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Ground-state solutions for fractional Kirchhoff-Choquard equations with critical growth

  • Jie Yang EMAIL logo and Haibo Chen
Published/Copyright: April 9, 2025

Abstract

We study the following fractional Kirchhoff-Choquard equation:

a + b R N ( Δ ) s 2 u 2 d x ( Δ ) s u + V ( x ) u = ( I μ * F ( u ) ) f ( u ) , x R N , u H s ( R N ) ,

where a , b are positive constants, N > 2 s , μ ( ( N 4 s ) + , N ) , s ( 0 , 1 ) , and I μ is the Riesz potential. Considering the case that nonlinearity f has critical growth, combining a monotonicity trick and global compactness lemma, we prove that the equation has a ground-state solution. Moreover, we study the regularity of ground-state solutions to the above equation, which extends the results in Moroz-Van Schaftingen [Existence of groundstates for a class of nonlinear Choquard equations, Trans. Amer. Math. Soc. 367 (2015), 6557–6579] to the fractional Laplacian case.

MSC 2010: 35R11; 49J35

1 Introduction

In this article, we study the following fractional Kirchhoff-Choquard equation:

(KC) a + b R N ( Δ ) s 2 u 2 d x ( Δ ) s u + V ( x ) u = ( I μ * F ( u ) ) f ( u ) , x R N , u H s ( R N ) ,

where a , b are positive constants, N > 2 s , μ ( ( N 4 s ) + , N ) , s ( 0 , 1 ) , and

I μ = Γ N μ 2 Γ μ 2 π N 2 2 μ x N μ , x R N \ { 0 } ,

is the Riesz potential. The operator ( Δ ) s is the fractional Laplacian defined by

( Δ ) s u = 1 ( ξ 2 s ( u ) ) , ξ R N ,

where denotes the Fourier transform on R N . It is well known that it can also be computed by

( Δ ) s v ( x ) = C N , s P . V . R N v ( x ) v ( y ) x y N + 2 s d x d y ,

if v is smooth enough, where C N , s is the normalization constant, and P.V. denotes a Cauchy principle value.

When s = 1 , a = 1 , b = 0 , N = 3 , μ = 2 , F ( u ) = u 2 , from equation (KC), we derive the nonlinear Choquard equation as follows:

(1.1) Δ u + u = ( I 2 * u 2 ) u , x R 3 .

Choquard used (1.1) to describe electrons trapped in their own holes. To some extent, it is similar to the Hartree Fock theory of single component plasma in 1976, as shown in [16].

The more general form of (1.1) is as follows:

(1.2) Δ u + V ( x ) u = ( I μ * F ( u ) ) f ( u ) , x R N .

When V ( x ) 1 , Moroz and Van Schaftingen [23] investigated the generalized Berestycki-Lions condition [4] in the sense of Hartree-type nonlinearity. They assumed N 3 and f C ( R , R ) satisfied the following conditions:

  1. There exists t 0 R \ { 0 } such that F ( t 0 ) 0 , where F : t R 0 t f ( τ ) d τ .

  2. There exists C > 0 such that for every t R , t f ( t ) C t N + μ N + t N + μ N 2 .

  3. F is subcritical

    lim t 0 F ( t ) t N + μ N = lim t F ( t ) t N + μ N 2 = 0 ,

where N + μ N , N + μ N 2 are called lower and upper H-L-S critical exponents. Under almost necessary conditions above, they derived the existence of ground-state solution, local regularity properties, and radial symmetry of solutions to (1.2). For more related work, we refer to [1,7,10,11,22,33,34,36,37,39].

When a = 1 , b = 0 , equation (KC) reduces to the following fractional Choquard equation:

(1.3) ( Δ ) s u + V ( x ) u = ( I μ * F ( u ) ) f ( u ) , x R N , u H s ( R N ) .

It originates from the study of fractional Laplacian with nonlocal Hartree-type nonlinearities. For the case V ( x ) = w , F ( u ) = u p , the regularity, existence, nonexistence, symmetry, and decay properties of solutions for equation (1.3) have been obtained by d’Avenia et al. [2]. Subsequently, these results were extended to more general potential V and nonlinearity f , see [14,17,18,20,21,2729,38]. For instance, Shen-Gao-Yang were concerned equation (1.3) with V = 1 . They assumed f C ( R , R ) satisfied the general Berestycki-Lions-type assumptions:

  1. There exists C > 0 such that for every t R ,

    t f ( t ) C t 2 + t N + μ N 2 s ;

  2. Let F : t R 0 t f ( τ ) d τ and suppose that

    lim t 0 F ( t ) t 2 = 0 , lim t F ( t ) t 2 μ , s * = 0 ,

    where 2 μ , s * = N + μ N 2 s is the fractional upper H-L-S critical exponent.

  3. There exists t 0 R such that F ( t 0 ) 0 .

The fractional Kirchhoff-type problem has been widely studied in recent years, a problem naturally arises: can we obtain a ground-state solution to problem (KC) involving the fractional upper H-L-S critical exponent? In this article, attention is paid to this case and a positive answer is provided. We assume that f C ( R , R ) satisfies the following conditions:

  1. lim t 0 f ( t ) t = 0 , lim t + f ( t ) t 2 μ , s * 1 = 1 ;

  2. there exist ν > 0 and r ( 2 , 2 μ , s * ) such that

    f ( t ) t 2 s + μ N 2 s + ν t r 1 , t R .

Because of the emergence of convolution non-linearity, the methods used in [20] cannot be directly applied to obtain the desired solution of equation (KC). On the other hand, due to the variable potential V ( x ) , the existing methods for handling Choquard-type equations in [5] fail in equation (KC). Encouraged by the aforementioned work, we were particularly inspired by the spirit of [35] and adopted similar techniques to achieve relevant results for (KC). First of all, we discuss the constant potential case. Concretely, through taking a minimizing sequence in a Pohozaev manifold, we obtain a minimizer. After that, we show the minimizer is a critical point. Therefore, we gain the existence of the ground-state solution of Pohozaev-type to (KC) with constant potential. In view of this result, we employed some analytical techniques to obtain the existence of Pohozaev-type ground-state solutions to (KC) with general potential.

Throughout the article, we assume that V satisfies

  1. V C 1 ( R N ) L ( R N ) and ( N 4 s ) V ( x ) ( V ( x ) , x ) 0 for any x R N .

  2. 0 < V ( x ) liminf y + V ( y ) = V R + for all x R N and the inequality is strict in a subset of positive Lebesgue measure.

We remark that in order to apply Jeanjean’s monotonicity technique [13] to equation (KC), the corresponding limit problem plays an important role. Hence, we investigate the limit problem

a + b R N ( Δ ) s 2 u 2 d x ( Δ ) s u + V u = ( I μ * F ( u ) ) f ( u ) , x R N , u H s ( R N ) . ( K C )

Due to the fact that the Nehari-type monotonicity condition of f is not satisfied, the Nehari manifold method used in [26] is no longer effective in our setup. To prove our results, we use the Pohozaev manifold. We define

= { u H s ( R N ) \ { 0 } : P ( u ) = 0 } ,

where

P ( u ) = a ( N 2 s ) 2 u D s , 2 ( R N ) 2 + N V 2 u L 2 ( R N ) 2 + b ( N 2 s ) 2 u D s , 2 ( R N ) 4 N + μ 2 R N ( I μ * F ( u ) ) F ( u ) d x .

We will obtain the next result.

Theorem 1.1

Let N > 2 s , μ ( ( N 4 s ) + , N ) , V > 0 , r > max μ + N 4 s N 2 s , N + μ 2 ( N 2 s ) , ( H 7 )–( H 8 ) hold, then there exists b 0 > 0 such that for b ( 0 , b 0 ) , equation ( K C ) has a Pohozaev-type ground-state solution.

The proof of Theorem 1.1 was enlightened by He and Rădulescu [12]. The author studied the existence of positive solutions for the fractional Choquard equation with pure critical nonlinearity

( Δ ) s u + a ( x ) u = ( I μ * u 2 μ , s * ) u 2 μ , s * , x R N , x D s , 2 ( R N ) , u ( x ) > 0 .

They proved that in the case of small perturbations of the potential, the above equation has at least one positive solution. However, we note that due to the appearance of the Kirchhoff term and the fact that the nonlinear term is a more general function with a critical index rather than a power function, the ideas of He and Rădulescu [12] cannot be immediately applied to the proof of Theorem 1.1. We need to make some precise estimates of this problem. for example, see Lemma 3.5.

When s = 1 , Cassani and Zhang [6] considered the following problem:

ε 2 Δ u + V ( x ) u = ε μ ( I μ * F ( u ) ) f ( u ) , x R N ,

where ε > 0 is the dimensionalized Planck constant. The non-locality caused by the Kirchhoff term and the non-local operator ( Δ ) s make the argument discussed in [6] impossible to proceed directly in our case, as some estimates become more subtle.

On this basis, we consider equation (KC) with general potential. Next is the second result.

Theorem 1.2

Let N > 2 s , μ ( ( N 4 s ) + , N ) , r > max μ + N 4 s N 2 s , N + μ 2 ( N 2 s ) . Assume that ( V 1 ) ( V 2 ) , ( H 7 ) ( H 8 ) hold, then there exists b 1 > 0 such that for b ( 0 , b 1 ) , equation (KC) has a ground-state solution.

Remark 1.1

Motivated by [6], we apply Jeanjean’s monotonicity technique [13] to prove Theorem 1.2. Compared with [6], the presence of fractional Laplacian results in some methods in [6] that not being applicable to our problem. The major difficulty is to construct a bounded (PS) sequence and prove that the sequence weakly converges to a critical point. By proving a global compactness lemma and utilizing Theorem 1.1, we can overcome the aforementioned difficulties.

Now, to obtain more information about solutions to equation (KC), we study the regularity of ground-state solutions.

Theorem 1.3

Let N > 2 s , μ N ( N 6 s ) N 2 s + , N . Assume that f C ( R , R ) satisfies ( H 4 ) , V L ( R N ) . Let u H s ( R N ) be a nontrivial solution to equation (KC).

  1. If s 1 2 , then u L ( R N ) C 0 , α ( R N ) , for any α < 2 s .

  2. If s > 1 2 , then u L ( R N ) C 1 , α ( R N ) , for any α < 2 s 1 .

Throughout the article, we use the following notations:

  • L q ( R N ) denotes the Lebesgue space with the norm

    u L q ( R N ) = R N u q d x 1 q .

  • For any x R N and R > 0 , B R ( x ) { y R N : y x < R } .

  • C , C 1 , C 2 , indicate positive numbers that may be different in different lines.

The rest of this article is organized as follows. In Section 2, we state some preliminary notations and the main lemmas. In Section 3, we study the limit problem. The proof of Theorem 1.2 is given in Section 4. In Section 5, we study the regularity of weak solutions to equation (KC).

2 Preliminaries

Let

H s ( R N ) u L 2 ( R N ) : ( Δ ) s 2 u L 2 ( R N )

be the Hilbert space endowed with an equivalent norm

u H s ( R N ) a u D s , 2 ( R N ) 2 + R N V ( x ) u 2 d x 1 2 ,

where u D s , 2 ( R N ) R N ( Δ ) s 2 u 2 d x 1 2 ,

D s , 2 ( R N ) u L 2 s * ( R N ) : ( Δ ) s 2 u L 2 ( R N ) ,

and 2 s * = 2 N N 2 s . It follows from [25, Propositions 3.4 and 3.6] that

(2.1) R 2 N u ( x ) u ( y ) 2 x y N + 2 s d x d y = 2 C ( N , s ) 1 R N ( Δ ) s 2 u 2 d x ,

where

C ( N , s ) = R N 1 cos ζ 1 ζ N + 2 s d ζ 1 .

To prove Theorem 1.2, we consider the next functional Φ : H s ( R N ) R defined by

Φ ( u ) = 1 2 R N a ( Δ ) s 2 u 2 + V ( x ) u 2 d x + b 4 R N ( Δ ) s 2 u 2 d x 2 1 2 R N ( I μ * F ( u ) ) F ( u ) d x .

Obviously, Φ is well defined in H s ( R N ) and Φ C 1 ( H s ( R N ) , R ) , and

(2.2) Φ ( u ) , v = R N a ( Δ ) s 2 u ( Δ ) s 2 v + V ( x ) u v d x + b R N ( Δ ) s 2 u 2 d x × R N ( Δ ) s 2 u ( Δ ) s 2 v d x R N ( I μ * F ( u ) ) f ( u ) v d x .

We define the Pohozaev manifold

= { u H s ( R N ) : P ( u ) = 0 } ,

where

P ( u ) = a ( N 2 s ) u D s , 2 ( R N ) 2 2 + 1 2 R N [ N V ( x ) + ( V ( x ) , x ) ] u 2 d x + b ( N 2 s ) u D s , 2 ( R N ) 4 2 N + μ 2 R N ( I μ * F ( u ) ) F ( u ) d x

is the Pohozaev identity associated with equation (KC). Throughout this article, we always assume that ( V 1 ) ( V 2 ) and ( H 7 ) ( H 8 ) hold. In order to investigate the non-local term R N ( I μ * F ( u ) ) F ( u ) d x , we recall the following Hardy-Littlewood-Sobolev inequality [15].

Lemma 2.1

[15] Let μ ( 0 , N ) , p , q > 1 be such that 1 p + 1 q = 1 + μ N . For any f L p ( R N ) and g L q ( R N ) , one has

R 2 N f ( x ) g ( y ) x y N μ d x d y C ( N , μ , p ) f L p ( R N ) g L q ( R N ) .

Lemma 2.2

[15] Let 1 β , g L γ 1 ( R N ) and h L γ 2 ( R N ) . Then, there exists a constant C > 0 such that

g * h L β ( R N ) C g L γ 1 ( R N ) h L γ 2 ( R N ) ,

where

1 γ 1 + 1 γ 2 = 1 + 1 β .

It follows from Lemma 2.2 that for any v L q ( R N ) , q ( 1 , N μ ) , I μ * v L N q N μ q ( R N ) , and

(2.3) I μ * v L N q N μ q ( R N ) C ( μ , N , q ) v L q ( R N ) .

Combining the Hardy-Littlewood-Sobolev inequality with Sobolev’s inequality, for any u D s , 2 ( R N ) , we obtain

R N I μ * u N + μ N 2 s u N + μ N 2 s d x C ( N , μ ) R N u 2 N N 2 s d x N + μ N .

Let S be the best Sobolev constant

S = inf u D s , 2 ( R N ) \ { 0 } R N ( Δ ) s 2 u 2 d x R N u 2 s * d x 2 2 s *

and

(2.4) S H , L = inf u D s , 2 ( R N ) \ { 0 } R N ( Δ ) s 2 u 2 d x R N ( I μ * u 2 μ , s * ) u 2 μ , s * d x 1 2 μ , s * .

Then,

(2.5) S H , L = S ( C ( N , μ ) ) N 2 s N + μ

and it is achieved by

U ˜ ( x ) = S μ ( 2 s N ) 4 s ( μ + 2 s ) ( C ( N , μ ) ) 2 s N 2 ( 2 s + μ ) U ( x ) ,

where U ( x ) = β ( 1 + x 2 ) ( N 2 s ) 2 is a minimizer for S with β = S N ( 2 s ) Γ ( N ) π N 2 Γ ( N 2 ) N 2 s 2 N . It follows from [12] that

R N ( Δ ) s 2 U 2 d x = R N U 2 s * d x = S N 2 s

and

R N ( Δ ) s 2 U ˜ 2 d x = R N ( I μ * U ˜ 2 μ , s * ) U ˜ 2 μ , s * d x = S H , L N + μ μ + 2 s .

By ( H 7 ) ( H 8 ) , Lemma 2.1, and the Sobolev embedding theorem, we have some essential characters of F .

Lemma 2.3

The following conclusions hold for F:

  1. For any ε > 0 , there exists C ε > 0 such that

    F ( t ) ε t μ N + 1 + C ε t 2 μ , s * .

  2. For all u H s ( R N ) , one has

    R N ( I μ * F ( u ) ) F ( u ) d x C u H s ( R N ) 2 + 2 μ N + u H s ( R N ) 2 2 μ , s * .

  3. For all u , v H s ( R N ) , we have

    R N ( I μ * F ( u ) ) f ( u ) v d x C u L 2 ( R N ) N + μ N + u L 2 s * ( R N ) 2 μ , s * × u L 2 ( R N ) μ N v L 2 ( R N ) + u L 2 s * ( R N ) 2 μ , s * 1 v L 2 s * ( R N ) .

Proof

(i) It follows immediately from ( H 7 ) .

(ii) From (i), Lemma 2.1, and the Sobolev embedding theorem, we can conclude that

R N ( I μ * F ( u ) ) F ( u ) d x C F ( u ) L 2 N N + μ ( R N ) 2 C R N u N + μ N + u 2 μ , s * 2 N N + μ d x N + μ N C u L 2 ( R N ) 2 ( N + μ ) N + u L 2 N N 2 s ( R N ) 2 2 μ , s * C u H s ( R N ) 2 + 2 μ N + u H s ( R N ) 2 2 μ , s * .

(iii) From ( H 7 ) , Lemma 2.1, and the Sobolev embedding theorem, we obtain

R N ( I μ * F ( u ) ) f ( u ) v d x C ( μ ) R N F ( u ) 2 N N + μ d x N + μ 2 N R N f ( u ) v 2 N N + μ d x N + μ 2 N C u L 2 ( R N ) N + μ N + u L 2 N N 2 s ( R N ) 2 μ , s * u L 2 ( R N ) μ N v L 2 ( R N ) + u L 2 N N 2 s ( R N ) 2 μ , s * 1 v L 2 N N 2 s ( R N ) .

Now, we recall a fractional version of Lions vanishing lemma [26].

Lemma 2.4

[26] Assume that { u n } is bounded in H s ( R N ) and

lim n + sup y R N B R ( y ) u n 2 d x = 0 ,

for some R > 0 . Then, u n 0 in L t ( R N ) for all t ( 2 , 2 s * ) .

At this point, let us introduce a nonlocal version of the Brezis-Lieb lemma [6, Lemma 2.2].

Lemma 2.5

[6, Lemma 2.2] Assume that μ ( 0 , N ) and there exists a constant C > 0 such that

f ( t ) C ( t μ N + t μ + 2 s N 2 s ) , t R .

Let { u n } H s ( R N ) be such that u n u weakly in H s ( R N ) and a.e. in R N as n . Then,

R N ( I μ * F ( u n ) ) F ( u n ) d x = R N ( I μ * F ( u n u ) ) F ( u n u ) d x + R N ( I μ * F ( u ) ) F ( u ) d x + o n ( 1 ) .

The next two lemmas are useful in proving the splitting property of the energy functional.

Lemma 2.6

[6, Lemma 2.6] Let μ ( 0 , N ) , q ( 1 , N μ ) and let { g n } L 1 ( R N ) L q ( R N ) be bounded and such that, up to a subsequence, for any bounded domain Ω R N , g n 0 strongly in L q ( Ω ) as n . Then, up to a subsequence, ( I μ * g n ) ( x ) 0 a.e. in R N as n .

Lemma 2.7

Let Ω R N be a domain and assume that { u n } H s ( Ω ) be such that u n u weakly in H s ( Ω ) as n . Then,

  1. For any 1 < γ q 2 N N 2 s and q > 2 ,

    lim n + Ω u n γ 1 u n u n u γ 1 ( u n u ) u γ 1 u q γ d x = 0 .

  2. Assume ζ C ( R , R ) and ζ ( t ) = o ( t μ N ) , as t 0 , ζ ( t ) c ( 1 + t γ ) for any t R , then,

    1. For any q [ γ + 1 , 2 N N 2 s ] ,

      lim n + Ω Ξ ( u n ) Ξ ( u n u ) Ξ ( u ) q γ + 1 d x = 0 ,

      where Ξ ( t ) = 0 t ζ ( τ ) d τ .

    2. If we further assume that Ω = R N , μ ( ( N 4 s ) + , N ) , and lim t + ζ ( t ) t μ + 2 s N 2 s = 0 , then,

      R N ζ ( u n ) ζ ( u n u ) ζ ( u ) 2 N N + μ v 2 N N + μ d x = o n ( 1 ) v H s ( R N ) 2 N N + μ ,

      for any v C 0 ( R N ) .

Proof

It can be proved similarly to [6, Lemma 2.5] and [24, Lemma 2.5]; thus, we omit the details for simplicity.□

The next result is a splitting property of the nonlocal energy functional for the fractional Kirchhoff-Choquard equation in R N with critical growth.

Lemma 2.8

Let μ ( ( N 4 s ) + , N ) , and assume that { u n } H s ( R N ) be such that u n u weakly in H s ( R N ) as n . Then, up to a subsequence,

R N [ ( I μ * F ( u n ) ) f ( u n ) ( I μ * F ( u n u ) ) f ( u n u ) ( I μ * F ( u ) ) f ( u ) ] v d x = o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) .

Proof

Let

(2.6) f ˜ ( t ) = f ( t ) t 4 s + μ N N 2 s t , F ˜ = 0 t f ˜ ( τ ) d τ , t R .

Note that for any v C 0 ( R N ) ,

R N ( I μ * F ( u n ) ) f ( u n ) v d x = R N ( I μ * F ( u n ) ) f ˜ ( u n ) v d x + R N ( I μ * F ( u n ) ) u n 4 s + μ N N 2 s u n v d x .

Step 1. We are going to show that for any v C 0 ( R N ) ,

(2.7) R N ( I μ * F ( u n ) ) u n 4 s + μ N N 2 s u n v d x = R N ( I μ * F ( u n u ) ) u n u 4 s + μ N N 2 s ( u n u ) v d x + R N ( I μ * F ( u ) ) u 4 s + μ N N 2 s u v d x + o n ( 1 ) v H s ( R N ) .

Recalling that μ > ( N 4 s ) + , by Lemma 2.7 (ii)(1), we can take ζ ( t ) = f ( t ) , γ = μ + 2 s N 2 s , q = 2 N N 2 s such that

(2.8) lim n + R N F ( u n ) F ( u n u ) F ( u ) 2 N N + μ d x = 0 .

Then, for ϕ n = u n 4 s + μ N N 2 s u n , and ϕ n = u n u 4 s + μ N N 2 s ( u n u ) , as well as ϕ n = u 4 s + μ N N 2 s u , there exists C > 0 such that

R N ϕ n v 2 N N + μ d x R N ϕ n 2 N 2 s + μ d x 2 s + μ N + μ R N v 2 N N 2 s d x N 2 s N + μ C R N v 2 N N 2 s d x N 2 s N + μ .

Together with Lemma 2.1, (2.8), we derive that

R N [ I μ * ( F ( u n ) F ( u n u ) F ( u ) ) ] ϕ n v d x C R N F ( u n ) F ( u n u ) F ( u ) 2 N N + μ d x N + μ 2 N R N ϕ n v 2 N N + μ d x N + μ 2 N = o n ( 1 ) v H s ( R N ) ,

uniformly for any v C 0 ( R N ) as n .

On the one hand, according to Lemma 2.7 (i) with γ = 2 s + μ N 2 s and q = 2 N N 2 s ,

lim n + R N u n 4 s + μ N N 2 s u n u n u 4 s + μ N N 2 s ( u n u ) u 4 s + μ N N 2 s u 2 N 2 s + μ d x = 0 .

For ψ n = F ( u n ) , and ψ n = F ( u n u ) , as well as ψ n = F ( u ) , we can obtain { ψ n } is bounded in L 2 N N + μ ( R N ) . By Lemma 2.1 and Hölder’s inequality, there exists C > 0 such that

R N ( I μ * ψ n ) ( u n 4 s + μ N N 2 s u n u n u 4 s + μ N N 2 s ( u n u ) u 4 s + μ N N 2 s u ) v d x C R N u n 4 s + μ N N 2 s u n u n u 4 s + μ N N 2 s ( u n u ) u 4 s + μ N N 2 s u 2 N N + μ v 2 N N + μ d x N + μ 2 N C R N u n 4 s + μ N N 2 s u n u n u 4 s + μ N N 2 s ( u n u ) u 4 s + μ N N 2 s u 2 N 2 s + μ d x 2 s + μ 2 N R N v 2 N N 2 s d x N 2 s 2 N = o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) . Then,

(2.9) R N [ I μ * F ( u n ) ] u n 4 s + μ N N 2 s u n v d x = R N [ I μ * F ( u n u ) ] u n u 4 s + μ N N 2 s ( u n u ) v d x + R N [ I μ * F ( u ) ] u 4 s + μ N N 2 s u v d x + R N [ I μ * F ( u n u ) ] u 4 s + μ N N 2 s u v d x + R N [ I μ * F ( u ) ] u n u 4 s + μ N N 2 s ( u n u ) v d x + o n ( 1 ) ,

for any v C 0 ( R N ) . Recalling that F ( u ) L 2 N N + μ ( R N ) , by (2.3), we conclude that I μ * F ( u ) 2 N N + 2 s L N + 2 s N μ ( R N ) . By [24, Lemma 2.6], u n u 2 N ( 2 s + μ ) ( N 2 s ) ( N + 2 s ) 0 weakly in L ( N + 2 s ) ( 2 s + μ ) ( R N ) as n . This implies

lim n R N I μ * F ( u ) 2 N N + 2 s u n u 2 N ( 2 s + μ ) ( N 2 s ) ( N + 2 s ) d x = 0 ,

which yields

(2.10) R N [ I μ * F ( u ) ] u n u 4 s + μ N N 2 s ( u n u ) v d x R N I μ * F ( u ) 2 N N + 2 s u n u 2 N ( 2 s + μ ) ( N 2 s ) ( N + 2 s ) d x N + 2 s 2 N R N v 2 N N 2 s d x N 2 s 2 N = o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) .

Since μ ( ( N 4 s ) + , N ) , for γ ( 1 , 2 N N + μ ) ( 1 , N μ ) , according to Rellich’s theorem, up to a subsequence, for any bounded domain Ω R N , F ( u n u ) 0 strongly in L γ ( Ω ) as n . It follows from Lemma 2.6 that up to a subsequence, I μ * F ( u n u ) 0 a.e. in R N as n . In view of (2.3), we obtain

sup n I μ * F ( u n u ) 2 N N + 2 s L N + 2 s N μ ( R N ) C sup n F ( u n u ) L 2 N N + μ ( R N ) < ,

which together with [24, Lemma 2.6] implies I μ * F ( u n u ) 2 N N + 2 s 0 weakly in L N + 2 s N μ ( R N ) as n . Using u μ + 2 s N 2 s 2 N N + 2 s L N + 2 s 2 s + μ ( R N ) ,

lim n + R N I μ * F ( u n u ) 2 N N + 2 s u 2 s + μ N 2 s 2 N N + 2 s d x = 0 ,

and Hölder’s inequality, we infer that

(2.11) R N [ I μ * F ( u n u ) ] u 4 s + μ N N 2 s u v d x R N I μ * F ( u n u ) 2 N N + 2 s u 2 N ( 2 s + μ ) ( N 2 s ) ( N + 2 s ) d x N + 2 s 2 N R N v 2 N N 2 s d x N 2 s 2 N = o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) . Therefore, together with (2.9)–(2.11), we deduce that (2.7) holds true.

Step 2. We claim

R N [ I μ * F ( u n ) ] f ˜ ( u n ) v d x = R N [ I μ * F ( u n u ) ] f ˜ ( u n u ) v d x + R N [ I μ * F ( u ) ] f ˜ ( u ) v d x + o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) . We can find that

R N [ I μ * ( F ( u n ) F ( u n u ) F ( u ) ) ] f ˜ ( u n ) v d x = o n ( 1 ) v H s ( R N ) , R N [ I μ * ( F ( u n ) F ( u n u ) F ( u ) ) ] f ˜ ( u n u ) v d x = o n ( 1 ) v H s ( R N ) , R N [ I μ * ( F ( u n ) F ( u n u ) F ( u ) ) ] f ˜ ( u ) v d x = o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) . Indeed, there exists η ( 0 , 1 ) and C ( η ) > 0 such that f ˜ ( t ) t μ N for t η and f ˜ ( t ) C ( η ) t 2 s + μ N 2 s for t η . Therefore, for any v C 0 ( R N ) , there exists C > 0 such that

R N f ˜ ( u n ) v 2 N N + μ d x = { x R N : u n ( x ) η } f ˜ ( u n ) v 2 N N + μ d x + { x R N : u n ( x ) η } f ˜ ( u n ) v 2 N N + μ d x { x R N : u n ( x ) η } u n 2 μ N + μ v 2 N N + μ d x + [ C ( η ) ] 2 N N + μ { x R N : u n ( x ) η } u n 2 N ( 2 s + μ ) ( N 2 s ) ( N + μ ) v 2 N N + μ d x R N u n 2 d x μ N + μ R N v 2 d x N N + μ + [ C ( η ) ] 2 N N + μ R N u n 2 N N 2 s d x 2 s + μ N + μ R N v 2 N N 2 s d x N 2 s N + μ C v H s ( R N ) 2 N N + μ ,

which implies

R N f ˜ ( u n ) v 2 N N + μ d x N + μ 2 N C v H s ( R N ) uniformly for all v C 0 ( R N ) , n = 1 , 2 , .

Together with Lemma 2.1 and (2.8), we deduce that

R N [ I μ * ( F ( u n ) F ( u n u ) F ( u ) ) ] f ˜ ( u n ) v d x C R N F ( u n ) F ( u n u ) F ( u ) 2 N N + μ d x N + μ 2 N R N f ˜ ( u n ) v 2 N N + μ d x N + μ 2 N = o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) .

By a similar measure, one obtains

R N ( I μ * F ( u n ) ) [ f ˜ ( u n ) f ˜ ( u n u ) f ˜ ( u ) ] v d x = o n ( 1 ) v H s ( R N ) , R N ( I μ * F ( u n u ) ) [ f ˜ ( u n ) f ˜ ( u n u ) f ˜ ( u ) ] v d x = o n ( 1 ) v H s ( R N ) , R N ( I μ * F ( u ) ) [ f ˜ ( u n ) f ˜ ( u n u ) f ˜ ( u ) ] v d x = o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) . It follows from Lemmas 2.1 and 2.7 (ii)(2) that there exists C > 0 such that

R N ( I μ * F ( u n ) ) [ f ˜ ( u n ) f ˜ ( u n u ) f ˜ ( u ) ] v d x C R N f ˜ ( u n ) f ˜ ( u n u ) f ˜ ( u ) 2 N N + μ v 2 N N + μ d x N + μ 2 N = o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) .

Next, we show that

(2.12) R N [ I μ * F ( u n u ) ] f ˜ ( u ) v d x = o n ( 1 ) v H s ( R N )

and

(2.13) R N [ I μ * F ( u ) ] f ˜ ( u n u ) v d x = o n ( 1 ) v H s ( R N ) ,

for any v C 0 ( R N ) . It is easy to see that for any ε ( 0 , 1 ) , there exists ϱ ( 0 , 1 ) and C ε > 0 such that f ˜ ( t ) ε t μ N for t ϱ and f ˜ ( t ) C ε t μ + 2 s N 2 s for t ϱ . Therefore, for any v C 0 ( R N ) , using Lemma 2.1 and Hölder’s inequality, we obtain

R N [ I μ * F ( u n u ) ] f ˜ ( u ) v d x ε { u ( x ) ϱ } I μ * F ( u n u ) u μ N v d x + C ε { u ( x ) ϱ } I μ * F ( u n u ) u 2 s + μ N 2 s v d x ε F ( u n u ) L 2 N N + μ ( R N ) { u ( x ) ϱ } u 2 μ N + μ v 2 N N + μ d x N + μ 2 N + C ε R N I μ * F ( u n u ) 2 N N + 2 s u 2 s + μ N 2 s 2 N N + 2 s d x N + 2 s 2 N R N v 2 N N 2 s d x N 2 s 2 N .

Together with (2.11), we infer that (2.12) holds. By a similar argument as (2.12), we can prove (2.13). This completes the proof.□

3 Proof of Theorem 1.1

In this section, we consider the limit equation associated with equation (KC):

a + b R N ( Δ ) s 2 u 2 d x ( Δ ) s u + V u = ( I μ * F ( u ) ) f ( u ) , x R N , u H s ( R N ) ( K C )

and denote the energy functional by

Φ ( u ) = a 2 u D s , 2 ( R N ) 2 + V 2 u L 2 ( R N ) 2 + b 4 u D s , 2 ( R N ) 4 1 2 R N ( I μ * F ( u ) ) F ( u ) d x .

We define

= { u H s ( R N ) \ { 0 } : P ( u ) = 0 } ,

where

(3.1) P ( u ) = a ( N 2 s ) 2 u D s , 2 ( R N ) 2 + N V 2 u L 2 ( R N ) 2 + b ( N 2 s ) 2 u D s , 2 ( R N ) 4 N + μ 2 R N ( I μ * F ( u ) ) F ( u ) d x .

Lemma 3.1

Φ is not bounded from below.

Proof

Set u θ = u ( θ 1 x ) . Then, from θ > 1 , we see that

γ ( θ ) = Φ ( u θ ) = a θ N 2 s 2 u D s , 2 ( R N ) 2 + V θ N 2 u L 2 ( R N ) 2 + b θ 2 N 4 s 4 u D s , 2 ( R N ) 4 θ N + μ 2 R N ( I μ * F ( u ) ) F ( u ) d x .

Since N > 2 s , μ > ( N 4 s ) + , we conclude that Φ ( u θ ) as θ + .□

Remark 3.1

By direct calculations, we see that P ( u ) = d Φ ( u θ ) d θ θ = 1 .

Lemma 3.2

For any u H s ( R N ) \ { 0 } , there exists a unique θ 0 > 0 such that u θ 0 . Moreover, Φ ( u θ 0 ) = max θ > 0 Φ ( u θ ) .

Proof

Let u H s ( R N ) \ { 0 } be fixed and we observe that

γ ( θ ) = 0 θ γ ( θ ) = 0 u θ , for θ > 0 .

By V R + , Lemma 2.3(ii), and μ > ( N 4 s ) + , we obtain lim θ 0 + γ ( θ ) = 0 , γ ( θ ) > 0 for θ > 0 small, and γ ( θ ) < 0 for θ large. Hence, max θ > 0 γ ( θ ) is achieved at θ = θ 0 ( u ) > 0 such that γ ( θ 0 ) = 0 and u θ 0 .

Next, we show θ 0 is unique for any u H s ( R N ) \ { 0 } . In fact, assume by contradiction that there exist 0 < θ 1 < θ 2 such that γ ( θ 1 ) = γ ( θ 2 ) = 0 , i.e.

a ( N 2 s ) 2 θ 1 μ + 2 s u D s , 2 ( R N ) 2 + V N 2 θ 1 μ u L 2 ( R N ) 2 + b ( N 2 s ) 2 θ 1 μ + 4 s N u D s , 2 ( R N ) 4 = N + μ 2 R N ( I μ * F ( u ) ) F ( u ) d x

and

a ( N 2 s ) 2 θ 2 μ + 2 s u D s , 2 ( R N ) 2 + V N 2 θ 2 μ u L 2 ( R N ) 2 + b ( N 2 s ) 2 θ 2 μ + 4 s N u D s , 2 ( R N ) 4 = N + μ 2 R N ( I μ * F ( u ) ) F ( u ) d x ,

which is a contradiction and the proof is complete.□

In light of Lemmas 2.3 and 3.1, we obtain that Φ possesses the mountain-pass geometry.

Lemma 3.3

The functional Φ satisfies the following properties:

  1. there exist ρ , α > 0 such that Φ ( u ) α for u H s ( R N ) = ρ ;

  2. there exists v 0 H s ( R N ) \ B ρ such that Φ ( v 0 ) 0 .

Proof

(i) From Lemma 2.3(ii) and Sobolev’s embedding theorem, we obtain

Φ ( u ) 1 2 u H s ( R N ) 2 C ( u H s ( R N ) 2 + 2 μ N + u H s ( R N ) 2 2 μ , s * ) ,

which implies that there exist α > 0 and ρ > 0 such that

Φ ( u ) α > 0 , u H s ( R N ) = ρ .

(ii) It follows immediately from Lemma 3.1.□

We now define

c = inf γ Γ max t [ 0 , 1 ] Φ ( γ ( t ) ) > 0 , and m = inf u Φ ( u ) ,

where

Γ = { γ C ( [ 0 , 1 ] , H s ( R N ) ) : γ ( 0 ) = 0 , Φ ( γ ( 1 ) ) < 0 } .

Similar to the arguments of [31, Chapter 4], we obtain c = m without proof.

Lemma 3.4

There exists ϱ > 0 such that

u H s ( R N ) ϱ , u .

Proof

From (3.1), Lemma 2.3 (ii) and (iii), and the Sobolev embedding theorem, we deduce

N 2 s 2 u H s ( R N ) 2 a ( N 2 s ) 2 u D s , 2 ( R N ) 2 + N 2 R N V u 2 d x + b ( N 2 s ) 2 u D s , 2 ( R N ) 4 = N + μ 2 R N ( I μ * F ( u ) ) F ( u ) d x C ( u H s ( R N ) 2 + 2 μ N + u H s ( R N ) 2 2 μ , s * ) ,

which implies

u H s ( R N ) ϱ ,

for any u .□

Lemma 3.5

Let μ ( ( N 4 s ) + , N ) and assume that r > max μ + N 4 s N 2 s , N + μ 2 ( N 2 s ) . Then, there exists b 0 > 0 such that for b ( 0 , b 0 ) ,

c < 2 s + μ 2 ( N + μ ) N + μ N 2 s N 2 s μ + 2 s ( a S H , L ) N + μ μ + 2 s .

Proof

Let η ( x ) C 0 ( R N ) be a cut-off function such that 0 η ( x ) 1 in R N , η 1 in B 1 ( 0 ) , and η 0 in R N \ B 2 ( 0 ) . For ε > 0 , let ϕ ε ( x ) = η ( x ) U ε ( x ) , where

U ε ( x ) = β μ , s ε N 2 s 2 ( ε 2 + x 2 ) N 2 s 2 ,

where β μ , s = S μ ( 2 s N ) 4 s ( μ + 2 s ) ( C ( N , μ ) ) 2 s N 2 ( 2 s + μ ) S N ( 2 s ) Γ ( N ) π N 2 Γ ( N 2 ) N 2 s 2 N . As in [12], we obtain

(3.2) R N ( Δ ) s 2 ϕ ε 2 d x = S N 2 s + O ( ε N 2 s ) ,

R N ϕ ε 2 N N 2 s = S N 2 s + O ( ε N ) , R N ϕ ε 2 d x = C ε 2 s + O ( ε N 2 s ) , N > 4 s , C ε 2 s ln ε + O ( ε 2 s ) , N = 4 s , C ε N 2 s + O ( ε 2 s ) , N < 4 s ,

where C > 0 . Thus, we obtain

(3.3) R N a ( Δ ) s 2 ϕ ε 2 + V ϕ ε 2 d x = a S N 2 s + C ε 2 s + O ( ε N 2 s ) , N > 4 s , C ε 2 s ln ε + O ( ε 2 s ) , N = 4 s , C ε N 2 s + O ( ε 2 s ) , N < 4 s .

By direct calculation, we conclude that

R N ϕ ε 2 N r N + μ d x N + μ N = C ε N + μ ( N 2 s ) r + o ( ε N + μ ( N 2 s ) r ) .

Together with Lemma 2.1, we see that

(3.4) R N ( I μ * ϕ ε N + μ N 2 s ) ϕ ε r d x C ( N , μ ) R N ϕ ε 2 N N 2 s N + μ 2 N R N ϕ ε 2 N r N + μ N + μ 2 N C ε N + μ ( N 2 s ) r 2 + o ( ε N + μ ( N 2 s ) r 2 )

and

(3.5) R 2 N ϕ ε ( x ) 2 μ , s * ϕ ε ( y ) 2 μ , s * x y N μ d x d y N 2 s N + μ C ( N , μ ) N 2 s N + μ ϕ ε L 2 s * ( R N ) 2 = C ( N , μ ) N 2 s N + μ S N 2 s + O ( ε N ) N 2 s N = C ( N , μ ) N 2 s N + μ ( C ( N , μ ) N 2 s N + μ S H , L ) N 2 s + O ( ε N ) N 2 s N = C ( N , μ ) N 2 s N + μ N 2 s S H , L N 2 s 2 s + O ( ε N 2 s ) .

Meanwhile,

(3.6) R 2 N ϕ ε ( x ) 2 μ , s * ϕ ε ( y ) 2 μ , s * x y N μ d x d y B 1 × B 1 ϕ ε ( x ) 2 μ , s * ϕ ε ( y ) 2 μ , s * x y N μ d x d y = B 1 × B 1 U ε ( x ) 2 μ , s * U ε ( y ) 2 μ , s * x y N μ d x d y = R 2 N U ε ( x ) 2 μ , s * U ε ( y ) 2 μ , s * x y N μ d x d y 2 ( R N \ B 1 ) × B 1 U ε ( x ) 2 μ , s * U ε ( y ) 2 μ , s * x y N μ d x d y ( R N \ B 1 ) × ( R N \ B 1 ) U ε ( x ) 2 μ , s * U ε ( y ) 2 μ , s * x y N μ d x d y [ C ( N , μ ) ] N 2 s S H , L N + μ 2 s 2 A ,

where

A = ( R N \ B 1 ) × B 1 U ε ( x ) 2 μ , s * U ε ( y ) 2 μ , s * x y N μ d x d y

and

= ( R N \ B 1 ) × ( R N \ B 1 ) U ε ( x ) 2 μ , s * U ε ( y ) 2 μ , s * x y N μ d x d y .

By computing directly, we obtain

(3.7) A = ( R N \ B 1 ) × B 1 β μ , s N + μ N 2 s ε N + μ ( ε 2 + x 2 ) N + μ 2 ( ε 2 + y 2 ) N + μ 2 x y N μ d x d y O ( ε N + μ ) R N \ B 1 1 ( ε 2 + x 2 ) N d x N + μ 2 N B 1 1 ( ε 2 + y 2 ) N d y N + μ 2 N O ( ε N + μ ) R N \ B 1 1 x 2 N d x N + μ 2 N 0 1 r N 1 ( ε 2 + r 2 ) N d r N + μ 2 N = O ( ε N + μ ) 0 1 ε z N 1 ( 1 + z 2 ) N d z N + μ 2 N = O ( ε N + μ 2 )

and

(3.8) = ( R N \ B 1 ) × ( R N \ B 1 ) β μ , s N + μ N 2 s ε N + μ ( ε 2 + x 2 ) N + μ 2 ( ε 2 + y 2 ) N + μ 2 x y N μ d x d y ( R N \ B 1 ) × ( R N \ B 1 ) β N + μ N 2 s ε N + μ x N + μ y N + μ x y N μ d x d y = O ( ε N + μ ) .

By (3.6)–(3.8), we deduce that

(3.9) R 2 N ϕ ε ( x ) 2 μ , s * ϕ ε ( y ) 2 μ , s * x y N μ d x d y [ C ( N , μ ) ] N 2 s S H , L N + μ 2 s O ( ε N + μ 2 ) .

Similarly, we note that

R N ( I μ * ϕ ε N + μ N 2 s ) ϕ ε r d x R 2 N U ε ( x ) N + μ N 2 s U ε ( y ) r x y N μ d x d y ( R N \ B 1 ) × B 1 U ε ( x ) N + μ N 2 s U ε ( y ) r x y N μ d x d y B 1 × ( R N \ B 1 ) U ε ( x ) N + μ N 2 s U ε ( y ) r x y N μ d x d y ( R N \ B 1 ) × ( R N \ B 1 ) U ε ( x ) N + μ N 2 s U ε ( y ) r x y N μ d x d y ,

where

R 2 N U ε ( x ) N + μ N 2 s U ε ( y ) r x y N μ d x d y = C ε N + μ ( N 2 s ) r 2 , ( R N \ B 1 ) × B 1 U ε ( x ) N + μ N 2 s U ε ( y ) r x y N μ d x d y C ε N + μ ( N 2 s ) r 2 + o ( ε N + μ ( N 2 s ) r 2 ) , B 1 × ( R N \ B 1 ) U ε ( x ) N + μ N 2 s U ε ( y ) r x y N μ d x d y C ε ( N 2 s ) r 2 + o ( ε ( N 2 s ) r 2 ) , ( R N \ B 1 ) × ( R N \ B 1 ) U ε ( x ) N + μ N 2 s U ε ( y ) r x y N μ d x d y C ε N + μ + ( N 2 s ) r 2 + o ( ε N + μ + ( N 2 s ) r 2 ) .

Therefore, we have

(3.10) R N ( I μ * ϕ ε N + μ N 2 s ) ϕ ε r d x C ε N + μ ( N 2 s ) r 2 + o ε N + μ ( N 2 s ) r 2 ,

recalling that r > N + μ 2 ( N 2 s ) . It follows from ( H 8 ) that for any t > 0 ,

Φ ( t ϕ ε ) t 2 2 R N ( a ( Δ ) s 2 ϕ ε 2 + V ϕ ε 2 ) d x + b t 4 4 R N ( Δ ) s 2 ϕ ε 2 d x 2 ν ( N 2 s ) t r + N + μ N 2 s r ( N + μ ) R N I μ * ϕ ε N + μ N 2 s ϕ ε r d x t 2 ( N + μ ) N 2 s 2 N 2 s N + μ 2 R N I μ * ϕ ε N + μ N 2 s ϕ ε N + μ N 2 s d x h ε ( t ) .

It is clear that h ε ( t ) as t + and h ε ( t ) > 0 for t > 0 small, and there exists a unique critical point t ε in ( 0 , + ) , which is the maximum point of h ε . In view of h ( t ε ) = 0 , we derive

(3.11) t ε R N a ( Δ ) s 2 ϕ ε 2 + V ϕ ε 2 d x + b t ε 3 R N ( Δ ) s 2 ϕ ε 2 d x 2 = r + N + μ N 2 s ν ( N 2 s ) t ε r + N + μ N 2 s 1 r ( N + μ ) R N I μ * ϕ ε N + μ N 2 s ϕ ε r d x + t ε 2 ( N + μ ) N 2 s 1 N 2 s N + μ R N I μ * ϕ ε N + μ N 2 s ϕ ε N + μ N 2 s d x .

Claim. There exist t 0 and t 1 (both independent of ε ) such that t 0 t ε t 1 for ε > 0 small.

By (3.3)–(3.5), (3.11), and r < N + μ N 2 s , there exist c 1 , c 2 > 0 (independent of ε ) such that for ε small,

c 1 t ε c 2 ε N + μ ( N 2 s ) r 2 t ε r + N + μ N 2 s 1 + t ε 2 ( N + μ ) N 2 s 1 2 t ε r + N + μ N 2 s 1 ,

which implies that t ε t 0 , noting that μ > ( N 4 s ) + . By (3.11), we can see that

R N a ( Δ ) s 2 ϕ ε 2 + V ϕ ε 2 d x + b t ε 2 R N ( Δ ) s 2 ϕ ε 2 2 t ε 2 ( N + μ ) N 2 s 2 N 2 s N + μ R N I μ * ϕ ε N + μ N 2 s ϕ ε N + μ N 2 s d x .

Combining (3.2), (3.3), and (3.9), the above inequality yields t ε t 1 for some t 1 > 0 and ε small.

To find the estimate of max t > 0 Φ ( t ϕ ε ) , we define

g ( t ) = t 2 2 R N a ( Δ ) s 2 ϕ ε 2 + V ϕ ε 2 d x t 2 ( N + μ ) N 2 s 2 N 2 s N + μ 2 R N I μ * ϕ ε N + μ N 2 s ϕ ε N + μ N 2 s d x .

By (2.5), (3.3), and (3.9), we obtain

(3.12) max t > 0 g ( t ) μ + 2 s 2 ( N + μ ) N + μ N 2 s N 2 s μ + 2 s ( a S H , L ) N + μ μ + 2 s + C ε 2 s + O ( ε N 2 s ) , N > 4 s , C ε 2 s ln ε + O ( ε 2 s ) , N = 4 s , C ε N 2 s + O ( ε min { 2 s , N + μ 2 } ) , N < 4 s .

Using (3.2), (3.10), and (3.12), we conclude that

max t > 0 Φ ( t ϕ ε ) = h ε ( t ε ) max t > 0 g ( t ) + C b + O ( ε N 2 s ) C ε N + μ ( N 2 s ) r 2 + o ε N + μ ( N 2 s ) r 2 μ + 2 s 2 ( N + μ ) N + μ N 2 s N 2 s μ + 2 s ( a S H , L ) N + μ μ + 2 s + C b C ε N + μ ( N 2 s ) r 2 + C ε 2 s + O ( ε N 2 s ) , N > 4 s , C b C ε N + μ ( N 2 s ) r 2 + C ε 2 s ln ε + O ( ε 2 s ) , N = 4 s , C b C ε N + μ ( N 2 s ) r 2 + C ε N 2 s + O ( ε min { 2 s , N + μ 2 } ) , N < 4 s μ + 2 s 2 ( N + μ ) N + μ N 2 s N 2 s μ + 2 s ( a S H , L ) N + μ μ + 2 s + C b C ε 0 N + μ ( N 2 s ) r 2 ,

noting the fact N + μ ( N 2 s ) r 2 < min { 2 s , N 2 s } . Thus, there exists b 0 > 0 such that for b ( 0 , b 0 ) ,

C b < C ε 0 N + μ ( N 2 s ) r 2 ,

which implies that the desired result holds.□

Lemma 3.6

Let μ ( ( N 4 s ) + , N ) and assume that r > max μ + N 4 s N 2 s , N + μ 2 ( N 2 s ) . Then, for b ( 0 , b 0 ) , m is achieved.

Proof

Let { u n } be such that Φ ( u n ) m . It follows from μ > ( N 4 s ) + and Lemma 2.3(ii) that, for n large enough,

(3.13) 1 + m > Φ ( u n ) = Φ ( u n ) 1 N + μ P ( u n ) = a ( μ + 2 s ) 2 ( N + μ ) u n D s , 2 ( R N ) 2 + μ V 2 ( N + μ ) u n L 2 ( R N ) 2 + b ( μ + 4 s N ) 2 ( N + μ ) u n D s , 2 ( R N ) 4 a ( μ + 2 s ) 2 ( N + μ ) u n D s , 2 ( R N ) 2 + μ V 2 ( N + μ ) u n L 2 ( R N ) 2 .

Hence, { u n D s , 2 ( R N ) 2 } is bounded, and there exists M 0 such that

u n D s , 2 ( R N ) 2 M 2 .

Therefore, { u n } is bounded in H s ( R N ) . Passing to a subsequence, we obtain

u n u , in H s ( R N ) , u n u , in L loc q ( R N ) , for q [ 1 , 2 s * ) , u n u , a.e. in R N .

We claim that there exists δ > 0 and y n R N such that

B 1 ( y n ) u n 2 d x > δ .

If not, by Lemma 2.4, we derive u n 0 strongly in L t ( R N ) for t ( 2 , 2 s * ) as n + . By virture of P ( u n ) = 0 , we obtain

(3.14) a ( N 2 s ) 2 u n D s , 2 ( R N ) 2 + N V 2 u n L 2 ( R N ) 2 + b ( N 2 s ) 2 u n D s , 2 ( R N ) 4 = N + μ 2 R N ( I μ * F ( u n ) ) F ( u n ) d x .

Let

f ˜ ( t ) = f ( t ) t 4 s + μ N N 2 s t , F ˜ = 0 t f ˜ ( s ) d s , t R .

Noting that

lim t 0 f ˜ ( t ) t μ N = lim t + f ˜ ( t ) t μ + 2 s N 2 s = 0 ,

we can infer that

lim n + R N F ˜ ( u n ) 2 N N + μ = 0 ,

which implies that

R N [ I μ * F ˜ ( u n ) ] F ˜ ( u n ) d x C R N F ˜ ( u n ) 2 N N + μ d x N + μ 2 N 0 , as n .

Similarly,

lim n + R N [ I μ * F ˜ ( u n ) ] u n N + μ N 2 s d x = 0 .

It follows from (3.14) that

(3.15) a ( N 2 s ) 2 u n D s , 2 ( R N ) 2 + N V 2 u n L 2 ( R N ) 2 + b ( N 2 s ) 2 u n D s , 2 ( R N ) 4 = N + μ 2 N 2 s N + μ 2 R N ( I μ * u n N + μ N 2 s ) u n N + μ N 2 s d x .

Set

lim n + u n D s , 2 ( R N ) 2 = l > 0 .

From (3.15) and (2.4), we obtain

a ( N 2 s ) 2 l N + μ 2 N 2 s N + μ 2 R N ( I μ * u n N + μ N 2 s ) u n N + μ N 2 s d x N + μ 2 N 2 s N + μ 2 S H , L 1 R N ( Δ ) s 2 u n 2 d x N + μ N 2 s = N + μ 2 N 2 s N + μ 2 ( S H , L ) N + μ N 2 s l N + μ N 2 s ,

which yields

l a ( N + μ ) N 2 s N 2 s μ + 2 s S H , L N + μ 2 s + μ .

According to μ > ( N 4 s ) + , (3.13) implies that

m = lim n + Φ ( u n ) a 2 μ + 2 s N + μ l μ + 2 s 2 ( N + μ ) N + μ N 2 s N 2 s μ + 2 s ( a S H , L ) N + μ 2 s + μ ,

which contradicts with Lemma 3.5. Thus, the Claim holds true.

Let v n ( x ) = u n ( x + y n ) . Then,

u n H s ( R N ) = v n H s ( R N ) , Φ ( v n ) m , P ( v n ) = 0

and

B 1 ( 0 ) v n 2 d x > δ .

Passing to a subsequence, there exists v H s ( R N ) such that

v n v in H s ( R N ) , v n v in L loc q ( R N ) , for q [ 1 , 2 s * ) , v n v a.e. in R N

and

v n D s , 2 ( R N ) 2 M 2 .

Define

G ( u ) = a + b M 2 2 u D s , 2 ( R N ) 2 + V 2 u L 2 ( R N ) 2 1 2 R N ( I μ * F ( u ) ) F ( u ) d x .

Note that P ( v n ) = 0 . From Remark 3.1, we deduce Φ ( v n ) 0 . Together with v n v in H s ( R N ) , we can see G ( v ) = 0 . Therefore, similar to (3.1), we obtain

P ˜ ( v ) = ( a + b M 2 ) ( N 2 s ) 2 v D s , 2 ( R N ) 2 + N 2 R N V v 2 d x N + μ 2 R N ( I μ * F ( v ) ) F ( v ) d x = 0 .

From the weak lower-semi-continuity of the norm, we have

v D s , 2 ( R N ) 2 M 2 .

We now claim v D s , 2 ( R N ) 2 = M 2 . Suppose by contradiction that v D s , 2 ( R N ) 2 < M 2 , then

P ( v ) < P ˜ ( v ) = 0 .

By Lemma 3.2, there exists θ 0 > 0 such that v θ 0 . We need to claim θ 0 < 1 . Assume by contradiction that θ 0 1 . It follows from P ( v ) < 0 and P ( v θ 0 ) = 0 that

a ( N 2 s ) 2 v D s , 2 ( R N ) 2 + N V 2 v L 2 ( R N ) 2 + b ( N 2 s ) 2 v D s , 2 ( R N ) 4 < N + μ 2 R N ( I μ * F ( v ) ) F ( v ) d x

and

a ( N 2 s ) 2 θ 0 μ + 2 s v D s , 2 ( R N ) 2 + ( N ) V 2 θ 0 μ v L 2 ( R N ) 2 + b ( N 2 s ) 2 θ 0 μ + 4 s N v D s , 2 ( R N ) 4 = N + μ 2 R N ( I μ * F ( v ) ) F ( v ) d x ,

which yields a contradiction. Then, θ 0 < 1 . Therefore, by (3.13), we obtain

c Φ ( v θ 0 ) = Φ ( v θ 0 ) 1 N + μ P ( v θ 0 ) = a ( μ + 2 s ) θ 0 N 2 s 2 ( N + μ ) v D s , 2 ( R N ) 2 + μ V θ 0 N 2 ( N + μ ) v L 2 ( R N ) 2 + b ( μ + 4 s N ) θ 0 2 N 4 s 2 ( N + μ ) v D s , 2 ( R N ) 4 < a ( μ + 2 s ) 2 ( N + μ ) v D s , 2 ( R N ) 2 + μ V 2 ( N + μ ) v L 2 ( R N ) 2 + b ( μ + 4 s N ) 2 ( N + μ ) v D s , 2 ( R N ) 4 lim n + Φ ( v n ) 1 N + μ P ( v n ) = m ,

which is a contradiction. Thus, v D s , 2 ( R N ) 2 = M 2 . Hence, we can see that P ( v ) = 0 , Φ ( v ) = m .□

In the same way as [19, Lemma 2.6], we can deduce the following lemma.

Lemma 3.7

If u ˆ and Φ ( u ˆ ) = m , then u ˆ is a critical point of Φ .

Proof of Theorem 1.1

Combining Lemmas 3.3, 3.6, and 3.7, we assume there exists a Pohozaev type ground-state solution u ˆ to ( K C ) .□

4 Proof of Theorem 1.2

Since V ( x ) is not a constant, it is difficult to obtain the boundedness of (PS) sequences, we introduce the following result developed by Jeanjean [13].

Theorem 4.1

[13] Let X be a Banach space and ϒ R + an interval. Consider a family of C 1 functional φ λ on X with the form

φ λ ( u ) = A ( u ) λ B ( u ) , λ ϒ ,

where B ( u ) 0 , u X , and such that either A ( u ) + or B ( u ) + as u X . If there exists v 1 , v 2 X such that

c λ = inf γ Γ max t [ 0 , 1 ] φ λ ( γ ( t ) ) > max { φ λ ( v 1 ) , φ λ ( v 2 ) } , λ ϒ ,

where Γ = { γ C ( [ 0 , 1 ] , X ) : γ ( 0 ) = v 1 , γ ( 1 ) = v 2 } .

Then, for almost every λ ϒ , there exists a sequence { v n } X such that

  1. { v n } is bounded;

  2. φ λ ( v n ) c λ ;

  3. φ λ ( v n ) 0 in the dual X of X .

Moreover, λ c λ is left continuous.

In order to explore Theorem 4.1 for λ ϒ = [ δ 0 , 1 ] , where δ 0 ( 0 , 1 ) is a constant. We study a family of functionals Φ λ : H s ( R N ) R given by

Φ λ ( u ) = 1 2 u H s ( R N ) 2 + b 4 u D s , 2 ( R N ) 4 λ 2 R N ( I μ * F ( u ) ) F ( u ) d x .

Then, set

Φ λ ( u ) = A ( u ) λ B ( u ) ,

where

A ( u ) = 1 2 u H s ( R N ) 2 + b 4 u D s , 2 ( R N ) 4 + ,

as u H s ( R N ) + and

B ( u ) = λ 2 R N ( I μ * F ( u ) ) F ( u ) d x 0 .

We can easily verify all the conditions of Theorem 4.1.

Lemma 4.1

The functional Φ λ satisfies the following properties:

  1. there exists e > 0 such that Φ λ ( e ) < 0 for all λ ϒ ;

  2. c λ = inf γ Γ max t [ 0 , 1 ] Φ λ ( γ ( t ) ) > max { Φ λ ( 0 ) , Φ λ ( e ) } for all λ ϒ , where

    Γ = { γ C ( [ 0 , 1 ] , H s ( R N ) ) : γ ( 0 ) = 0 , γ ( 1 ) = e } ;

  3. the map λ c λ is non-increasing and left continuous.

By Theorem 1.1, we deduce that for any λ ϒ , there exists b λ > 0 , for b ( 0 , b λ ) , the limit problem

a + b R N ( Δ ) s 2 u 2 d x ( Δ ) s u + V u = λ ( I μ * F ( u ) ) f ( u ) , in R N , u H s ( R N ) ,

admits a ground-state solution u λ H s ( R N ) , i.e., for any λ ϒ , there exists

u λ , λ = { u H s ( R N ) \ { 0 } : P , λ ( u ) = 0 }

such that

Φ , λ ( u λ ) = 0 , Φ , λ ( u λ ) = m , λ = inf u , λ Φ , λ ( u )

and

m , λ < 2 s + μ 2 ( N + μ ) N + μ λ ( N 2 s ) N 2 s μ + 2 s ( a S H , L ) N + μ μ + 2 s ,

where

Φ , λ ( u ) = a u D s , 2 ( R N ) 2 2 + V 2 u L 2 ( R N ) 2 + b u D s , 2 ( R N ) 4 4 λ 2 R N ( I μ * F ( u ) ) F ( u ) d x

and

P , λ ( u ) = a ( N 2 s ) 2 u D s , 2 ( R N ) 2 + N V 2 u L 2 ( R N ) 2 + b ( N 2 s ) 2 u D s , 2 ( R N ) 4 λ ( N + μ ) 2 R N ( I μ * F ( u ) ) F ( u ) d x .

Lemma 4.2

Let μ ( ( N 4 s ) + , N ) and assume that r > max μ + N 4 s N 2 s , N + μ 2 ( N 2 s ) . Then, for any λ ϒ , b ( 0 , b λ ) , c λ < m , λ .

Proof

Let u λ be the minimizer of m , λ . From Lemma 3.2, Lemma 4.1(i), and ( V 2 ) , we conclude that for any λ ϒ ,

c λ max t > 0 Φ λ ( u λ ( t 1 x ) ) < max t > 0 Φ , λ ( u λ ( t 1 x ) ) = Φ , λ ( u λ ) = m , λ .

The proof is completed.□

From Lemma 4.1, we obtain that for almost every λ ϒ , there exists a sequence { u n } H s ( R N ) such that

Φ λ ( u n ) c λ , Φ λ ( u n ) 0 .

Next, we aim to show the convergence of the above sequence { u n } in H s ( R N ) .

It follows from c λ > 0 that there exists M λ > 0 such that

u n D s , 2 ( R N ) 2 M λ 2 , as n .

For any u H s ( R N ) , define

(4.1) G λ ( u ) a + b M λ 2 2 u D s , 2 ( R N ) 2 + 1 2 R N V ( x ) u 2 d x λ 2 R N ( I μ * F ( u ) ) F ( u ) d x

and

(4.2) G , λ ( u ) a + b M λ 2 2 u D s , 2 ( R N ) 2 + 1 2 R N V u 2 d x λ 2 R N ( I μ * F ( u ) ) F ( u ) d x .

They are corresponding energy functionals of the following problems:

( a + b M λ 2 ) ( Δ ) s u + V ( x ) u = λ ( I μ * F ( u ) ) f ( u ) , ( a + b M λ 2 ) ( Δ ) s u + V u = λ ( I μ * F ( u ) ) f ( u ) .

Next, we give the global compactness lemma, which plays a crucial role in getting compactness.

Lemma 4.3

Let μ ( ( N 4 s ) + , N ) and assume that r > max μ + N 4 s N 2 s , N + μ 2 ( N 2 s ) . Then, for λ ϒ , b ( 0 , b λ ) , and c λ > 0 , let { u n } H s ( R N ) be a bounded ( P S ) c λ sequence for Φ λ . Then, there exists a u λ H s ( R N ) and M λ R such that G λ ( u λ ) = 0 . Moreover, there exist integer k N { 0 } , sequence { y n j } R N , and nontrivial positive critical points w j H s ( R N ) of G , λ for 1 j k such that

  1. y n j + and y n i y n j + for i j , as n ;

  2. c λ + b M λ 4 4 = G λ ( u λ ) + j = 1 k G , λ ( w j ) ;

  3. u n u λ j = 1 k w j ( y n j ) H s ( R N ) 0 ;

  4. M λ 2 = u λ D s , 2 ( R N ) 2 + j = 1 k w j D s , 2 ( R N ) 2 .

Proof

Since { u n } H s ( R N ) is a bounded sequence satisfying

Φ λ ( u n ) c λ > 0 , Φ λ ( u n ) 0 .

Then, we obtain

G λ ( u n ) = c λ + b M λ 4 4 + o n ( 1 ) , G λ ( u n ) 0 in H s ( R N ) .

Hence, there exists u λ H s ( R N ) \ { 0 } satisfying

u n u λ in H s ( R N ) , u n u λ in L loc q ( R N ) for q [ 1 , 2 s * ) , u n u λ a.e. in R N .

Using standard arguments, we can obtain that G λ ( u λ ) ϕ = 0 for all ϕ H s ( R N ) . Meanwhile, u λ satisfies the Pohozaev identity

P ˜ λ ( u λ ) = ( a + b M λ 2 ) ( N 2 s ) 2 u λ D s , 2 ( R N ) 2 + 1 2 R N [ N V ( x ) + ( V ( x ) , x ) ] u λ 2 d x ( N + μ ) λ 2 R N ( I μ * F ( u λ ) ) F ( u λ ) d x = 0 .

By ( V 1 ) and μ > N 4 s , we deduce

G λ ( u λ ) = G λ ( u λ ) 1 N + μ P ˜ λ ( u λ ) ( a + b M λ 2 ) ( μ + 2 s ) 2 ( N + μ ) u λ D s , 2 2 + 1 2 ( N + μ ) R N [ μ V ( x ) ( V ( x ) , x ) ] u λ 2 d x b M λ 2 4 u λ D s , 2 ( R N ) 2 .

Set v n 1 = u n u λ . Then, we have v n 1 0 . We will show that there exists R 0 > 0 , σ > 0 and a sequence { y n 1 } R N such that

(4.3) liminf n B R 0 ( y n 1 ) v n 1 2 d x σ > 0 .

Indeed, assume by contradiction, for any R > 0 ,

lim n sup y R N B R ( y n 1 ) v n 1 2 d x = 0 .

Then, Lemma 2.4 yields that v n 1 0 in L q ( R N ) , q ( 2 , 2 s * ) . It follows from (2.6) that

(4.4) lim n + R N ( I μ * F ˜ ( v n 1 ) ) F ˜ ( v n 1 ) d x = 0

and

lim n + R N ( I μ * F ˜ ( v n 1 ) ) v n 1 N + μ N 2 s d x = 0 , lim n + R N ( I μ * F ˜ ( v n 1 ) ) f ˜ ( v n 1 ) v n 1 d x = 0 .

By Lemmas 2.5 and 2.8, we obtain

(4.5) G λ ( u n ) = G λ ( u λ ) + a + b M λ 2 2 v n 1 D s , 2 ( R N ) + 1 2 R N V ( x ) v n 1 2 d x λ 2 N 2 s N + μ 2 R N [ I μ * v n 1 N + μ N 2 s ] v n 1 N + μ N 2 s d x + o n ( 1 )

and

(4.6) ( a + b M λ 2 ) v n 1 D s , 2 ( R N ) 2 + R N V ( x ) v n 1 2 d x = λ N 2 s N + μ R N ( I μ * v n 1 N + μ N 2 s ) v n 1 N + μ N 2 s d x + o n ( 1 ) .

Set

lim n + v n 1 D s , 2 ( R N ) 2 = l > 0 .

Then, by (4.6) and (2.4), we deduce that

a l λ N 2 s N + μ R N ( I μ * v n 1 N + μ N 2 s ) v n 1 N + μ N 2 s d x λ N 2 s N + μ S H , L 1 R N ( Δ ) s 2 v n 1 2 d x N + μ N 2 s λ N 2 s N + μ S H , L N + μ N 2 s l N + μ N 2 s ,

which yields

l a ( N + μ ) λ ( N 2 s ) N 2 s μ + 2 s S H , L N + μ μ + 2 s .

Together with (4.5) and (4.6), we derive that

c λ μ + 2 s 2 ( N + μ ) ( N + μ ) λ ( N 2 s ) N 2 s μ + 2 s ( a S H , L ) N + μ μ + 2 s .

This is a contradiction. Then, (4.3) holds.

Then, combining with the boundedness of { v n 1 } , we can see that

v n 1 ( + y n 1 ) w 1 0 .

It follows from v n 1 0 that { y n 1 } is unbounded. Up to a sub-sequence, we assume that y n 1 + . Now, we will prove that G , λ ( w 1 ) = 0 . It suffices to show that

G , λ ( v n 1 ( + y n 1 ) ) , φ 0 , φ C 0 ( R N ) .

From (2.1), we obtain

(4.7) G λ ( v n 1 ) , φ ( y n 1 ) = C ( N , s ) 2 ( a + b M λ 2 ) R 2 N ( v n 1 ( x ) v n 1 ( y ) ) ( φ ( x y n 1 ) φ ( y y n 1 ) ) x y N + 2 s d x d y + R N V ( x ) v n 1 ( x ) φ ( x y n 1 ) d x λ R N ( I μ * F ( v n 1 ( x ) ) f ( v n 1 ( x ) ) φ ( x y n 1 ) ) d x = C ( N , s ) 2 ( a + b M λ 2 ) R 2 N ( v n 1 ( x + y n 1 ) v n 1 ( y + y n 1 ) ) ( φ ( x ) φ ( y ) ) x y N + 2 s d x d y + R N V ( x + y n 1 ) v n 1 ( x + y n 1 ) φ ( x ) d x λ R N ( I μ * F ( v n 1 ( x + y n 1 ) ) f ( v n 1 ( x + y n 1 ) ) ) φ ( x ) d x .

Recalling v n 1 0 in H s ( R N ) , similar to the proof of (4.6), we deduce that

G λ ( v n 1 ) , φ ( y n 1 ) G λ ( 0 ) , φ ( y n 1 ) 0 ,

which yields

G λ ( v n 1 ) , φ ( y n 1 ) 0 ,

together with (4.7), we obtain

(4.8) C ( N , s ) 2 ( a + b M λ 2 ) R 2 N ( v n 1 ( x + y n 1 ) v n 1 ( y + y n 1 ) ) ( φ ( x ) φ ( y ) ) x y N + 2 s d x d y + R N V ( x + y n 1 ) v n 1 ( x + y n 1 ) φ ( x ) d x λ R N ( I μ * F ( v n 1 ( x + y n 1 ) ) f ( v n 1 ( x + y n 1 ) ) ) φ ( x ) d x 0 .

From ( V 2 ) , for any ε > 0 , there exists R > 0 such that

V ( x ) V < ε , x R .

We assume that supp φ B R 0 ( 0 ) with some R 0 > 0 . Hence, choosing n large enough such that x + y n 1 y n 1 x y n 1 R 0 R , which implies, for n large enough,

V ( x + y n 1 ) V < ε .

Thus, by Hölder’s inequality, we obtain

(4.9) R N V ( x + y n 1 ) v n 1 ( x + y n 1 ) φ ( x ) d x R N V v n 1 ( x + y n 1 ) φ ( x ) d x supp φ V ( x + y n 1 ) V v n 1 ( x + y n 1 ) φ ( x ) d x C ε .

Combining (4.9) and (4.8), we obtain

G , λ ( v n 1 ( + y n 1 ) ) , φ 0 , φ C 0 ( R N ) .

Next, we will prove

(4.10) G λ ( u n ) G λ ( u λ ) G , λ ( v n 1 ) 0 .

In view of ( V 2 ) , we conclude

R N [ V ( x ) V ] u n u λ 2 d x 0 .

Using Brezis-Lieb lemma [3] and Lemma 2.5, we obtain (4.10) holds.

Set v n 2 ( ) v n 1 ( ) w 1 ( y n 1 ) , then v n 2 0 in H s ( R N ) . By Brezis-Lieb lemma [3], Lemmas 2.5 and 2.8, we obtain

(4.11) v n 2 D s , 2 ( R N ) 2 = u n u λ w 1 ( y n 1 ) D s , 2 ( R N ) 2 = u n u λ D s , 2 ( R N ) 2 w 1 ( y n 1 ) D s , 2 ( R N ) 2 + o n ( 1 ) = u n D s , 2 ( R N ) 2 u λ D s , 2 ( R N ) 2 w 1 ( y n 1 ) D s , 2 ( R N ) 2 + o n ( 1 ) ,

and

(4.12) R N V ( x ) v n 2 2 d x = R N V ( x ) v n 1 2 d x R N V ( x ) w 1 ( x y n 1 ) 2 d x + o n ( 1 ) = R N V ( x ) u n 2 d x R N V ( x ) u λ 2 d x R N V ( x ) w 1 ( x y n 1 ) 2 d x + o n ( 1 )

and

(4.13) R N ( I μ * F ( v n 2 ) ) F ( v n 2 ) d x = R N ( I μ * F ( v n 1 ) ) F ( v n 1 ) d x R N ( I μ * F ( w 1 ( x y n 1 ) ) ) F ( w 1 ( x y n 1 ) ) d x + o n ( 1 ) = R N ( I μ * F ( u n ) ) F ( u n ) d x R N ( I μ * F ( u λ ) ) F ( u λ ) d x R N ( I μ * F ( w 1 ( x y n 1 ) ) ) F ( w 1 ( x y n 1 ) ) d x + o n ( 1 ) ,

and

(4.14) R N ( I μ * F ( v n 2 ) ) f ( v n 2 ) φ d x = R N ( I μ * F ( v n 1 ) ) f ( v n 1 ) φ d x R N ( I μ * F ( w 1 ( x y n 1 ) ) ) f ( w 1 ( x y n 1 ) ) φ d x + o n ( 1 ) = R N ( I μ * F ( u n ) ) f ( u n ) φ d x R N ( I μ * F ( u λ ) ) f ( u λ ) φ d x R N ( I μ * F ( w 1 ( x y n 1 ) ) ) f ( w 1 ( x y n 1 ) ) φ d x + o n ( 1 ) ,

for any φ H V s ( R N ) . From (4.11)–(4.14), we can similarly deduce that

(4.15) G λ ( v n 2 ) = G λ ( u n ) G λ ( u λ ) G , λ ( w 1 ) + o n ( 1 ) , G , λ ( v n 2 ) = G , λ ( v n 1 ) G , λ ( w 1 ) + o n ( 1 ) , G λ ( v n 2 ) , φ = G λ ( u n ) , φ G λ ( u λ ) , φ G , λ ( w 1 ) , φ + o n ( 1 ) = o n ( 1 ) .

Combining with (4.10), we obtain

G λ ( u n ) = G λ ( u λ ) + G , λ ( v n 1 ) + o n ( 1 ) = G λ ( u λ ) + G λ ( v n 2 ) + G , λ ( w 1 ) + o n ( 1 ) .

From the fact G , λ ( w 1 ) = 0 , we see that G , λ ( w 1 ) 0 , together with (4.10), it follows that

G λ ( v n 2 ) = G λ ( u n ) G λ ( u λ ) G , λ ( w 1 ) + o n ( 1 ) < c λ + b M λ 4 4 .

If v n 2 H s ( R N ) 0 , i.e.

u n u λ w 1 ( y n 1 ) H s ( R N ) 0 ,

and thus Lemma 4.3 holds with j = 1 . If v n 2 0 strongly in H s ( R N ) , then there exists a sequence y n 2 R N and w 2 H s ( R N ) such that v n 2 ( x + y n 2 ) w 2 in H s ( R N ) . From (4.15), we obtain that G , λ ( w 2 ) = 0 . Moreover, v n 2 0 in H s ( R N ) yields that y n 2 + and y n 1 y n 2 + . By iterating this procedure, we obtain sequences of points { y n j } R N such that y n j + and y n j y n i + if i j as n + and v n j = v n j 1 w j 1 ( x y n j 1 ) with j 2 such that v n j 0 in H s ( R N ) , G , λ ( w j ) = 0 . From the properties of the weak convergence, we obtain

(4.16) u n H s ( R N ) 2 u λ H s ( R N ) 2 j = 1 k w j ( x y n j ) H s ( R N ) 2 = o n ( 1 ) , G λ ( u n ) G λ ( u λ ) j = 1 k G , λ ( w j ( x y n j ) ) G , λ ( v n k + 1 ) = o n ( 1 ) .

Since { u n } is bounded in H s ( R N ) , it follows from (4.16) that the iteration stops at some finite index k , that is v n k + 1 0 in H s ( R N ) . By (4.16), it is easy to check that conclusions (ii)–(iv). The proof is complete.□

In the following lemma, we conclude the convergence of the bounded ( P S ) c λ sequence { u n } for Φ λ .

Lemma 4.4

Let μ ( ( N 4 s ) + , N ) and assume that r > max μ + N 4 s N 2 s , N + μ 2 ( N 2 s ) . Then, for fixed λ ϒ , b ( 0 , b λ ) , let { u n } H s ( R N ) be a bounded ( P S ) c λ sequence for Φ λ . Then, there exists a nontrivial u λ H s ( R N ) such that u n u λ .

Proof

It follows from Theorem 4.1 and Lemma 4.1, for almost all λ ϒ , there exists a bounded ( P S ) c λ sequence { u n } H s ( R N ) such that

Φ λ ( u n ) c λ , Φ λ ( u n ) 0 .

By Lemma 4.3, we can conclude that there exist u λ H s ( R N ) , k N { 0 } , a sequence { y n j } R N , w j H s ( R N ) for 1 j k and M λ > 0 such that

u n u λ in H s ( R N ) , u n D s , 2 ( R N ) 2 M λ 2 , G λ ( u λ ) = 0 , G , λ ( w j ) = 0

and

(4.17) u n u λ j = 1 k w j ( + y n j ) H s ( R N ) 0 , c λ + b M λ 4 4 = G λ ( u λ ) + j = 1 k G , λ ( w j )

and

M λ 2 = u λ D s , 2 ( R N ) 2 + j = 1 k w j D s , 2 ( R N ) 2 ,

where G λ ( ) and G , λ ( ) are given in (4.1) and (4.2). It is easy to check that

(4.18) G λ ( u λ ) b M λ 2 4 u λ D s , 2 ( R N ) 2 .

Note that G , λ ( w j ) = 0 . We obtain

P ˜ , λ ( w j ) = ( a + b M λ 2 ) ( N 2 s ) 2 w j D s , 2 ( R N ) 2 + N 2 R N V ( w j ) 2 d x ( N + μ ) λ 2 R N ( I μ * F ( w j ) ) F ( w j ) d x = 0 .

By (4.2), we have

0 = P ˜ , λ ( w j ) P , λ ( w j ) , 1 j k .

From Lemma 3.2, there exists t j ( 0 , 1 ] such that P , λ ( ( w j ) t j ) = 0 . By a direct calculation, we deduce that

G , λ ( w j ) = G , λ ( w j ) 1 2 ( N 2 s ) P ˜ , λ ( w j ) = ( a + b M λ 2 ) 4 w j D s , 2 ( R N ) 2 + ( N 4 s ) V 2 ( N 2 s ) w j L 2 ( R N ) 2 + μ + 4 s N 2 ( N 2 s ) R N ( I μ * F ( w j ) ) F ( w j ) d x a t j N 2 s 4 w j D s , 2 ( R N ) 2 + ( N 4 s ) V t j N 2 ( N 2 s ) w j L 2 ( R N ) 2 + ( μ + 4 s N ) t j N + μ 2 ( N 2 s ) R N ( I μ * F ( w j ) ) F ( w j ) d x + b M λ 2 4 w j D s , 2 ( R N ) 2 = Φ , λ ( ( w j ) t j ) 1 2 ( N 2 s ) P , λ ( ( w j ) t j ) + b M λ 2 4 w j D s , 2 ( R N ) 2 m , λ + b M λ 2 4 w j D s , 2 ( R N ) 2 , 1 j k .

From (4.17) and (4.18), if k 0 , we can deduce that

c λ + b M λ 4 4 = G λ ( u λ ) + j = 1 k G , λ ( w j ) b M λ 2 4 u λ D s , 2 ( R N ) 2 + k m λ , + j = 1 k b M λ 2 4 w j D s , 2 ( R N ) 2 m λ , ,

which is a contradiction according to Lemma 4.2. Thus, k = 0 and (4.17) yields u n u λ .□

Remark 4.1

From the proof of Lemma 3.4, we see that the map λ b λ is increasing. Then, choosing an arbitrary sequence { λ n } ϒ such that λ n 1 , for b ( 0 , b λ n ) ( 0 , b 1 ) , by Lemma 4.4, we obtain a sequence { u λ n } H s ( R N ) such that Φ λ n ( u λ n ) = c λ n and Φ λ n ( u λ n ) = 0 , as λ n 1 .

Proof of Theorem 1.2

Combining Theorem 4.1 and Lemma 4.1, for λ ϒ , we conclude that there exists a bounded ( P S ) c λ sequence { u n } H s ( R N ) such that

Φ λ ( u n ) c λ , Φ λ ( u n ) 0 .

Applying Lemma 4.4, for λ ϒ , b ( 0 , b λ ) , we deduce that Φ λ has a nontrivial critical point u λ H s ( R N ) with Φ λ ( u λ ) = c λ . According to Remark 4.1, choosing an arbitrary sequence { λ n } ϒ such that λ n 1 , for b ( 0 , b λ n ) ( 0 , b 1 ) , we obtain the existence of a sequence { u λ n } H s ( R N ) such that

Φ λ n ( u λ n ) = c λ n and Φ λ n ( u λ n ) = 0 , as λ n 1 .

Now, we need to show { u λ n } is bounded. By ( V 1 ) and Lemma 2.3(ii), we obtain

c δ 0 c λ n = Φ λ n ( u λ n ) 1 2 ( N 2 s ) P λ n ( u λ n ) = a 4 u λ n D s , 2 ( R N ) 2 + 1 4 ( N 2 s ) R N [ ( N 4 s ) V ( x ) ( V ( x ) , x ) ] u λ n 2 d x + λ n ( μ + 4 s N ) 4 ( N 2 s ) R N ( I μ * F ( u λ n ) ) F ( u λ n ) d x b λ n ( μ + 4 s N ) 4 ( N 2 s ) R N ( I μ * F ( u λ n ) ) F ( u λ n ) d x .

Together with Φ λ n ( u λ n ) = c λ n , we conclude

u λ n H s ( R N ) 2 2 c δ 0 + λ n 2 R N ( I μ * F ( u λ n ) ) F ( u λ n ) d x C .

Therefore by Theorem 4.1(iii), we obtain

lim n Φ ( u λ n ) = lim n Φ λ n ( u λ n ) + λ n 1 2 R N ( I μ * F ( u λ n ) ) F ( u λ n ) d x = lim n c λ n = c 1

and

lim n Φ ( u λ n ) , φ = lim n Φ ( u λ n ) , φ + ( λ n 1 ) R N ( I μ * F ( u λ n ) ) f ( u λ n ) φ d x = 0 ,

which imply that { u λ n } is a bounded ( P S ) c 1 sequence for Φ . Therefore, by Lemma 4.4, there exists a nontrivial critical point u 0 for Φ with Φ ( u 0 ) = c 1 .

At last, we prove the existence of a ground-state solution to equation (KC). Set

m = inf { Φ ( u ) : u 0 , Φ ( u ) = 0 } .

It is easy to see that 0 m Φ ( u 0 ) = c 1 < + . For any u satisfying Φ ( u ) = 0 , by standard arguments, we see that u H s ( R N ) ϱ for some ϱ > 0 . While, by ( V 1 ) and P ( u ) = 0 , we obtain

Φ ( u ) = Φ ( u ) 1 2 ( N 2 s ) P ( u ) a 4 u D s , 2 ( R N ) 2 ,

which yields m 0 . If m = 0 , then there exists a critical point sequence { u n } of Φ with Φ ( u n ) 0 which implies lim n u n D s , 2 ( R N ) 2 = 0 . Combining with Φ ( u n ) , u n = 0 , we obtain lim n u n H s ( R N ) = 0 , which contradicts u n H s ( R N ) ϱ . Therefore, 0 < m < + . Then, let { u n } be a sequence satisfying Φ ( u n ) = 0 , Φ ( u n ) m . Applying the same argument as above, we can obtain that { u n } is bounded. Similar to the proof of Lemma 4.4, there exists u H s ( R N ) such that Φ ( u ) = 0 , Φ ( u ) = m .□

5 Regularity of solutions

In this section, we prove the regularity of solutions to equation (KC). We present a preliminary result, which is crucial for the subsequent proof.

Lemma 5.1

[23] Let ϑ , r , ν , t [ 1 , + ) and w [ 0 , 2 ] such that

1 + μ N 1 ν 1 t = w ϑ + 2 w r .

If ι ( 0 , 2 ) satisfies

min ( ϑ , r ) μ N 1 ν < ι < max ( ϑ , r ) 1 1 ν ,

min ( ϑ , r ) μ N 1 t < 2 ι < max ( ϑ , r ) 1 1 t ,

then for every H L ν ( R N ) , K L t ( R N ) and u L ϑ ( R N ) L r ( R N ) ,

R N ( I μ * ( H u ι ) ) K u 2 ι d x C R N H ν d x 1 ν R N K t d x 1 t R N u ϑ d x w ϑ R N u r d x 2 w r .

Lemma 5.2

If H , K L 2 N μ ( R N ) + L 2 N μ + 2 s ( R N ) and μ N < ι < 2 μ N , then for any ε > 0 , there exists C ε , ι R such that for each u H s ( R N ) ,

R N ( I μ * ( H u ι ) ) K u 2 ι d x ε 2 R 2 N u ( x ) u ( y ) 2 x y N + 2 s d x d y + C ε , ι R N u 2 d x .

Proof

Set H = H 1 + H 2 and K = K 1 + K 2 with H 1 , K 1 L 2 N μ ( R N ) and H 2 , K 2 L 2 N μ + 2 s ( R N ) . Using Lemma 5.1 with ϑ = r = 2 N N 2 s , ν = t = 2 N μ + 2 s and w = 0 , we observe that

R N ( I μ * ( H 2 u ι ) ) K 2 u 2 ι d x C R N H 2 2 N μ + 2 s d x μ + 2 s 2 N R N K 2 2 N μ + 2 s d x μ + 2 s 2 N R N u 2 s * d x N 2 s N .

Taking ϑ = r = 2 , ν = t = 2 N μ and w = 2 , we see that

R N ( I μ * ( H 1 u ι ) ) K 1 u 2 ι d x C R N H 1 2 N μ d x μ 2 N R N K 1 2 N μ d x μ 2 N R N u 2 d x .

In a similar vein, applying Lemma 5.1 with ν = 2 N μ + 2 s , t = 2 N μ , ϑ = 2 , r = 2 s * , and w = 1 , we obtain

R N ( I μ * ( H 2 u ι ) ) K 1 u 2 ι d x C R N H 2 2 N μ + 2 s d x μ + 2 s 2 N R N K 1 2 N μ d x μ 2 N R N u 2 d x 1 2 R N u 2 s * d x 1 2 s * ,

with ν = 2 N μ , t = 2 N μ + 2 s , ϑ = 2 , r = 2 s * , and w = 1 ,

R N ( I μ * ( H 1 u ι ) ) K 2 u 2 ι d x C R N H 1 2 N μ d x μ 2 N R N K 2 2 N μ + 2 s d x μ + 2 s 2 N R N u 2 d x 1 2 R N u 2 s * d x 1 2 s * .

From the Sobolev inequality, we conclude that for each u H s ( R N ) ,

R N ( I μ * ( H u ι ) ) K u 2 ι d x C R N H 1 2 N μ d x μ 2 N R N K 1 2 N μ d x μ 2 N R N u 2 d x + C R N H 2 2 N μ + 2 s d x μ + 2 s 2 N R N K 2 2 N μ + 2 s d x μ + 2 s 2 N R 2 N u ( x ) u ( y ) 2 x y N + 2 s d x d y .

We can choose H 2 and K 2 such that

C R N H 2 2 N μ + 2 s d x R N K 2 2 N μ + 2 s d x μ + 2 s 2 N ε 2 .

This completes the proof.□

Lemma 5.3

If H , K L 2 N μ ( R N ) + L 2 N μ + 2 s ( R N ) and u H s ( R N ) is a solution of

(5.1) a + b R N ( Δ ) s 2 u 2 d x ( Δ ) s u + V ( x ) u = ( I μ * H u ) K .

Then, u L p ( R N ) for p 2 , N μ 2 N N 2 s . Moreover, if μ < N 2 s , then u L p ( R N ) for p [ 2 , 2 s * N N 2 s ] .

Proof

Using Lemma 5.2 with ι = 1 , we derive that there exists Θ > 0 such that for each ϕ H s ( R N ) ,

(5.2) R N ( I μ * H ϕ ) K ϕ d x a 2 R 2 N ϕ ( x ) ϕ ( y ) 2 x y N + 2 s d x d y + Θ 2 R N ϕ 2 d x .

Take sequences { H n } n N and { K n } n N in L 2 N μ + 2 s ( R N ) such that H n H and K n K , and H n H , K n K almost everywhere in R N . For every n N , we define the form E n : H s ( R N ) × H s ( R N ) R by

E n ( u , v ) a + b R N ( Δ ) s u 2 d x R 2 N ( u ( x ) u ( y ) ) ( v ( x ) v ( y ) ) x y N + 2 s d x d y + Θ R N u v d x R N ( I μ * H n u ) K n v d x ,

which is bilinear and coercive. It follows from the Minty-Browder theorem [8] that there exists a unique solution u n H s ( R N ) for

(5.3) a + b R N ( Δ ) s u n 2 d x ( Δ ) s u n + Θ u n = ( I μ * ( H n u n ) ) K n + ( Θ V ( x ) ) u ,

where u H s ( R N ) is the solution of (5.1). We obseve that the sequence { u n } converges weakly to u in H s ( R N ) as n . Indeed, thanks to (5.3), we deduce that

(5.4) a + b R N ( Δ ) s u n 2 d x u n D s , 2 ( R N ) 2 + Θ u n L 2 ( R N ) 2 = R N ( I μ * ( H n u n ) ) K n u n d x + R N ( Θ V ( x ) ) u u n d x .

Combining with (5.2), ( V 1 ) , by Hölder and Young inequalities, we obtain that

a 2 u n D s , 2 ( R N ) + Θ 2 u n L 2 ( R N ) R N ( Θ V ( x ) ) u u n d x R N Θ V ( x ) Θ u 2 d x + Θ 4 R N u n 2 d x ,

which yields

a 2 u n D s , 2 ( R N ) + Θ 4 u n L 2 ( R N ) R N Θ V ( x ) Θ u 2 d x C u L 2 ( R N ) 2 .

This means { u n } is bounded in H s ( R N ) . Therefore, there exists M > 0 , u ¯ H s ( R N ) such that

u n u ¯ in H s ( R N ) , u n u ¯ , a.e. in R N , u n D s , 2 ( R N ) M 2 .

It follows from H n L 2 N μ + 2 s ( R N ) that H n u n is bounded in L 2 N μ + N ( R N ) . It is easy to see that H n u n H u ¯ in L 2 N μ + N ( R N ) . Since K n K , by the Lebesgue dominated convergence theorem, we obtain that for any ϕ C 0 ( R N ) , K n ϕ K ϕ in L 2 N N + μ ( R N ) . Then, there holds

R N ( I μ * ( H n u n ) ) K n ϕ d x R N ( I μ * ( H u ¯ ) ) K ϕ d x , ϕ C 0 ( R N ) .

Therefore, u ¯ is a weak solution to the equation

( a + b M 2 ) ( Δ ) s u ¯ + Θ u ¯ = ( I μ * ( H u ¯ ) ) K + ( Θ V ( x ) ) u ,

so that

(5.5) ( a + b M 2 ) u ¯ D s , 2 ( R N ) 2 + Θ u ¯ L 2 ( R N ) 2 = R N ( I μ * ( H u ¯ ) ) K u ¯ d x + R N ( Θ V ( x ) ) u u ¯ d x .

Set v n = u n u ¯ . Gathering (5.4) and (5.5), we conclude that

( a + b M 2 ) v n D s , 2 ( R N ) 2 + Θ v n L 2 ( R N ) 2 = o n ( 1 ) ,

which implies v n D s , 2 ( R N ) 2 = o n ( 1 ) . Obviously, u ¯ is a weak solution to the equation

a + b R N ( Δ ) s u ¯ 2 d x ( Δ ) s u ¯ + Θ u ¯ = ( I μ * ( H u ¯ ) ) K + ( Θ V ( x ) ) u .

u ¯ = u follows directly from the fact that there exists a unique solution to (5.3).

For δ > 0 , let

u n , δ ( x ) = δ , u n δ , u n , δ < u n < δ , δ , u n δ .

We note that u n , δ p 2 u n , δ H s ( R N ) . Indeed, we set D 1 { x R N : u n δ } and D 2 { x R N : u n < δ } . A direct computation shows that

R 2 N u n , δ ( x ) p 2 u n , δ ( x ) u n , δ ( y ) p 2 u n , δ ( y ) 2 x y N + 2 s d x d y = 2 D 1 × D 2 u n , δ ( x ) p 2 u n , δ ( x ) u n , δ ( y ) p 2 u n , δ ( y ) 2 x y N + 2 s d x d y + D 2 × D 2 u n , δ ( x ) p 2 u n , δ ( x ) u n , δ ( y ) p 2 u n , δ ( y ) 2 x y N + 2 s d x d y C D 1 × D 2 u n ( x ) u n ( y ) 2 x y N + 2 s d x d y + D 2 × D 2 u n ( x ) u n ( y ) 2 x y N + 2 s d x d y C R 2 N u n ( x ) u n ( y ) 2 x y N + 2 s d x d y

and

R N u n , δ p 2 u n , δ 2 d x = D 1 u n , δ p 2 u n , δ 2 d x + D 2 u n , δ p 2 u n , δ 2 d x δ 2 p 4 R N u n 2 d x .

Let χ ( u n , δ ) u n , δ p 2 u n , δ and we obtain that χ is increasing according to

0 ( t 1 t 2 ) ( χ ( t 1 ) χ ( t 2 ) ) , t 1 , t 2 R .

Denoting Λ ( t ) 0 t ( χ ( ξ ) ) 1 2 d ξ and using the Jensen inequality, we see that

(5.6) ( t 1 t 2 ) ( χ ( t 1 ) χ ( t 2 ) ) = ( t 1 t 2 ) t 2 t 1 χ ( t ) d t = ( t 1 t 2 ) t 2 t 1 ( Λ ( t ) ) 2 d t t 2 t 1 ( Λ ( t ) ) d t 2 = Λ ( t 1 ) Λ ( t 2 ) 2 , t 1 , t 2 R .

It follows that

(5.7) Λ ( u n , δ ( x ) ) Λ ( u n , δ ( y ) ) 2 ( u n , δ ( x ) u n , δ ( y ) ) ( u n , δ ( x ) p 2 u n , δ ( x ) u n , δ ( y ) p 2 u n , δ ( y ) ) .

Taking u n , δ p 2 u n , δ as a test function in (5.3), we have

(5.8) a R 2 N Λ ( u n , δ ) ( x ) Λ ( u n , δ ) ( y ) 2 x y N + 2 s d x d y + Θ R N u n , δ p 2 2 d x a + b R N ( Δ ) s u n 2 d x × R 2 N ( u n , δ ( x ) u n , δ ( y ) ) ( u n , δ ( x ) p 2 u n , δ ( x ) u n , δ ( y ) p 2 u n , δ ( y ) ) x y N + 2 s d x d y + Θ R N u n , δ p 2 u n , δ u n d x = R N ( I μ * ( H n u n ) ) ( K n u n , δ p 2 u n , δ ) d x + R N ( Θ V ( x ) ) u n , δ p 2 u n , δ u d x { u n > δ } ( I μ * ( H n u n ) ) ( K n u n p 2 u n ) d x + { u n δ } ( I μ * ( H n u n ) ) ( K n u n , δ p 2 u n , δ ) d x + R N ( Θ V ( x ) ) u n , δ p 2 u n , δ u d x

If p < 2 N μ , applying Lemma 5.2 with ι = 2 p , we derive that there exists C > 0 such that

{ u n δ } ( I μ * ( H n u n ) ) ( K n u n , δ p 2 u n , δ ) d x = R N ( I μ * H n u n , δ ) ( K n u n , δ p 2 u n , δ ) d x a 2 R 2 N u n , δ ( x ) p 2 u n , δ ( y ) p 2 2 x y N + 2 s d x d y + C R N u n , δ p 2 2 d x a 2 R 2 N Λ ( u n , δ ) ( x ) Λ ( u n , δ ) ( y ) 2 x y N + 2 s d x d y + C R N u n , δ p 2 2 d x .

Combining with (5.8), recalling Lemma 2.1, we deduce that

(5.9) a 2 R 2 N Λ ( u n , δ ) ( x ) Λ ( u n , δ ) ( y ) 2 x y N + 2 s d x d y C R N ( u n p + u p ) d x + { u n > δ } ( I μ * ( H n u n ) ) ( K n u n p 2 u n ) d x C R N ( u n p + u p ) d x + C R N K n u n p 1 r d x 1 r { u n > δ } H n u n t d x 1 t ,

where 1 r = μ 2 N + 1 1 p , 1 t = μ 2 N + 1 p . According to Hölder inequality, if u n L p ( R N ) , we obtain K n u n p 1 L r ( R N ) and H n u n L t ( R N ) . Hence,

lim δ + { u n > δ } ( I μ * ( H n u n ) ) ( K n u n p 2 u n ) d x = 0 .

Combining with (5.9), using the Sobolev embedding theorem, we conclude that

(5.10) limsup n + R N u n p N N 2 s d x N 2 s N C limsup n + R N u n p d x ,

which yields u L p ( R N ) for p 2 , N μ 2 N N 2 s .

From N 2 s > μ , we can see that

2 < 2 s * < 2 N μ .

Taking p = 2 s * in (5.10) and recalling (5.2), Sobolev embedding theorem and Fatou’s lemma, we conclude that

limsup n + R N u n 2 s * N N 2 s d x N 2 s N C limsup n + R N u n 2 s * d x C limsup n + R 2 N u n ( x ) u n ( y ) 2 x y N + 2 s d x d y .

Then, we have u L p ( R N ) , where p [ 2 , 2 s * N N 2 s ]

Lemma 5.4

Suppose that all conditions of Theorem 1.3 hold, and u H s ( R N ) is a nontrivial solution of equation (KC). Then, there holds

I μ * F ( u ) L ( R N ) C .

Proof

Define H : R N R and K : R N R by H ( x ) = F ( u ( x ) ) u ( x ) and K ( x ) = f ( u ( x ) ) for x R N . Note that

K ( x ) C ( u ( x ) μ N + u ( x ) μ + 2 s N 2 s )

and

H ( x ) C ( u ( x ) μ N + u ( x ) μ + 2 s N 2 s ) .

Therefore, H , K L 2 N μ ( R N ) + L 2 N μ + 2 s ( R N ) . It follows from Lemma 5.3 that u L p ( R N ) for each p 2 , N μ 2 N N 2 s . According to Lemma 2.3-(i), F ( u ) L γ ( R N ) for γ 2 N N + μ , 2 N 2 ( N + μ ) μ . Since μ ( 0 , N ) , we conclude

2 N N + μ < N μ < 2 N 2 ( N + μ ) μ ,

which implies I μ * F ( u ) L ( R N ) C , applying Lemma 2.2.□

Proof of Theorem 1.3

For τ 1 and T > 0 , define

u T ( x ) = T , u ( x ) < T , u ( x ) , u ( x ) T , T , u ( x ) > T

and u ˜ T = u u T 2 ( τ 1 ) . First, we show u ˜ T H s ( R N ) . We define Ω 1 { x R N : u ( x ) T } and Ω 2 { x R N : u ( x ) > T } . It follows from the definition of u ˜ T that

R 2 N u ˜ T ( x ) u ˜ T ( y ) 2 x y N + 2 s d x d y = Ω 1 × Ω 1 u ˜ T ( x ) u ˜ T ( y ) 2 x y N + 2 s d x d y + 2 Ω 1 × Ω 2 u ˜ T ( x ) u ˜ T ( y ) 2 x y N + 2 s d x d y + Ω 2 × Ω 2 u ˜ T ( x ) u ˜ T ( y ) 2 x y N + 2 s d x d y C T 4 ( τ 1 ) R 2 N u ( x ) u ( y ) 2 x y N + 2 s d x d y ,

which yields that u ˜ T H s ( R N ) . Consider the function

Γ ( t ) 0 t [ u ˜ T ( ξ ) ] 1 2 d ξ .

Similar to (5.6) and (5.7), we obtain

(5.11) u ˜ T H s ( R N ) C R 2 N Γ ( u ) ( x ) Γ ( u ) ( y ) 2 x y N + 2 s d x d y + u ˜ T L 2 ( R N ) 2 C R 2 N ( u ( x ) u ( y ) ) ( u ˜ T ( x ) u ˜ T ( y ) ) x y N + 2 s d x d y + R N u u ˜ T d x .

From Φ ( u ) , v = 0 , we take v = u ˜ T in (2.2). Together with (5.11), Lemma 5.4, and the Sobolev embedding theorem, we obtain that for any γ [ 2 , 2 s * ] ,

(5.12) R N u u T τ 1 γ d x 2 γ C R 2 N ( u ( x ) u ( y ) ) ( u ˜ T ( x ) u ˜ T ( y ) ) x y N + 2 s d x d y + R N u u ˜ T d x C τ 2 R N u μ N 1 u u T τ 1 2 d x + R N u 2 μ , s * 2 u u T τ 1 2 d x .

Now, we provide an L -estimate of u . We prove it in two different cases.

Case 1. 2 μ , s * > 2 N < 4 s + μ .

We will divide the proof into three steps.

Step 1. Let τ 1 2 N + μ 2 N , N μ 2 N N 2 s N + μ N 2 . We claim that

(5.13) 1 + R N u 2 μ , s * τ 1 d x 2 2 μ , s * ( τ 1 1 ) < .

According to the definition of u T , we have

R N u μ N 1 u u T τ 1 1 2 d x R N u 2 τ 1 + N + μ N 2 d x .

Let L > 0 be fixed later. By Hölder’s inequality, we obtain

R N u 2 μ , s * 2 u u T τ 1 1 2 d x L 2 μ , s * N + μ N { x : u ( x ) L } u N + μ N 2 u u T τ 1 1 2 d x + { x : u ( x ) > L } u 2 μ , s * 2 u u T τ 1 1 2 d x L 2 μ , s * N + μ N { x : u ( x ) L } u N + μ N + 2 τ 1 2 d x + { x : u ( x ) > L } u 2 μ , s * d x 2 μ , s * 2 2 μ , s * R N u u T τ 1 1 2 μ , s * d x 2 2 μ , s * .

By u L p ( R N ) for any p [ 2 , N μ 2 N N 2 s ) , we can fix L such that

{ x : u > L } u 2 μ , s * d x 2 μ , s * 2 2 μ , s * 1 2 C τ 1 2 .

The above two inequalities together with (5.12) imply that

R N u u T τ 1 1 2 μ , s * d x 2 2 μ , s * 2 C τ 1 2 R N u N + μ N 2 u u T τ 1 1 2 d x + L 2 μ , s * N + μ N R N u N + μ N + 2 ( τ 1 1 ) d x .

Letting T in the above inequality, we see that

(5.14) R N u 2 μ , s * τ 1 d x 2 2 μ , s * 2 C τ 1 2 1 + L 2 μ , s * N + μ N R N u N + μ N + 2 ( τ 1 1 ) d x .

Since N + μ N + 2 ( τ 1 1 ) [ 2 , N μ 2 N N 2 s ) and (5.14), we can deduce that (5.13) holds. Then, u L p ˆ 1 ( R N ) , where

p ˆ 1 2 , N μ 2 N N 2 s 2 μ , s * 2 N + μ 2 N , 2 μ , s * N μ 2 N N 2 s N + μ N 2 .

Step 2. Let τ 2 = 1 + 2 μ , s * ( τ 1 1 ) 2 . We claim that

1 + R N u 2 μ , s * τ 2 d x 2 2 μ , s * ( τ 2 1 ) ( C τ 2 ) 2 τ 2 1 1 + R N u 2 μ , s * τ 1 d x 2 2 μ , s * ( τ 1 1 ) .

Choosing τ [ τ 1 , τ 2 ] , we obtain that

2 N + μ N + 2 ( τ 1 ) < 2 μ , s * + 2 ( τ 1 ) 2 μ , s * τ 1 .

It follows from (5.12) and (5.13) that

(5.15) R N u 2 μ , s * τ d x 2 2 μ , s * C τ 2 R N u N + μ N + 2 ( τ 1 ) d x + R N u 2 μ , s * + 2 ( τ 1 ) d x < .

Letting τ = τ 2 in (5.15) and using the Young inequality, we obtain that

R N u 2 μ , s * τ 2 d x 2 2 μ , s * C τ 2 2 R N u N + μ N + 2 ( τ 2 1 ) d x + R N u 2 μ , s * + 2 ( τ 2 1 ) d x C τ 2 2 1 + R N u 2 μ , s * + 2 ( τ 2 1 ) d x .

Noting that 2 μ , s * > 2 , we can observe

( x 1 + x 2 ) 2 2 μ , s * x 1 2 2 μ , s * + x 2 2 2 μ , s * , x 1 , x 2 > 0 .

Then, we can see that

1 + R N u 2 μ , s * τ 2 d x 2 2 μ , s * 1 + R N u 2 μ , s * τ 2 d x 2 2 μ , s * C τ 2 2 1 + R N u 2 μ , s * + 2 ( τ 2 1 ) d x ,

which yields

1 + R N u 2 μ , s * τ 2 d x 2 2 μ , s * ( τ 2 1 ) C τ 2 2 τ 2 1 1 + R N u 2 μ , s * τ 1 d x 1 τ 2 1 = C τ 2 2 τ 2 1 1 + R N u 2 μ , s * τ 1 d x 2 2 μ , s * ( τ 1 1 ) .

Step 3. Iterating the above steps and setting

(5.16) τ i + 1 1 = 2 μ , s * 2 ( τ i 1 ) , i N ,

we obtain that

1 + R N u 2 μ , s * τ i + 1 d x 2 2 μ , s * ( τ i + 1 1 ) C τ 2 2 τ i + 1 1 1 + R N u 2 μ , s * τ i d x 1 τ i + 1 1 = ( C τ i + 1 ) 2 τ i + 1 1 1 + R N u 2 μ , s * τ i d x 2 2 μ , s * ( τ i 1 ) ,

which implies that

u L 2 μ , s * τ n + 1 ( R N ) i = 1 n ( C τ i + 1 ) 2 τ i + 1 1 1 + R N u 2 μ , s * τ 1 d x 2 2 μ , s * ( τ 1 1 ) τ n + 1 1 2 τ n + 1 .

Together with (5.16), we obtain that

(5.17) u L 2 μ , s * τ n + 1 ( R N ) i = 1 n ( C τ i + 1 ) 2 τ i + 1 1 1 + R N u 2 μ , s * τ 1 d x 2 2 μ , s * ( τ 1 1 ) ( 2 μ , s * ) n ( τ 1 1 ) 2 [ 2 n + ( 2 μ , s * ) n ( τ 1 1 ) ] .

It is easy to see that Π i = 1 ( C τ i + 1 ) 2 τ i + 1 1 < . Taking n in (5.17), we have u L ( R N ) < .

Case 2. 2 μ , s * 2 N 4 s + μ .

In this case, we take τ = 2 s * 2 .

Step 1. It follows from Lemma 5.2 that u L p ( R N ) for p 2 , 2 s * N N 2 s . Since μ > N ( N 6 s ) + N 2 s , one sees that

2 < N + μ N + 2 ( τ 1 ) < 2 μ , s * + 2 ( τ 1 ) < 2 s * N N 2 s ,

which implies

R N u N + μ N 2 u u T τ 1 2 d x < +

and

R N u 2 μ , s * 2 u u T τ 1 2 d x < + .

For any 0 < l < + , define

A τ R N u N + μ N 2 u u T τ 1 2 d x = { x : u ( x ) l } u N + μ N 2 u u T τ 1 2 d x + { x : u ( x ) > l } u N + μ N 2 u u T τ 1 2 d x = A τ ( l ) + A τ c ( l )

and

B τ R N u 2 μ , s * 2 u u T τ 1 2 d x = { x : u ( x ) l } u 2 μ , s * 2 u u T τ 1 2 d x + { x : u ( x ) > l } u 2 μ , s * 2 u u T τ 1 2 d x = B τ ( l ) + B τ c ( l ) .

Clearly,

lim l A τ ( l ) = A τ , lim l 0 A τ ( l ) = 0

and

lim l B τ ( l ) = B τ , lim l 0 B τ ( l ) = 0 .

If A τ = A τ ( l ) , or B τ = B τ ( l ) where l < . Then, u L ( R N ) . Thus, we only need to consider the case

A τ ( l ) < A τ and B τ ( l ) < B τ , l < .

Without loss of generality, we take l = 1 , and there exist 0 < C 1 , C 2 < 1 such that

(5.18) A τ ( l ) = C 1 A τ , and B τ ( l ) = C 2 B τ .

Noting that μ < N , we obtain

A τ c = { x : u ( x ) 1 } u N + μ N 2 u u T τ 1 2 d x { x : u ( x ) 1 } u u T τ 1 2 d x .

Together with (5.18), we deduce that

(5.19) A τ = A τ ( 1 ) + A τ c ( 1 ) = 1 1 C 1 A τ c ( 1 ) 1 1 C 1 { x : u ( x ) 1 } u u T τ 1 2 d x .

In the same way, due to 2 μ , s * 2 , it is also easy to obtain

(5.20) B τ = B τ ( 1 ) + B τ c ( 1 ) = 1 1 C 2 B τ c ( 1 ) 1 1 C 2 { x : u ( x ) 1 } u u T τ 1 2 d x .

Using (5.19), (5.20), and (5.12), we obtain that

(5.21) R N u u T τ 1 γ d x 2 γ C 1 C 1 + C 1 C 2 R N u u T τ 1 2 d x .

Taking T in (5.21), we can see that

(5.22) R N u τ γ d x 2 γ C 1 C 1 + C 1 C 2 R N u τ 2 d x ,

which yields that

u L γ 2 s * 2 ( R N ) C 1 C 1 + C 1 C 2 1 2 s * u L 2 s * ( R N ) < .

Recalling that γ [ 2 , 2 s * ] , we deserve that u L p 1 ( R N ) , where p 1 [ 2 , 2 s * ] [ 2 s * , ( 2 s * ) 2 2 ] .

Particularly, taking γ = 2 s * , we obtain that

(5.23) u L ( 2 s * ) 2 2 ( R N ) C 1 C 1 + C 1 C 2 1 2 s * u L 2 s * ( R N ) < .

Step 2. From

2 s * < N + μ N + 2 ( τ 2 1 ) < 2 μ , s * + 2 ( τ 2 1 ) < ( 2 s * ) 2 2 ,

and the definition of A τ and B τ , we have

A τ 2 < , B τ 2 < .

Similar to step 1, we only need to present the case

A τ 2 ( 1 ) < A τ 2 and B τ 2 ( 1 ) < B τ 2 , l < .

Furthermore, through simple calculations, we can infer that

(5.24) A τ ( 1 ) = { x : u ( x ) 1 } u N + μ N 2 u u T τ 1 2 d x { x : u ( x ) 1 } u N + μ N 2 u u T τ 1 ( u T τ 2 τ ) 2 d x = A τ 2 ( 1 )

and

(5.25) A τ c ( 1 ) = { x : u ( x ) > 1 } u N + μ N 2 u u T τ 1 2 d x { x : u ( x ) 1 } u N + μ N 2 u u T τ 1 ( u T τ 2 τ ) 2 d x = A τ 2 c ( 1 ) ,

Similarly, we can see that

Combining (5.19), (5.20), (5.24), and (5.25), we conclude that

A τ 2 ( 1 ) A τ ( 1 ) = C 1 1 C 1 A τ c ( 1 ) C 1 1 C 1 A τ 2 c ( 1 ) ,

which yields

A τ 2 1 1 C 1 { x : u ( x ) 1 } u u T τ 2 1 2 d x .

Similarly, we can see that

B τ 2 1 1 C 2 { x : u ( x ) 1 } u u T τ 2 1 2 d x .

From (5.26), τ 2 = 2 s * 2 2 > 1 and (5.12), we deduce that

(5.26) u L γ 2 s * 2 2 ( R N ) C 1 C 1 + C 1 C 2 1 2 2 s * 2 2 u L ( 2 s * ) 2 2 ( R N ) < ,

which implies that u L p 2 ( R N ) , where p 2 [ 2 , 2 s * ] 2 s * , ( 2 s * ) 2 2 ( 2 s * ) 2 2 , 2 s * 2 s * 2 2 .

Particularly, taking γ = 2 s * , we obtain that

(5.27) u L ( 2 s * ) 3 2 2 ( R N ) C 1 C 1 + C 1 C 2 1 2 s * + 1 2 2 s * 2 2 u L 2 s * ( R N ) < .

Step 3. By repeating the above process, for n N , we can deduce that

(5.28) u L ( 2 s * ) n + 1 2 n ( R N ) C 1 C 1 + C 1 C 2 i = 1 n 1 2 2 s * 2 i u L 2 s * ( R N ) < .

It is easy to check that

i = 1 n 1 2 2 s * 2 i < .

Therefore, taking the limit as n , (5.28) implies that

u L ( R N ) C u L 2 s * ( R N ) < .

Combining V L ( R N ) , Lemmas 2.3 and 5.4, there exists C > 0 such that

(5.29) ( Δ ) s u = h ( u ) C ( u + u μ N + u q 1 ) ,

where h ( u ) a + b R N ( Δ ) s 2 u 2 d x 1 [ V ( x ) u + ( I μ * F ( u ) ) f ( u ) ] . Together with (5.29) and Lemma 5.4, we conclude that h L ( R N ) . Then, by [30, Proposition 2.1.9], the proof is complete.□

Acknowledgements

The authors thank the anonymous referees for their valuable comments and nice suggestions to improve the results. J. Yang is supported by the Natural Science Foundation of Hunan Province of China (2023JJ30482, 2022JJ30463), Research Foundation of Education Bureau of Hunan Province (23A0558, 22A0540), and Aid Program for Science and Technology Innovative Research Team in Higher Educational Institutions of Hunan Province.

  1. Author contributions: All authors contributed equally to this work.

  2. Conflict of interest: On behalf of all authors, the corresponding author states that there is no conflict of interest.

  3. Data availability statement: Not applicable.

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Received: 2024-06-07
Revised: 2024-09-23
Accepted: 2025-02-25
Published Online: 2025-04-09

© 2025 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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