) f (t_{2} u (x)) u (x)}{t_{2}}, a.e. (x, y) \in R^{N} \times R^{N} . \end{matrix}$ Since $\begin{matrix} 0 & = \frac{F (t_{2} u (y)) f (t_{2} u (x)) u (x)}{t_{2}} - \frac{F (t_{1} u (y)) f (t_{1} u (x)) u (x)}{t_{1}} \\ = \frac{F (t_{2} u (y))}{t_{2}} (f (t_{2} u (x)) u (x) - f (t_{1} u (x)) u (x)) + (\frac{F (t_{2} u (y))}{t_{2}} - \frac{F (t_{1} u (y))}{t_{1}}) f (t_{1} u (x)) u (x), \end{matrix}$ and $\begin{matrix} \frac{F (t_{2} u (y))}{t_{2}} (f (t_{2} u (x)) u (x) - f (t_{1} u (x)) u (x)) ⩾ 0, \end{matrix}$ so that $\begin{matrix} (2.9) & (\frac{F (t_{2} u (y))}{t_{2}} - \frac{F (t_{1} u (y))}{t_{1}}) f (t_{1} u (x)) u (x) = 0, a.e. (x, y) \in R^{N} \times R^{N} . \end{matrix}$ By (F2) and (F3), there exist two constants $β_{1}$ , $β_{2} > 0$ such that $\begin{matrix} (2.10) & f^{- 1} ({0}) = [- β_{1}, β_{2}] . \end{matrix}$ We claim that $\begin{matrix} meas ({x : t_{1} u (x) > β_{2}} \cup {x : t_{1} u (x) < - β_{1}}) > 0 . \end{matrix}$ Arguing by contradiction, suppose that $\begin{matrix} meas ({x : t_{1} u (x) > β_{2}} \cup {x : t_{1} u (x) < - β_{1}}) = 0 \end{matrix}$ which, recalling (2.10), means that $\begin{matrix} f (t_{1} u (x)) = 0, a.e. (x, y) \in R^{N} \times R^{N} . \end{matrix}$ Moreover, $\begin{matrix} 0 = \frac{g^{'} (t_{1})}{t_{1}} = ‖ u ‖_{μ}^{2} - \int_{R^{N}} \int_{R^{N}} \frac{F (t_{1} u (y)) f (t_{1} u (x)) u (x)}{t_{1} | x - y |^{α}} d x d y = ‖ u ‖_{μ}^{2} \end{matrix}$ which is a contradiction. Without loss of generality, we assume $\begin{matrix} meas ({x : t_{1} u (x) > β_{2}}) > 0 . \end{matrix}$ By $t_{2} > t_{1}$ , we have $\begin{matrix} {x : t_{2} u (x) > β_{2}} \supset {x : t_{1} u (x) > β_{2}} = : E . \end{matrix}$ Thus, by (F2), we know $\begin{matrix} \frac{F (t_{1} u (x))}{t_{1}} < \frac{F (t_{2} u (x))}{t_{2}}, \forall x \in E . \end{matrix}$ Hence, $\begin{matrix} (\frac{F (t_{2} u (y))}{t_{2}} - \frac{F (t_{1} u (y))}{t_{1}}) f (t_{1} u (x)) u (x) > 0, \forall (x, y) \in E \times E \end{matrix}$ which contradicts (2.9) and the proof is complete. □
Lemma 2.6.
Assume that (F1), (F2), (F3) and (V) hold. If $e \in H^{1} (R^{N})$ and $‖ e ‖_{μ} = 1$ , then there exist two constants $r_{0} > ρ_{0} > 0$ such that $\begin{matrix} κ : = inf_{‖ u ‖_{μ} = ρ_{0}} I_{μ} (u) > I_{μ} (0) ⩾ I_{μ} (r_{0} e) . \end{matrix}$
Proof.
By Lemma 2.5, there exists $r_{0} > 0$ such that $\begin{matrix} I_{μ} (r_{0} e) ⩽ 0 = I_{μ} (0) . \end{matrix}$ From Hardy–Littlewood–Sobolev inequality and (F1), we know $\begin{matrix} I_{μ} (u) & = \frac{1}{2} ‖ u ‖_{μ}^{2} - \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (u (x)) F (u (y))}{| x - y |^{α}} d x d y \\ ⩾ \frac{1}{2} ‖ u ‖_{μ}^{2} - C (‖ u ‖_{μ}^{2 p_{1}} + ‖ u ‖_{μ}^{2 p_{2}}) . \end{matrix}$ Then there exists $ρ_{0} > 0$ such that ${inf}_{‖ u ‖_{μ} = ρ_{0}} I_{μ} (u) > 0$ . □
Lemma 2.7.
Assume that (F1), (F2), (F3) and (V) hold. Then
there exists $ρ_{0} > 0$ such that $\begin{matrix} (2.11) & m_{μ} : = inf_{N_{μ}} I_{μ} ⩾ κ : = inf_{‖ u ‖_{μ} = ρ_{0}} I_{μ} (u) > 0 . \end{matrix}$

there exists $d_{0} > 0$ such that $\begin{matrix} (2.12) & inf_{u \in N_{μ}} ‖ u ‖_{μ} ⩾ d_{0} . \end{matrix}$

Proof.
By Lemma 2.5 and Lemma 2.6, it is clear that (2.11) is hold. Let $u \in N_{μ}$ , then $\begin{matrix} ‖ u ‖_{μ}^{2} = \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) f (u (x)) u (x)}{| x - y |^{α}} d x d y . \end{matrix}$ By (F1) and Hardy–Littlewood–Sobolev inequality, we know $\begin{matrix} \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) f (u (x)) u (x)}{| x - y |^{α}} d x d y ⩽ C (‖ u ‖_{μ}^{2 p_{1}} + ‖ u ‖_{μ}^{2 p_{2}}) \end{matrix}$ and then $\begin{matrix} ‖ u ‖_{μ}^{2} ⩽ C (‖ u ‖_{μ}^{2 p_{1}} + ‖ u ‖_{μ}^{2 p_{2}}) . \end{matrix}$ Hence, $\begin{matrix} inf_{u \in N_{μ}} ‖ u ‖_{μ} ⩾ d_{0} > 0 . \end{matrix}$ The proof is thus finished. □
Lemma 2.8.
Assume that (F1), (F2), (F3) and (V) hold. Then there exist a constant $c \in [κ, m_{μ}]$ and a sequence ${u_{n}} \subset H^{1} (R^{N})$ satisfying $\begin{matrix} I_{μ} (u_{n}) \to c, ‖ I_{μ}^{'} (u_{n}) ‖ (1 + ‖ u_{n} ‖_{μ}) \to 0, \end{matrix}$ where $m_{μ}$ is defined by ( 2.11 ).
Proof.
Choosing $v_{i} \in N_{μ}$ such that $\begin{matrix} m_{μ} ⩽ I_{μ} (v_{i}) < m_{μ} + \frac{1}{i}, i \in N_{+} . \end{matrix}$ By Lemma 2.6, there exists $t_{i} > ρ_{0}$ such that $\begin{matrix} ‖ t_{i} v_{i} ‖_{μ} > ρ_{0} \end{matrix}$ and $\begin{matrix} I_{μ} (t_{i} v_{i}) < 0 = I_{μ} (0) < inf_{‖ u ‖_{μ} = ρ_{0}} I_{μ} (u) . \end{matrix}$ By Lemma 2.1, for every $i \in N_{+}$ , there exists a sequence ${u_{n}^{i}} \subset H^{1} (R^{N})$ satisfying $\begin{matrix} I_{μ} (u_{n}^{i}) \to c_{i}, ‖ I_{μ}^{'} (u_{n}^{i}) ‖ (1 + {‖ u_{n}^{i} ‖}_{μ}) \to 0, \end{matrix}$ where $\begin{matrix} c_{i} : = sup_{O \in O_{i}} inf_{\partial O} I_{μ} \end{matrix}$ and $O_{i}$ be the set of all bounded open sets $O \subset H^{1} (R^{N})$ such that $0 \in O$ but $t_{i} v_{i} \notin \bar{O}$ . It is clear that $\begin{matrix} c_{i} \in [κ, sup_{0 ⩽ s ⩽ 1} I_{μ} (s t_{i} v_{i})] = [κ, I_{μ} (v_{i})] \subseteq [κ, m_{μ} + \frac{1}{i}) . \end{matrix}$ For every i, there exists $n_{i}$ such that $\begin{matrix} ‖ I_{μ}^{'} (u_{n_{i}}^{i}) ‖ (1 + {‖ u_{n_{i}}^{i} ‖}_{μ}) < \frac{1}{i}, I_{μ} (u_{n_{i}}^{i}) \in [κ - \frac{1}{i}, m_{μ} + \frac{1}{i}) . \end{matrix}$ Write $w_{i} = u_{n_{i}}^{i}$ , then $\begin{matrix} ‖ I_{μ}^{'} (w_{i}) ‖ (1 + ‖ w_{i} ‖_{μ}) \to 0, I_{μ} (w_{i}) \in [κ - \frac{1}{i}, m_{μ} + \frac{1}{i}) . \end{matrix}$ Up to a subsequence, $\begin{matrix} ‖ I_{μ}^{'} (w_{i}) ‖ (1 + ‖ w_{i} ‖_{μ}) \to 0, I_{μ} (w_{i}) \to c \in [κ, m_{μ}] . \end{matrix}$ □
Lemma 2.9.
Assume that (F1), (F2), (F3) and (V) hold. If ${u_{n}} \subset H^{1} (R^{N})$ and ${μ_{n}} \subset R$ satisfy $\begin{matrix} 0 ⩽ μ_{n} ⩽ μ_{0} < \frac{{(N - 2)}^{2}}{4} \end{matrix}$ and $\begin{matrix} sup_{n} | I_{μ_{n}} (u_{n}) | < \infty, ‖ I_{μ_{n}}^{'} (u_{n}) ‖ (1 + ‖ u_{n} ‖_{μ_{n}}) \to 0, \end{matrix}$ then ${‖ u_{n} ‖_{μ_{n}}}$ is bounded. In particular, if $\begin{matrix} sup_{n} | I_{μ} (u_{n}) | < \infty, ‖ I_{μ}^{'} (u_{n}) ‖ (1 + ‖ u_{n} ‖_{μ}) \to 0, \end{matrix}$ then ${‖ u_{n} ‖_{μ}}$ is bounded.
Proof.
Arguing indirectly, suppose that $‖ u_{n} ‖_{μ_{n}} \to \infty$ . Let $v_{n} = u_{n} / ‖ u_{n} ‖_{μ_{n}}$ . If $δ : = {lim^{‾}}_{n \to \infty} {sup}_{y \in R^{N}} \int_{B_{1} (y)} | v_{n} |^{2} d x = 0$ , then by Lions’ concentration compactness principle, $v_{n} \to 0$ in $L^{s} (R^{N})$ for $2 < s < 2^{}$ . Write $M : = {sup}_{n} | I_{μ_{n}} (u_{n}) |$ , $R : = \sqrt{4 M + 1}$ and $t_{n} : = R / ‖ u_{n} ‖_{μ_{n}}$ . By (F1), for any $ε > 0$ , there exists $C_{ε} > 0$ such that $\begin{matrix} F (s) ⩽ ε | s |^{\frac{2 N - α}{N}} + C_{ε} | s |^{p_{3}}, \forall s \in R, \end{matrix}$ where $p_{3} : = \frac{(N - 2) p_{2} + N + α}{2 N - 4}$ . Notice that $2 N p_{3} / (2 N - α) \in (2, 2^{})$ , it follows from Lemma 2.2 and Hardy–Littlewood–Sobolev inequality that $\begin{matrix} \int_{R^{N}} \int_{R^{N}} \frac{F (t_{n} u_{n} (y)) F (t_{n} u_{n} (x))}{| x - y |^{α}} d x d y \\ ⩽ C (N, α) {| F (t_{n} u_{n}) |}_{2 N / (2 N - α)}^{2} \\ ⩽ C (N, α) {(\int_{R^{N}} {(ε | t_{n} u_{n} |^{\frac{2 N - α}{N}} + C_{ε} | t_{n} u_{n} |^{p_{3}})}^{\frac{2 N}{2 N - α}} d x)}^{\frac{2 N - α}{N}} \\ ⩽ 2 C (N, α) ε^{2} t_{n}^{\frac{4 N - 2 α}{N}} | u_{n} |_{2}^{\frac{4 N - 2 α}{N}} + 2 C (N, α) C_{ε}^{2} t_{n}^{2 p_{3}} | u_{n} |_{\frac{2 N p_{3}}{2 N - α}}^{2 p_{3}} \\ = 2 C (N, α) ε^{2} R^{\frac{4 N - 2 α}{N}} | v_{n} |_{2}^{\frac{4 N - 2 α}{N}} + 2 C (N, α) C_{ε}^{2} R^{2 p_{3}} | v_{n} |_{\frac{2 N p_{3}}{2 N - α}}^{2 p_{3}} \\ ⩽ 2 C (N, α) ε^{2} R^{\frac{4 N - 2 α}{N}} θ_{2}^{\frac{4 N - 2 α}{N}} + o (1), \end{matrix}$ where $θ_{2}$ be the constant in Lemma 2.2. Choosing $ε > 0$ satisfying $2 C (N, α) ε^{2} R^{\frac{4 N - 2 α}{N}} θ_{2}^{\frac{4 N - 2 α}{N}} < R^{2} / 2$ , by Lemma 2.4 we have $\begin{array}{rcl} M & ⩾ & I_{μ_{n}} (u_{n}) \\ ⩾ & I_{μ_{n}} (t_{n} u_{n}) + \frac{1 - t_{n}^{2}}{2} ⟨ I_{μ_{n}}^{'} (u_{n}), u_{n} ⟩ \\ = & I_{μ_{n}} (t_{n} u_{n}) + o (1) \\ = & \frac{R^{2}}{2} - \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (t_{n} u_{n} (y)) F (t_{n} u_{n} (x))}{| x - y |^{α}} d x d y + o (1) \\ ⩾ & \frac{R^{2}}{2} - \frac{R^{2}}{4} + o (1) \\ = & M + \frac{1}{4} + o (1) . \end{array}$ This contradiction shows $δ > 0$ . Going if necessary to a subsequence, we may assume the existence of $k_{n} \in Z^{N}$ such that $\int_{B_{1 + \sqrt{N}} (k_{n})} | v_{n} |^{2} d x ⩾ \frac{δ}{2}$ . Let $\begin{matrix} {\tilde{v}}_{n} (\cdot) = v_{n} (\cdot + k_{n}), {\tilde{u}}_{n} (\cdot) = u_{n} (\cdot + k_{n}), \end{matrix}$ then $\begin{matrix} \int_{B_{1 + \sqrt{N}} (0)} | {\tilde{v}}_{n} |^{2} d x ⩾ \frac{δ}{2} . \end{matrix}$ By Lemma 2.2, ${{\tilde{v}}_{n}}$ is bounded in $H^{1} (R^{N})$ , we may assume ${\tilde{v}}_{n} ⇀ \tilde{v}$ in $H^{1} (R^{N})$ , ${\tilde{v}}_{n} \to \tilde{v}$ in $L_{loc}^{s} (R^{N})$ , $2 ⩽ s < 2^{}$ , ${\tilde{v}}_{n} \to \tilde{v}$ a.e. on $R^{N}$ . Then $\int_{B_{1 + \sqrt{N}} (0)} | \tilde{v} |^{2} d x > 0$ , there exist a set $A_{1} \subset B_{1 + \sqrt{N}} (0)$ with $meas (A_{1}) > 0$ and a constant $β_{0} > 0$ such that $| \tilde{v} | ⩾ β_{0}$ on $A_{1}$ . By Egoroff’s theorem, we can find a set $A_{2} \subset A_{1}$ with $meas (A_{2}) > 0$ such that ${\tilde{v}}_{n} \to \tilde{v}$ uniformly on $A_{2}$ and we may assume $| {\tilde{v}}_{n} | ⩾ \frac{β_{0}}{2}$ on $A_{2}$ . Hence, $\begin{matrix} {\tilde{u}}_{n} (x) = ‖ u_{n} ‖_{μ_{n}} {\tilde{v}}_{n} (x) \to \infty, \forall x \in A_{2} . \end{matrix}$ By (F3), we have $\begin{matrix} \frac{F ({\tilde{u}}_{n} (x))}{‖ u_{n} ‖_{μ_{n}}} = \frac{F ({\tilde{u}}_{n} (x))}{{\tilde{u}}_{n} (x)} {\tilde{v}}_{n} (x) \to \infty, \forall x \in A_{2} . \end{matrix}$ Finally, by Fatou’s Lemma, one can has $\begin{matrix} {lim_{_}}_{n \to \infty} \iint_{R^{2 N}} \frac{F (u_{n} (x)) F (u_{n} (y))}{‖ u_{n} ‖_{μ_{n}}^{2} | x - y |^{α}} d x d y \\ ⩾ {lim_{_}}_{n \to \infty} \iint_{A_{2} \times A_{2}} \frac{F ({\tilde{u}}_{n} (x)) F ({\tilde{u}}_{n} (y))}{‖ u_{n} ‖_{μ_{n}}^{2} | x - y |^{α}} d x d y \\ ⩾ \iint_{A_{2} \times A_{2}} {lim_{_}}_{n \to \infty} \frac{F ({\tilde{u}}_{n} (x)) F ({\tilde{u}}_{n} (y))}{‖ u_{n} ‖_{μ_{n}}^{2} | x - y |^{α}} d x d y \\ = \infty . \end{matrix}$ Then $\begin{matrix} o (1) = \frac{I_{μ_{n}} (u_{n})}{‖ u_{n} ‖_{μ_{n}}^{2}} = \frac{1}{2} - \frac{1}{2} \iint_{R^{2 N}} \frac{F (u_{n} (x)) F (u_{n} (y))}{‖ u_{n} ‖_{μ_{n}}^{2} | x - y |^{α}} d x d y \to - \infty \end{matrix}$ which is a contradiction. □
Lemma 2.10.
Assume that (F1), (F2), (F3) and (V) hold. If* $u_{n} ⇀ u$ in $H^{1} (R^{N})$ and $\begin{matrix} (2.13) & \begin{matrix} lim_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (y)) [f (u_{n} (x)) u_{n} (x) - F (u_{n} (x))]}{| x - y |^{α}} d x d y \\ = \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) [f (u (x)) u (x) - F (u (x))]}{| x - y |^{α}} d x d y, \end{matrix} \end{matrix}$ then $\begin{matrix} (2.14) & lim_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (x)) F (u_{n} (y))}{| x - y |^{α}} d x d y = \int_{R^{N}} \int_{R^{N}} \frac{F (u (x)) F (u (y))}{| x - y |^{α}} d x d y \end{matrix}$ and $\begin{matrix} (2.15) & lim_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (y)) f (u_{n} (x)) u (x)}{| x - y |^{α}} d x d y = \int_{R^{N}} \int_{R^{N}} \frac{F (u (y)) f (u (x)) u (x)}{| x - y |^{α}} d x d y . \end{matrix}$
Proof.
For any $ε > 0$ , there exists $R > 0$ such that $\begin{array}{l} (2.16) & \iint_{R^{2 N} ∖ (B_{R} (0) \times B_{R} (0))} \frac{F (u (x)) F (u (y))}{| x - y |^{α}} d x d y < ε, \\ (2.17) & \iint_{R^{2 N} ∖ (B_{R} (0) \times B_{R} (0))} \frac{F (u (y)) [f (u (x)) u (x) - F (u (x))]}{| x - y |^{α}} d x d y < ε, \end{array}$ and $\begin{matrix} (2.18) & {(\int_{R^{N} ∖ B_{R} (0)} | u |^{\frac{2 N p_{i}}{2 N - α}} d x)}^{\frac{2 N - α}{2 N p_{i}}} < ε, i = 1, 2 . \end{matrix}$ Passing to a subsequence, we have $u_{n} \to u$ in $L^{\frac{2 N p_{2}}{2 N - α}} (B_{R} (0))$ , $u_{n} \to u$ a.e. on $R^{N}$ , $| u_{n} | ⩽ w \in L^{\frac{2 N p_{2}}{2 N - α}} (B_{R} (0))$ . By Lebesgue’s dominated convergence theorem, one can get $\begin{matrix} (2.19) & \begin{matrix} lim_{n \to \infty} \iint_{B_{R} (0) \times B_{R} (0)} \frac{F (u_{n} (x)) F (u_{n} (y))}{| x - y |^{α}} d x d y \\ = \iint_{B_{R} (0) \times B_{R} (0)} \frac{F (u (x)) F (u (y))}{| x - y |^{α}} d x d y, \end{matrix} \end{matrix}$ and $\begin{matrix} lim_{n \to \infty} \iint_{B_{R} (0) \times B_{R} (0)} \frac{F (u_{n} (y)) [f (u_{n} (x)) u_{n} (x) - F (u_{n} (x))]}{| x - y |^{α}} d x d y \\ = \iint_{B_{R} (0) \times B_{R} (0)} \frac{F (u (y)) [f (u (x)) u (x) - F (u (x))]}{| x - y |^{α}} d x d y . \end{matrix}$ Then it follows from (2.13) that $\begin{matrix} (2.20) & \begin{matrix} lim_{n \to \infty} \iint_{R^{2 N} ∖ (B_{R} (0) \times B_{R} (0))} \frac{F (u_{n} (y)) [f (u_{n} (x)) u_{n} (x) - F (u_{n} (x))]}{| x - y |^{α}} d x d y \\ = \iint_{R^{2 N} ∖ (B_{R} (0) \times B_{R} (0))} \frac{F (u (y)) [f (u (x)) u (x) - F (u (x))]}{| x - y |^{α}} d x d y . \end{matrix} \end{matrix}$ By (2.17) and (2.20), there exists $n_{0} > 0$ such that $\begin{matrix} \iint_{R^{2 N} ∖ (B_{R} (0) \times B_{R} (0))} \frac{F (u_{n} (y)) [f (u_{n} (x)) u_{n} (x) - F (u_{n} (x))]}{| x - y |^{α}} d x d y < ε, \forall n ⩾ n_{0} . \end{matrix}$ By (F4), there exists $β > 0$ such that $\begin{matrix} F (s) ⩽ β [f (s) s - F (s)], \forall s \in R . \end{matrix}$ Thus, we have $\begin{matrix} (2.21) & \begin{matrix} \iint_{R^{2 N} ∖ (B_{R} (0) \times B_{R} (0))} \frac{F (u_{n} (x)) F (u_{n} (y))}{| x - y |^{α}} d x d y \\ ⩽ β \iint_{R^{2 N} ∖ (B_{R} (0) \times B_{R} (0))} \frac{F (u_{n} (y)) [f (u_{n} (x)) u_{n} (x) - F (u_{n} (x))]}{| x - y |^{α}} d x d y < β ε . \end{matrix} \end{matrix}$ Combining (2.19) with (2.21), one has $\begin{matrix} \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (x)) F (u_{n} (y))}{| x - y |^{α}} d x d y \to \int_{R^{N}} \int_{R^{N}} \frac{F (u (x)) F (u (y))}{| x - y |^{α}} d x d y . \end{matrix}$ The proof of (2.14) is now complete.

Set $Φ_{n} (x) = \int_{R^{N}} \frac{F (u_{n} (y))}{| x - y |^{α}} d y$ , $Φ (x) = \int_{R^{N}} \frac{F (u (y))}{| x - y |^{α}} d y$ . Since ${F (u_{n})}$ is bounded in $L^{\frac{2 N}{2 N - α}} (R^{N})$ and $F (u_{n}) \to F (u)$ a.e. on $R^{N}$ , we may assume $\begin{matrix} F (u_{n}) ⇀ F (u) in L^{\frac{2 N}{2 N - α}} (R^{N}) . \end{matrix}$ Using Hardy–Littlewood–Sobolev inequality, we have $\begin{matrix} Φ_{n} ⇀ Φ in L^{\frac{2 N}{α}} (R^{N}) . \end{matrix}$ Then $\begin{matrix} (2.22) & \int_{B_{R} (0)} Φ_{n} f (u) u d x \to \int_{B_{R} (0)} Φ f (u) u d x . \end{matrix}$ From Hölder inequality and Lebesgue’s dominated convergence theorem, we obtain $\begin{array}{l} \int_{B_{R} (0)} | Φ_{n} f (u_{n}) u - Φ_{n} f (u) u | d x \\ ⩽ {(\int_{B_{R} (0)} | Φ_{n} |^{\frac{2 N}{α}})}^{\frac{α}{2 N}} {(\int_{B_{R} (0)} {| f (u_{n}) - f (u) |}^{\frac{2 N p_{2}}{(2 N - α) (p_{2} - 1)}})}^{\frac{(2 N - α) (p_{2} - 1)}{2 N p_{2}}} {(\int_{B_{R} (0)} | u |^{\frac{2 N p_{2}}{2 N - α}})}^{\frac{2 N - α}{2 N p_{2}}} \\ (2.23) & = o (1) . \end{array}$ Combining (2.22) with (2.23), we obtain $\begin{matrix} (2.24) & \int_{B_{R} (0)} Φ_{n} f (u_{n}) u d x \to \int_{B_{R} (0)} Φ f (u) u d x . \end{matrix}$ From (F1), Hölder inequality and (2.18), we know $\begin{matrix} (2.25) & \begin{matrix} \int_{R^{N} ∖ B_{R} (0)} | Φ_{n} f (u_{n}) u | d x \\ ⩽ C_{1} | Φ_{n} |_{\frac{2 N}{α}} (| u_{n} |_{\frac{2 N p_{1}}{2 N - α}}^{p_{1} - 1} | u |_{\frac{2 N p_{1}}{2 N - α}, R^{N} ∖ B_{R} (0)} + | u_{n} |_{\frac{2 N p_{2}}{2 N - α}}^{p_{2} - 1} | u |_{\frac{2 N p_{2}}{2 N - α}, R^{N} ∖ B_{R} (0)}) \\ ⩽ C_{2} ε . \end{matrix} \end{matrix}$ Similar to (2.25), we have $\begin{matrix} (2.26) & \int_{R^{N} ∖ B_{R} (0)} | Φ f (u) u | d x ⩽ C_{3} ε . \end{matrix}$ By (2.24), (2.25) and (2.26), we get $\begin{matrix} | \int_{R^{N}} Φ_{n} f (u_{n}) u d x - \int_{R^{N}} Φ f (u) u d x | \to 0 . \end{matrix}$ This ends the proof. □

3. Proofs of Theorem 1.1 and Theorem 1.2

In this section, our main goal is to give the proofs of Theorem 1.1 and Theorem 1.2.

Proof of Theorem 1.1.

For the case $μ = 0$ , see the proof of Theorem 1.2. In what follows, we only consider the case when $0 < μ < \frac{{(N - 2)}^{2}}{4}$ . By Lemma 2.8 and Lemma 2.9, there exist a constant $c \in [κ, m_{μ}]$ and a sequence ${u_{n}} \subset H^{1} (R^{N})$ satisfying $I_{μ} (u_{n}) \to c$ , $‖ I_{μ}^{'} (u_{n}) ‖ (1 + ‖ u_{n} ‖_{μ}) \to 0$ and ${‖ u_{n} ‖_{μ}}$ is bounded. Similar to the proof of Lemma 2.9, ${u_{n}}$ is non-vanishing sequence. Passing to a subsequence, we may assume the existence of $y_{n} \in Z^{N}$ such that $\int_{B_{1 + \sqrt{N}} (y_{n})} | u_{n} |^{2} d x ⩾ δ > 0$ .

Next, we prove that ${| y_{n} |}$ is bounded. Arguing by contradiction, suppose that $| y_{n} | \to \infty$ . Let ${\tilde{u}}_{n} (\cdot) = u_{n} (\cdot + y_{n})$ , then $\int_{B_{1 + \sqrt{N}} (0)} | {\tilde{u}}_{n} |^{2} d x ⩾ δ > 0$ . By Lemma 2.2, we know ${‖ {\tilde{u}}_{n} ‖_{μ}}$ is bounded. We may assume ${\tilde{u}}_{n} ⇀ \tilde{u}$ in $H^{1} (R^{N})$ , ${\tilde{u}}_{n} \to \tilde{u}$ in $L_{loc}^{s} (R^{N})$ , $2 ⩽ s < 2^{*}$ and ${\tilde{u}}_{n} \to \tilde{u}$ a.e. on $R^{N}$ . Therefore, $\tilde{u} \neq 0$ . For any $ϕ \in C_{0}^{\infty} (R^{N})$ , let $ϕ_{n} (\cdot) = ϕ (\cdot - y_{n})$ , then ${‖ ϕ_{n} ‖_{μ}}$ is bounded. By $| y_{n} | \to \infty$ , one can have $\begin{matrix} (3.1) & \begin{matrix} \int_{R^{N}} \frac{| u_{n} ϕ_{n} |}{| x |^{2}} d x & = \int_{| x | > R} \frac{| u_{n} ϕ_{n} |}{| x |^{2}} d x + o (1) \\ ⩽ \frac{1}{R^{2}} | u_{n} |_{2} | ϕ_{n} |_{2} + o (1) \\ = o (1) . \end{matrix} \end{matrix}$ From $‖ I_{μ}^{'} (u_{n}) ‖ \to 0$ and (3.1), we have $\begin{matrix} ⟨ I_{0}^{'} ({\tilde{u}}_{n}), ϕ ⟩ & = \int_{R^{N}} (\nabla {\tilde{u}}_{n} \cdot \nabla ϕ + V (x) {\tilde{u}}_{n} ϕ) d x - \int_{R^{N}} \int_{R^{N}} \frac{F ({\tilde{u}}_{n} (y)) f ({\tilde{u}}_{n} (x)) ϕ (x)}{| x - y |^{α}} d x d y \\ = \int_{R^{N}} (\nabla u_{n} \cdot \nabla ϕ_{n} + V (x) u_{n} ϕ_{n}) d x - \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (y)) f (u_{n} (x)) ϕ_{n} (x)}{| x - y |^{α}} d x d y \\ = ⟨ I_{μ}^{'} (u_{n}), ϕ_{n} ⟩ + μ \int_{R^{N}} \frac{| u_{n} ϕ_{n} |}{| x |^{2}} d x \\ = o (1) . \end{matrix}$ Then it follows from Lemma 2.3 that $⟨ I_{0}^{'} (\tilde{u}), ϕ ⟩ = 0$ . Notice that $C_{0}^{\infty} (R^{N})$ is dense in $H^{1} (R^{N})$ , we have $⟨ I_{0}^{'} (\tilde{u}), \tilde{u} ⟩ = 0$ and $\tilde{u} \in N_{0}$ . By Lemma 2.5, there exists a unique $\tilde{t} > 0$ such that $\tilde{t} \tilde{u} \in N_{μ}$ . Then $\begin{array}{rcl} I_{0} (\tilde{u}) & ⩾ & I_{0} (\tilde{t} \tilde{u}) \\ = & \frac{1}{2} ‖ \tilde{t} \tilde{u} ‖_{0} - \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (\tilde{t} \tilde{u} (x)) F (\tilde{t} \tilde{u} (y))}{| x - y |^{α}} d x d y \\ > & \frac{1}{2} ‖ \tilde{t} \tilde{u} ‖_{μ} - \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (\tilde{t} \tilde{u} (x)) F (\tilde{t} \tilde{u} (y))}{| x - y |^{α}} d x d y \\ = & I_{μ} (\tilde{t} \tilde{u}) \\ (3.2) & ⩾ & m_{μ} . \end{array}$ From $\tilde{u} \in N_{0}$ and Fatou’s lemma, one can get $\begin{array}{rcl} I_{0} (\tilde{u}) & = & \frac{1}{2} ‖ \tilde{u} ‖_{0}^{2} - \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (\tilde{u} (x)) F (\tilde{u} (y))}{| x - y |^{α}} d x d y \\ = & \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (\tilde{u} (y)) [f (\tilde{u} (x)) \tilde{u} (x) - F (\tilde{u} (x))]}{| x - y |^{α}} d x d y \\ ⩽ & \frac{1}{2} {lim_{_}}_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F ({\tilde{u}}_{n} (y)) [f ({\tilde{u}}_{n} (x)) {\tilde{u}}_{n} (x) - F ({\tilde{u}}_{n} (x))]}{| x - y |^{α}} d x d y \\ = & \frac{1}{2} {lim_{_}}_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (y)) [f (u_{n} (x)) u_{n} (x) - F (u_{n} (x))]}{| x - y |^{α}} d x d y \\ = & {lim_{_}}_{n \to \infty} [I_{μ} (u_{n}) - \frac{1}{2} ⟨ I_{μ}^{'} (u_{n}), u_{n} ⟩] \\ (3.3) & = & {lim_{_}}_{n \to \infty} I_{μ} (u_{n}) = c ⩽ m_{μ}, \end{array}$ which is a contradiction. Then ${| y_{n} |}$ is bounded. There exists $R_{0} > 0$ such that $\begin{matrix} \int_{B_{R_{0}} (0)} | u_{n} |^{2} d x ⩾ δ > 0 . \end{matrix}$ We may assume $u_{n} ⇀ u_{0}$ in $H^{1} (R^{N})$ , and then a standard argument shows that $u_{0} \neq 0$ and $I_{μ}^{'} (u_{0}) = 0$ . By Lemma 2.3, we get $\begin{matrix} 0 < m_{μ} ⩽ I_{μ} (u_{0}) ⩽ {lim_{_}}_{n \to \infty} I_{μ} (u_{n}) = c ⩽ m_{μ} . \end{matrix}$ The proof is thus finished. □

Proof of Theorem 1.2.

Let $0 < μ_{1} ⩽ μ_{2} < \frac{{(N - 2)}^{2}}{4}$ . By Theorem 1.1, there exists $u_{1} \in N_{μ_{1}}$ such that $I_{μ_{1}} (u_{1}) = m_{μ_{1}}$ . By Lemma 2.5, there exists $t_{1} > 0$ such that $t_{1} u_{1} \in N_{μ_{2}}$ . Thus, $\begin{matrix} m_{μ_{1}} = I_{μ_{1}} (u_{1}) ⩾ I_{μ_{1}} (t_{1} u_{1}) ⩾ I_{μ_{2}} (t_{1} u_{1}) ⩾ m_{μ_{2}} . \end{matrix}$ For any $u \in N_{0}$ , there exists $t_{2} > 0$ such that $t_{2} u \in N_{μ_{1}}$ , then $\begin{matrix} I_{0} (u) ⩾ I_{0} (t_{2} u) ⩾ I_{μ_{1}} (t_{2} u) ⩾ m_{μ_{1}}, \end{matrix}$ which implies $m_{0} ⩾ m_{μ_{1}}$ . By Theorem 1.1, for any n, there exists $u_{n} \neq 0$ such that $\begin{matrix} I_{μ_{n}} (u_{n}) = m_{μ_{n}}, I_{μ_{n}}^{'} (u_{n}) = 0 . \end{matrix}$ Hence, ${sup}_{n} | I_{μ_{n}} (u_{n}) | ⩽ m_{0} < \infty$ . By Lemma 2.9, ${‖ u_{n} ‖_{μ_{n}}}$ is bounded. By a standard argument, we can prove that ${u_{n}}$ is non-vanishing sequence. Going if necessary to a subsequence, we may assume the existence of $y_{n} \in Z^{N}$ such that $\int_{B_{1 + \sqrt{N}} (y_{n})} u_{n}^{2} d x ⩾ δ > 0$ . Let ${\tilde{u}}_{n} (\cdot) = u_{n} (\cdot + y_{n})$ , then ${‖ {\tilde{u}}_{n} ‖_{μ_{n}}}$ is bounded and $\int_{B_{1 + \sqrt{N}} (x_{n})} {\tilde{u}}_{n}^{2} d x ⩾ δ > 0$ . Passing to a subsequence, ${\tilde{u}}_{n} ⇀ \bar{u}$ in $H^{1} (R^{N})$ , we may assume ${\tilde{u}}_{n} \to \bar{u}$ in $L_{loc}^{s} (R^{N})$ , $2 ⩽ s < 2^{*}$ and ${\tilde{u}}_{n} \to \bar{u}$ a.e. on $R^{N}$ . Then $\bar{u} \neq 0$ . By a standard argument, we can prove that $I_{0}^{'} (\bar{u}) = 0$ . By Fatou’s Lemma, we have $\begin{matrix} m_{0} & ⩾ {lim^{‾}}_{n \to \infty} m_{μ_{n}} \\ = {lim^{‾}}_{n \to \infty} I_{μ_{n}} (u_{n}) \\ = {lim^{‾}}_{n \to \infty} [I_{μ_{n}} (u_{n}) - \frac{1}{2} ⟨ I_{μ_{n}}^{'} (u_{n}), u_{n} ⟩] \\ = \frac{1}{2} {lim^{‾}}_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F (u_{n} (y)) [f (u_{n} (x)) u_{n} (x) - F (u_{n} (x))]}{| x - y |^{α}} d x d y \\ = \frac{1}{2} {lim^{‾}}_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F ({\tilde{u}}_{n} (y)) [f ({\tilde{u}}_{n} (x)) {\tilde{u}}_{n} (x) - F ({\tilde{u}}_{n} (x))]}{| x - y |^{α}} d x d y \\ ⩾ \frac{1}{2} {lim_{_}}_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F ({\tilde{u}}_{n} (y)) [f ({\tilde{u}}_{n} (x)) {\tilde{u}}_{n} (x) - F ({\tilde{u}}_{n} (x))]}{| x - y |^{α}} d x d y \\ ⩾ \frac{1}{2} \int_{R^{N}} \int_{R^{N}} \frac{F (\bar{u} (y)) [f (\bar{u} (x)) \bar{u} (x) - F (\bar{u} (x))]}{| x - y |^{α}} d x d y \\ = I_{0} (\bar{u}) - \frac{1}{2} ⟨ I_{0}^{'} (\bar{u}), \bar{u} ⟩ = I_{0} (\bar{u}) ⩾ m_{0} . \end{matrix}$ This implies $\bar{u}$ is a ground state of $I_{0}$ . Moreover, $\begin{matrix} lim_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F ({\tilde{u}}_{n} (y)) [f ({\tilde{u}}_{n} (x)) {\tilde{u}}_{n} (x) - F ({\tilde{u}}_{n} (x))]}{| x - y |^{α}} d x d y \\ = \int_{R^{N}} \int_{R^{N}} \frac{F (\bar{u} (y)) [f (\bar{u} (x)) \bar{u} (x) - F (\bar{u} (x))]}{| x - y |^{α}} d x d y . \end{matrix}$ It follows from Lemma 2.10 that $\begin{matrix} (3.4) & lim_{n \to \infty} \int_{R^{N}} \int_{R^{N}} \frac{F ({\tilde{u}}_{n} (y)) f ({\tilde{u}}_{n} (x)) [{\tilde{u}}_{n} (x) - \bar{u} (x)]}{| x - y |^{α}} d x d y = 0 . \end{matrix}$ Notice that $\begin{matrix} (3.5) & \begin{matrix} ‖ {\tilde{u}}_{n} - \bar{u} ‖_{μ_{n}}^{2} = & ⟨ I_{μ_{n}}^{'} ({\tilde{u}}_{n}), {\tilde{u}}_{n} - \bar{u} ⟩ - {(\bar{u}, {\tilde{u}}_{n} - \bar{u})}_{μ_{n}} \\ + \int_{R^{N}} \int_{R^{N}} \frac{F ({\tilde{u}}_{n} (y)) f ({\tilde{u}}_{n} (x)) [{\tilde{u}}_{n} (x) - \bar{u} (x)]}{| x - y |^{α}} d x d y . \end{matrix} \end{matrix}$ It is easy to see that ${\tilde{u}}_{n} \to \bar{u}$ in $H^{1} (R^{N})$ . □

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