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Integral inequalities with an extended Poisson kernel and the existence of the extremals

  • Chunxia Tao EMAIL logo and Yike Wang
Published/Copyright: August 28, 2023
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Advanced Nonlinear Studies
From the journal Advanced Nonlinear Studies Volume 23 Issue 1

Abstract

In this article, we first apply the method of combining the interpolation theorem and weak-type estimate developed in Chen et al. to derive the Hardy-Littlewood-Sobolev inequality with an extended Poisson kernel. By using this inequality and weighted Hardy inequality, we further obtain the Stein-Weiss inequality with an extended Poisson kernel. For the extremal problem of the corresponding Stein-Weiss inequality, the presence of double-weighted exponents not being necessarily nonnegative makes it impossible to obtain the desired existence result through the usual technique of symmetrization and rearrangement. We then adopt the concentration compactness principle of double-weighted integral operator, which was first used by the authors in Chen et al. to overcome this difficulty and obtain the existence of the extremals. Finally, the regularity of the positive solution for integral system related with the extended kernel is also considered in this article. Our regularity result also avoids the nonnegativity condition of double-weighted exponents, which is a common assumption in dealing with the regularity of positive solutions of the double-weighted integral systems in the literatures.

Keywords: Hardy-Littlewood-Sobolev inequality; Stein-Weiss inequality; Poisson kernel; concentration compactness principle
MSC 2010: 35A01; 39B72

1 Introduction

Hardy-Littlewood-Sobolev (HLS) inequality and Stein-Weiss inequality play a crucial role in many important problems (such as isoperimetric inequality, prescribing-curvature problem, etc.) from analysis and geometry. The classical HLS inequality, which was first established in [34,46], states that

∫ R n ∫ R n f ( y ) g ( x ) ∣ x − y ∣ λ d x d y ≤ C n , p , q ′ ‖ f ‖ L p ( R n ) ‖ g ‖ L q ′ ( R n ) ,

where 1 < q ′ , p < ∞ , 0 < λ < n , 1 q ′ + 1 p + λ n = 2 . Throughout this article, we always let p ′ and q ′ denote the conjugate of p and q , respectively, and we use the notation f ( x ) ≲ g ( x ) to mean that there exists some constant C such that f ( x ) ≤ C g ( x ) .

In 1950s, Stein and Weiss [47] proved the following double-weighted HLS inequality,

(1.1) ∫ R n ∫ R n f ( y ) g ( x ) ∣ y ∣ α ∣ x − y ∣ λ ∣ x ∣ β d x d y ≤ C n , α , β , p , q ′ ‖ f ‖ L p ‖ g ‖ L q ′ ,

where p , q ′ , α , β , and λ satisfy the following conditions:

1 q ′ + 1 p + α + β + λ n = 2 , 1 q ′ + 1 p ≥ 1 , α + β ≥ 0 , β < n q , α < n p ′ , 0 < λ < n .

For HLS inequality, Lieb [38] utilized the rearrangement argument to obtain the existence of the extremals and classified the extremal function in the conformal case of q ′ = p = 2 n 2 n − λ . More methods about the classification of the extremal functions in the conformal case have also been discussed in [6,7,26,27]. The authors in [10,11,37] classified all the positive solutions of integral equations which extremal functions of HLS inequality satisfy through the method of moving plane or moving sphere. For more applications of the moving plane method in partial differential equations, one can also see [18,24]. For the Stein-Weiss inequality (1.1), Lieb [38] also established the existence of the extremals under the assumption of α ≥ 0 , β ≥ 0 , and p < q . By using the concentration compactness principle for a double-weighted integral operator, Chen et al. [14] removed the assumption of α and β being nonnegative and further established the existence of extremals for the Stein-Weiss inequality on the Heisenberg group. For more results about HLS and Stein-Weiss inequalities on the Heisenberg group, we refer the reader to the works [25,28,30,31,36]. When p = q , Lieb [38] pointed out that the extremals of Stein-Weiss inequality cannot be expected to exist from the result in [35]. The reverse HLS inequality, the reverse Stein-Weiss inequality, and the existence of their extremals have also been established in [2,12,15,21,43,44].

It is interesting to establish the sharp HLS inequality on the upper half space, which is a form of trace inequality and related to the problem of isoperimetric inequality on the manifold with the nonpositive scalar curvature. Hang et al. [33] derived the following HLS inequality involving Poisson kernel on the upper half space,

(1.2) ∫ R + n ∫ ∂ R + n f ( y ) x n g ( x ) ∣ x − y ∣ n 2 d y d x ≤ C n ‖ f ‖ L 2 ( n − 1 ) n − 2 ( ∂ R + n ) ‖ g ‖ L 2 n n + 2 ( R + n ) .

Through conformal transformation between R + n and B 1 (see, e.g., [20,48]), inequality (1.2) is equivalent to an integral inequality on the ball [32]. This kind of integral inequality can be viewed as the isoperimetric inequality of high dimension for the domain, which is flat and isometric to ( B ¯ 1 n , w 4 n − 2 d x 2 ) on the manifold with the nonpositive scalar curvature. In two-dimensional case, from the Riemann mapping theorem, the classical Carleman inequality [5] is in fact the isoperimetric inequality for manifold ( M 2 , g ) with the nonpositive scalar curvature. It should be noted that the Poisson kernel

(1.3) P ( x ′ , x n ) = x n ( ∣ x ′ ∣ 2 + x n 2 ) n 2

appearing in inequality (1.2) is a fundamental solution of Laplacian operator on the upper half space R + n , that is, P ( x ′ , x n ) satisfies the following equation:

− Δ P ( x ′ , x n ) = 0 , ( x ′ , x n ) ∈ R + n , P ( x ′ , 0 ) = δ 0 ( x ′ ) , ( x ′ , 0 ) ∈ ∂ R + n .

The fundamental solution for the corresponding high-order equation on the upper half space

( − Δ ) m + 1 P m ( x ′ , x n ) = 0 , ( x ′ , x n ) ∈ R + n , P m ( x ′ , 0 ) = δ 0 ( x ′ ) , ( x ′ , 0 ) ∈ ∂ R + n , ∂ ∂ x n ( Δ k P m ( x ′ , x n ) ) = 0 , ( x ′ , x n ) ∈ ∂ R + n , 0 ≤ k ≤ m − 1 2 , Δ k ( P m ( x ′ , x n ) ) = Γ ( m + 1 ) Γ m + 1 2 − k Γ ( m − k − 1 ) Γ m + 1 2 Δ x ′ k ( P m ( x ′ , 0 ) ) , ( x ′ , x n ) ∈ ∂ R + n , 1 ≤ k ≤ m 2

is high-order Poisson kernel, which takes the form as follows:

(1.4) P m ( x ′ , x n ) = x n 1 + 2 m ( ∣ x ′ ∣ 2 + x n 2 ) n + 2 m 2 .

We also note that P m ( x ′ , x n ) naturally arises in the research of the sharp Sobolev trace inequalities for the higher order derivatives (see [1,17,45,49]). Moreover, to define the general fractional Laplacian operator, Caffarelli and Silvestre [4] introduced an extension operator L on the half space R + n by giving operator L = Δ + 1 − α x n . It is not difficult to verify that the fundamental solution for the equation

L P ( x ′ , x n ) = 0 , x = ( x ′ , x n ) ∈ R + n , P ( x ′ , 0 ) = δ 0 ( x ′ ) , ( x ′ , 0 ) ∈ ∂ R + n

is

(1.5) P b ( x ′ , x n ) = C 1 x n 1 − b ( ∣ x ′ ∣ 2 + x n 2 ) n − b 2 ,

where b = 1 − α .

We can generalize the kernel (1.3), (1.4), and (1.5) by taking the form as follows:

(1.6) P θ , λ ( x ′ , x n ) = x n θ ( ∣ x ′ ∣ 2 + x n 2 ) λ 2

with θ ≥ 0 , λ > 0 . There are many explorations about the integral inequalities with the kernel (1.6). Conformal HLS inequality (1.2) involving Poisson kernel in [33] is the inequality with the kernel (1.6) by taking θ = 1 , λ = n . For θ = 1 − α , λ = n − α 2 , Chen [8] derived the conformal integral inequality involving the kernel (1.5). When θ = 0 , 0 < λ < n − 1 , the integral inequality with the kernel (1.6) reduces to the sharp HLS inequality on the upper half space, which was proved by Dou and Zhu in [22]. Dou [19] further extended this inequality to the double weighted case. Dou et al. [20] also studied the conformal HLS inequality involving the fractional Poisson kernel ( θ = 1 , λ = n − α + 2 ). Chen et al. [16] extended this inequality to the nonconformal case. Moreover, for fractional Poisson kernel, Chen et al. [13] also established the Stein-Weiss inequality and existence of their extremals. For the general kernel (1.6), Gluck [29] studied the integral inequality (1.8) in the conformal case and Tao [48] considered the reverse integral inequality. It is interesting to establish the aforementioned integral inequalities on the upper half space with kernel (1.6) to nonconformal case. To be specific, we first derive the following HLS inequality with an extended Poisson kernel:

Theorem 1.1

For 0 ≤ θ < ∞ , θ < λ ≤ n − 1 + θ , and 1 < p , q ′ < ∞ satisfying

(1.7) n − 1 n p + 1 q ′ + λ − θ + 1 n = 2 ,

there exists some constant C n , p , q ′ , λ > 0 such that for all f ∈ L p ( ∂ R + n ) , g ∈ L q ′ ( R + n ) we have

(1.8) ∫ R + n ∫ ∂ R + n f ( y ) x n θ g ( x ) ∣ x − y ∣ λ d y d x ≤ C n , p , q ′ , λ ‖ f ‖ L p ( ∂ R + n ) ‖ g ‖ L q ′ ( R + n ) .

By using the above inequality (1.8) and weighted Hardy inequality with integral form on the upper half space, we can establish the following Stein-Weiss inequality with an extended Poisson kernel:

Theorem 1.2

For 0 ≤ θ < ∞ , θ < λ ≤ α + β + λ ≤ n − 1 + θ , α < n − 1 p ′ , β < n q + θ , 1 < p , q ′ < ∞ satisfying

(1.9) n − 1 n p + 1 q ′ + α + β + λ − θ + 1 n = 2 ,

then there exists some constant C n , p , q ′ , α , β , λ such that for all f ∈ L p ( ∂ R + n ) , g ∈ L q ′ ( R + n ) , we have

(1.10) ∫ R + n ∫ ∂ R + n f ( y ) x n θ g ( x ) ∣ y ∣ α ∣ x − y ∣ λ ∣ x ∣ β d y d x ≤ C n , p , q ′ , α , β , λ ‖ f ‖ L p ( ∂ R + n ) ‖ g ‖ L q ′ ( R + n ) .

A natural question is whether there exist extremal functions for inequality (1.10). Define

(1.11) P ( f ) ( x ) = ∫ ∂ R + n f ( y ) x n θ ∣ x − y ∣ λ d y .

It is easy to verify that inequality (1.10) is equivalent to the following inequality:

‖ P ( f ) ∣ x ∣ − β ‖ L q ( R + n ) ≤ C n , α , β , p , q ′ , λ ‖ f ∣ y ∣ α ‖ L p ( ∂ R + n ) .

Consider the following maximizing problem,

(1.12) C n , p , q ′ , α , β , λ ≔ sup { ‖ P ( f ) ( x ) ∣ x ∣ − β ‖ L q ( R + n ) : f ≥ 0 , ‖ f ( y ) ∣ y ∣ α ‖ L p ( ∂ R + n ) = 1 } .

Then we can prove that the sharp constant C n , p , q ′ , α , β , λ could actually be achieved.

Theorem 1.3

Under the hypothesis of Theorem 1.2, there exists some nonnegative function f such that ‖ f ( y ) ∣ y ∣ α ‖ L p ( ∂ R + n ) = 1 and ‖ P ( f ) ( x ) ∣ x ∣ − β ‖ L q ( R + n ) = C n , p , q ′ , α , β , λ .

Theorem 1.3 implies the existence of the extremals for inequality (1.8). In the study by Lieb [38], because of the use of Riesz rearrangement inequality [3] in proving the existence of the extremals, the assumptions α ≥ 0 , β ≥ 0 are necessary. However, through the concentration compactness principle of double-weighted integral operator, we remove this assumption so that the indices just satisfy α + β ≥ 0 .

Once we have obtained the inequality (1.10) and established the existence of extremal function, it is natural to consider the corresponding Euler-Lagrange system. Denote by

P ( x , y , θ , λ ) = x n θ ∣ x − y ∣ λ , ( x , y ) ∈ R + n × ∂ R + n .

By maximizing the following functional

(1.13) J ( f , g ) = ∫ R + n ∫ ∂ R + n ∣ y ∣ − α f ( y ) P ( x , y , θ , λ ) g ( x ) ∣ x ∣ − β d y d x

under the constraint ‖ f ‖ L p ( ∂ R + n ) = ‖ g ‖ L q ′ ( R + n ) = 1 , we can obtain that the extremal functions of inequality (1.10) satisfy the following Euler-Lagrange system up to some constants:

(1.14) J ( f , g ) f p − 1 ( y ) = ∫ R + n ∣ y ∣ − α P ( x , y , θ , λ ) g ( x ) ∣ x ∣ − β d x , y ∈ ∂ R + n , J ( f , g ) g q ′ − 1 ( x ) = ∫ ∂ R + n ∣ x ∣ − β P ( x , y , θ , λ ) f ( y ) ∣ y ∣ − α d y , x ∈ R + n .

Let u = c 1 f p − 1 , v = c 2 g q ′ − 1 , p 0 = p ′ − 1 , q 0 = q − 1 and pick two suitable constants c 1 and c 2 , then system (1.14) is simplified as follows:

(1.15) u ( y ) = ∫ R + n ∣ y ∣ − α P ( x , y , θ , λ ) v q 0 ( x ) ∣ x ∣ − β d x , y ∈ ∂ R + n , v ( x ) = ∫ ∂ R + n ∣ x ∣ − β P ( x , y , θ , λ ) u p 0 ( y ) ∣ y ∣ − α d y , x ∈ R + n .

We have the following theorem.

Theorem 1.4

Assume that ( u ( y ) , v ( x ) ) ∈ L p 0 + 1 ( ∂ R + n ) × L q 0 + 1 ( R + n ) is a pair of positive solutions of the integral system (1.15), and α , β , λ , θ , p 0 , and q 0 satisfy

(1.16) 0 ≤ θ < ∞ , θ < λ ≤ α + β + λ ≤ n − 1 + θ ,

(1.17) p 0 , q 0 > 1 , β − θ n < 1 q 0 + 1 < β + λ − θ n , n − 1 n 1 p 0 + 1 + 1 q 0 + 1 = α + β + λ − θ n .

Then ( u , v ) ∈ L r ( ∂ R + n ) × L s ( R + n ) for all r and s such that

(1.18) 1 r ∈ max max { α , 0 } n − 1 , 1 p 0 + 1 − n n − 1 1 q 0 + 1 + max { β − θ , 0 } n − 1 , min α + λ − max { β − θ , 0 } + β − θ n − 1 , 1 p 0 + 1 − n n − 1 1 q 0 + 1 + n n − 1 min β + λ + α − max { α , 0 } − θ n , 2 q 0 + 1 − max { β − θ , 0 } n , 1 ,

and

(1.19) 1 s ∈ max max { β − θ , 0 } n , 1 q 0 + 1 − n − 1 n 1 p 0 + 1 + max { α , 0 } n , min β + λ + α − max { α , 0 } − θ n , 1 q 0 + 1 − n − 1 n 1 p 0 + 1 + n − 1 n min α + λ − max { β − θ , 0 } + β − θ n − 1 , 2 p 0 + 1 − max { α , 0 } n − 1 , 1 .

2 The proof of Theorem 1.1

Proof

By duality, it is easy to prove that inequality (1.8) is equivalent to the following inequality:

(2.1) ‖ P ( f ) ( x ) ‖ L q ( R + n ) ≤ C n , p , q ′ , λ ‖ f ‖ L p ( ∂ R + n ) ,

where P ( f ) ( x ) is defined as (1.11). By Hölder’s inequality, we have

(2.2) P ( f ) ( x ) = ∫ ∂ R + n f ( y ) x n θ ( ∣ x ′ − y ∣ 2 + x n 2 ) λ 2 d y ≤ ‖ f ‖ L p ( ∂ R + n ) x n θ ∫ ∂ R + n 1 ( ∣ x ′ − y ∣ 2 + x n 2 ) λ p ′ 2 d y 1 p ′ = ‖ f ‖ L p ( ∂ R + n ) x n θ − λ + n − 1 p ′ ∫ ∂ R + n 1 ( ∣ y ∣ 2 + 1 ) λ p ′ 2 d y 1 p ′ ≤ C n , λ , p ‖ f ‖ L p ( ∂ R + n ) x n θ − λ + n − 1 p ′ .

The last step holds since the conditions 0 < 1 q = λ − θ n − n − 1 n 1 p ′ and θ < λ deduce λ p ′ > n − 1 . According to the condition (1.9), we can pick r and s such that 1 < r < q = n p ′ ( λ − θ ) p ′ − ( n − 1 ) , s > max { 1 , n − 1 λ } and 1 r + 1 = 1 p + 1 s . Then for all a > 0 , we have

(2.3) ∫ { x ∈ R + n : 0 < x n < a } P ( f ) r ( x ) d x = ∫ 0 a ∫ ∂ R + n ∣ P θ , λ ∗ f ∣ r d x ′ d x n ≤ ∫ x ∈ R + n : 0 < x n < a ‖ P θ , λ ‖ L s ( ∂ R + n ) r ‖ f ‖ L p ( ∂ R + n ) r d x n = ‖ f ‖ L p ( ∂ R + n ) r ∫ 0 a ∫ ∂ R + n x n θ s ( ∣ x ′ ∣ 2 + x n 2 ) λ s 2 d x ′ r s d x n = ‖ f ‖ L p ( ∂ R + n ) r ∫ 0 a x n ( θ − λ ) r + ( n − 1 ) r s d x n ∫ ∂ R + n 1 ( ∣ x ′ ∣ 2 + 1 ) λ s 2 d x ′ r s ≤ C n , λ , θ , r , s ‖ f ‖ L p ( ∂ R + n ) r a ( θ − λ ) r + ( n − 1 ) r s + 1 .

For any t > 0 , applying formula (2.2) and (2.3), we have

∣ { x ∈ ∂ R + n : P ( f ) ( x ) > t } ∣ = ∣ { x ∈ ∂ R + n : 0 < x n < C n , λ , p ′ ‖ f ‖ L p ( ∂ R + n ) ⋅ 1 t p ′ ( λ − θ ) p ′ − ( n − 1 ) , P ( f ) ( x ) > t } ∣ ≤ ∫ x ∈ ∂ R + n : 0 < x n < C n , λ , p ′ ‖ f ‖ L p ( ∂ R + n ) ⋅ 1 t p ′ ( λ − θ ) p ′ − ( n − 1 ) P f ( x ) t r d x ≤ C n , λ , p , q ′ ‖ f ‖ L p ( ∂ R + n ) q ⋅ 1 t q .

Therefore, for 1 < p , q < ∞ , we have

‖ P ( f ) ( x ) ‖ L w q ( R + n ) ≤ C n , p , q ′ , λ ‖ f ‖ L p ( ∂ R + n ) ,

where ‖ P ( f ) ( x ) ‖ L w q ( R + n ) is the weak L q norm of P ( f ) ( x ) defined as:

sup t > 0 t ∣ { x ∈ ∂ R + n : P ( f ) ( x ) > t } ∣ 1 q .

According to the Marcinkiewicz interpolation theorem, we can deduce that the inequality (2.1) holds for 1 < p , q < ∞ .□

3 The proof of Theorem 1.2

Throughout this section, we shall establish Stein-Weiss inequality with an extended Poisson kernel. We first obtain the following notation.

B R ( x ) = { y ∈ R n : ∣ y − x ∣ < R , x ∈ R n } , B R n − 1 ( x ) = { y ∈ ∂ R + n : ∣ y − x ∣ < R , x ∈ ∂ R + n } , B R + ( x ) = { y = ( y 1 , y 2 , … , y n ) ∈ B R ( x ) : y n > 0 , x ∈ ∂ R + n } .

For x = 0 , we write B R = B R ( 0 ) , B R n − 1 = B R n − 1 ( 0 ) , B R + = B R + ( 0 ) .

We need the following weighted Hardy inequality on the upper half space which established in [19]. The idea of proving Lemma 3.1 is from that of weighted Hardy inequality on whole space proved by Drábek et al. [23].

Lemma 3.1

Let W ( x ) and U ( y ) be nonnegative locally integrable functions defined on R + n and ∂ R + n , respectively, for 1 < p ≤ q < ∞ and f ≥ 0 on ∂ R + n ,

(3.1) ∫ R + n W ( x ) ∫ B ∣ x ∣ n − 1 f ( y ) d y q d x 1 q ≤ C 0 ( p , q ) ∫ ∂ R + n f p ( y ) U ( y ) d y 1 p ,

holds if and only if

(3.2) A 0 = sup R > 0 ∫ ∣ x ∣ ≥ R W ( x ) d x 1 q ∫ ∣ y ∣ ≤ R U 1 − p ′ ( y ) d y 1 p ′ < ∞ .

On the other hand,

(3.3) ∫ R + n W ( x ) ∫ ∂ R + n \ B ∣ x ∣ n − 1 f ( y ) d y q d x 1 q ≤ C 0 ( p , q ) ∫ ∂ R + n f p ( y ) U ( y ) d y 1 p ,

holds if and only if

(3.4) A 1 = sup R > 0 ∫ ∣ x ∣ ≤ R W ( x ) d x 1 q ∫ ∣ y ∣ ≥ R U 1 − p ′ ( y ) d y 1 p ′ < ∞ .

We now continue with the proof of the Theorem 1.2.

Proof

Without loss of generality, we may assume that f is nonnegative. Since q > 1 , then

‖ P ( f ) ∣ x ∣ − β ‖ L q ( R + n ) q ≲ P 1 + P 2 + P 3 ,

where

P 1 = ∫ R + n ∣ x ∣ − β ∫ B ∣ x ∣ ∕ 2 n − 1 x n θ f ( y ) ∣ x − y ∣ λ d y q d x , P 2 = ∫ R + n ∣ x ∣ − β ∫ B 2 ∣ x ∣ n − 1 \ B ∣ x ∣ ∕ 2 n − 1 x n θ f ( y ) ∣ x − y ∣ λ d y q d x , P 3 = ∫ R + n ∣ x ∣ − β ∫ ∂ R + n \ B 2 ∣ x ∣ n − 1 x n θ f ( y ) ∣ x − y ∣ λ d y q d x .

Thus, we just need to show

P j ≤ C n , α , β , p , q ′ , λ ‖ f ∣ y ∣ α ‖ L p ( ∂ R n + ) q , j = 1 , 2 , 3 .

First, let us examine P 1 . Since ∣ y ∣ ≤ ∣ x ∣ 2 in this case, we derive that

(3.5) P 1 ≤ C λ ∫ R + n ∣ x ∣ − β q − ( λ − θ ) q ∫ B ∣ x ∣ ∕ 2 n − 1 f ( y ) d y q d x .

Take W ( x ) = ∣ x ∣ − β q − ( λ − θ ) q and U ( y ) = ∣ y ∣ α p in (3.1). If W ( x ) and U ( y ) satisfy (3.2), we conclude that

P 1 ≤ C n , α , β , p , q ′ , λ ‖ f ∣ y ∣ α ‖ L p ( ∂ R n + ) q .

Indeed, since α < n − 1 p ′ , then for any R > 0 , we can derive

(3.6) ∫ ∣ x ∣ ≥ R W ( x ) d x = ∫ ∣ x ∣ ≥ R ∣ x ∣ − β q − ( λ − θ ) q d x = ∫ ∂ B 1 + d y ∫ R ∞ t − β q − ( λ − θ ) q + n − 1 d t = C n , λ , β , q , θ R − β q − ( λ − θ ) q + n

and

(3.7) ∫ ∣ y ∣ ≤ R U 1 − p ′ ( y ) d y = ∫ ∣ y ∣ ≤ R ( ∣ y ∣ α p ) 1 − p ′ d y = ∫ S n − 2 d η ∫ 0 R r α p ( 1 − p ′ ) + n − 2 d r = C n , λ , α , p R α p ( 1 − p ′ ) + n − 1 .

Combining (1.9), (3.6), and (3.7), we derive that

∫ ∣ x ∣ ≥ R W ( x ) d x 1 q ∫ ∣ y ∣ ≤ R U 1 − p ′ ( y ) d y 1 p ′ < C n , α , β , λ , p R − β − ( λ − θ ) + n q + α p ( 1 − p ′ ) + n − 1 p ′ = C n , α , β , λ , p .

Next we estimate P 3 . Since ∣ y ∣ ≥ 2 ∣ x ∣ in this case, it follows that ∣ y − x ∣ ≥ ∣ y ∣ 2 . Taking W ( x ) = ∣ x ∣ ( − β + θ ) q and U ( y ) = ∣ y ∣ ( λ + α ) p in (3.3), we obtain

P 3 ≤ C λ ∫ R + n ∣ x ∣ ( − β + θ ) q ∫ ∂ R + n \ B 2 ∣ x ∣ n − 1 f ( y ) ∣ y ∣ − λ d y q d x ≤ C n , α , β , λ , p ‖ f ∣ y ∣ α ‖ L p ( ∂ R n + ) q

if the condition (3.4) is satisfied. In fact, since β < n q + θ , then for any R > 0 , there holds

(3.8) ∫ ∣ x ∣ ≤ R W ( x ) d x = ∫ ∣ x ∣ ≤ R ∣ x ∣ ( − β + θ ) q d x = ∫ ∂ B 1 + d y ∫ 0 R t ( − β + θ ) q + n − 1 d t = C n , λ , β , q , θ R ( − β + θ ) q + n ,

(3.9) ∫ ∣ y ∣ ≥ R U 1 − p ′ ( y ) d y = ∫ ∣ y ∣ ≥ R ( ∣ y ∣ ( λ + α ) p ) 1 − p ′ d y = ∫ S n − 2 d η ∫ R ∞ r ( λ + α ) p ( 1 − p ′ ) + n − 2 d r = C n , λ , α , p R ( λ + α ) p ( 1 − p ′ ) + n − 1 .

Combining (3.8) and (3.9), we verify that condition (3.4) holds.

We are left to estimate P 2 . Since ∣ x ∣ 2 < ∣ y ∣ < 2 ∣ x ∣ and α + β ≥ 0 , it is easy to check

∣ x − y ∣ α + β < 3 α + β ∣ y ∣ α + β ≤ 3 α + β 2 β ∣ x ∣ β ∣ y ∣ α .

By using the HLS inequality with an extended kernel (1.8), we can derive that

P 2 = ∫ R + n ∣ x ∣ − β ∫ B 2 ∣ x ∣ n − 1 \ B ∣ x ∣ ∕ 2 n − 1 x n θ f ( y ) ∣ x − y ∣ λ d y q d x ≤ C α , β ∫ R + n ∫ B 2 ∣ x ∣ n − 1 \ B ∣ x ∣ ∕ 2 n − 1 x n θ f ( y ) ∣ y ∣ α ∣ x − y ∣ α + β + λ d y q d x ≤ C α , β ∫ R + n ∫ ∂ R + n x n θ f ( y ) ∣ y ∣ α ∣ x − y ∣ α + β + λ d y q d x . ≤ C n , α , β , p , q ′ , λ ‖ f ∣ y ∣ α ‖ L p ( ∂ R + n ) q .

Thus, we accomplish the proof of Theorem 1.2.□

4 The proof of Theorem 1.3

Proof

The condition “ α + β ≥ 0 ” implies that one of α and β is nonnegative. Without loss of generality, we can assume that β ≥ 0 . We further assume that { f i } is a nonnegative maximizing sequence for problem (1.12), that is,

‖ f i ∣ y ∣ α ‖ L p ( ∂ R + n ) = 1 and lim i → ∞ ‖ P ( f i ) ∣ x ∣ − β ‖ L q ( R + n ) = C n , α , β , p , q ′ , λ .

Then there exist { r i } > 0 such that

∫ B r i n − 1 ∣ f i ( y ) ∣ p ∣ y ∣ p α d y = 1 2 .

Let f i ˜ ( y ) = r i α p + n − 1 p f i ( r i y ) , we can calculate that

‖ f i ˜ ∣ y ∣ α ‖ L P ( ∂ R + n ) = 1 and lim i → ∞ ‖ P ( f i ˜ ) ∣ x ∣ − β ‖ L q ( R + n ) = C n , α , β , p , q ′ , λ .

Therefore, { f i ˜ } is also a nonnegative maximizing sequence, which satisfies

(4.1) ∫ B 1 n − 1 ∣ f i ˜ ( y ) ∣ p ∣ y ∣ p α d y = 1 2 .

We still denote the new sequence { f i ˜ } by { f i } . We divide the proof into three steps.

Step 1: We first use the following Lions’ first concentration compactness lemma (see [39,40]) to prove that { ∣ f i ( y ) ∣ p ∣ y ∣ p α } is tight in ∂ R + n , that is, for any ε > 0 , there exists a R ε > 0 such that

∫ ∂ R + n \ B R ε n − 1 ( 0 ) ∣ f i ( y ) ∣ p ∣ y ∣ p α d y ≤ ε .

Lemma 4.1

Let { ∣ f i ( y ) ∣ p ∣ y ∣ p α } be a nonnegative sequence satisfying:

∫ ∂ R + n ∣ f i ( y ) ∣ p ∣ y ∣ p α d y = m ,

where m > 0 is fixed. Then there exists a subsequence still denoted by { ∣ f i ( y ) ∣ p ∣ y ∣ p α } such that one of the following conditions holds:

  1. (Compactness) There exists { y i } in ∂ R + n such that for all ε > 0 , there exists R ε > 0 such that

    ∫ B R ε n − 1 ( y i ) ∣ f i ( y ) ∣ p ∣ y ∣ p α d y ≥ m − ε for a l l i ;

  2. (Vanishing) For all R > 0 , there holds:

    lim i → ∞ sup y ∈ ∂ R + n ∫ B R n − 1 ( y ) ∣ f i ( y ) ∣ p ∣ y ∣ p α d y = 0 ;

  3. (Dichotomy) There exists a k ∈ ( 0 , m ) such that for all ε > 0 , there exist R large enough, a point sequence { y i } and a sequence { R i } satisfying R i → + ∞ as i → ∞ such that

    ∣ f i 1 ( y ) ∣ = ∣ f i ( y ) ∣ χ B R n − 1 ( y i ) , ∣ f i 2 ( y ) ∣ = ∣ f i ( y ) ∣ χ ∂ R + n \ B R i n − 1 ( y i ) ,

    limsup i → ∞ ∣ k − ∫ ∂ R + n ∣ f i 1 ( y ) ∣ p ∣ y ∣ p α d y ∣ + ∣ ( m − k ) − ∫ ∂ R + n ∣ f i 2 ( y ) ∣ p ∣ y ∣ p α d y ∣ ≤ ε .

For the maximizing sequence { f i } , we take m = 1. Since { f i } satisfies (4.1), case (b) cannot occur. Now we exclude dichotomy. If case (c) occurs, we have

(4.2) lim i → ∞ ‖ f i 1 ∣ y ∣ α ‖ L p ( ∂ R + n ) p = k , lim i → ∞ ‖ f i 2 ∣ y ∣ α ‖ L p ( ∂ R + n ) p = 1 − k .

We claim that

(4.3) lim i → ∞ ‖ P ( f i ) ∣ x ∣ − β ‖ L q ( R + n ) q = lim i → ∞ ‖ P ( f i 1 ) ∣ x ∣ − β ‖ L q ( R + n ) q + lim i → ∞ ‖ P ( f i 2 ) ∣ x ∣ − β ‖ L q ( R + n ) q ,

then

lim i → ∞ ‖ P ( f i ) ∣ x ∣ − β ‖ L q ( R + n ) q ≤ lim i → ∞ C n , α , β , p , q ′ , λ q ‖ f i 1 ∣ y ∣ α ‖ L p ( ∂ R + n ) q + lim i → ∞ C n , α , β , p , q ′ , λ q ‖ f i 2 ∣ y ∣ α ‖ L p ( ∂ R + n ) q ≤ C n , α , β , p , q ′ , λ q k q p + C n , α , β , p , q ′ , λ q ( 1 − k ) q p < C n , α , β , p , q ′ , λ q ,

which contradicts that

lim i → ∞ ‖ P ( f i ) ∣ x ∣ − β ‖ L q ( R + n ) q = C n , α , β , p , q ′ , λ q .

Now we prove formula (4.3) in detail. Let ∣ f i 3 ( y ) ∣ = ∣ f i ( y ) ∣ χ B R i n − 1 ( y i ) \ B R n − 1 ( y i ) , then

lim i → ∞ ‖ f i 3 ∣ y ∣ α ‖ L p ( ∂ R + n ) = 0 .

According to the inequality

(4.4) ( a + b ) q ≤ a q + b q + 2 q − 1 q ( a q − 1 b + a b q − 1 ) ,

where q > 1 , we can obtain

lim i → ∞ ‖ P ( f i ) ∣ x ∣ − β ‖ L q ( R + n ) q = lim i → ∞ ‖ P ( f i 1 ) ∣ x ∣ − β + P ( f i 2 ) ∣ x ∣ − β + P ( f i 3 ) ∣ x ∣ − β ‖ L q ( R + n ) q ≤ lim i → ∞ ∫ R + n ( P ( f i 1 ) ∣ x ∣ − β + P ( f i 2 ) ∣ x ∣ − β ) q + ( P ( f i 3 ) ∣ x ∣ − β ) q + 2 q − 1 q ( P ( f i 1 ) ∣ x ∣ − β + P ( f i 2 ) ∣ x ∣ − β ) ( P ( f i 3 ) ∣ x ∣ − β ) q − 1 + 2 q − 1 q ( P ( f i 1 ) ∣ x ∣ − β + P ( f i 2 ) ∣ x ∣ − β ) q − 1 ( P ( f i 3 ) ∣ x ∣ − β ) d x ≤ lim i → ∞ ( ‖ P ( f i 1 ) ∣ x ∣ − β + P ( f i 2 ) ∣ x ∣ − β ‖ L q ( R + n ) q + C n , α , β , p , q ′ q ‖ f i 3 ∣ y ∣ α ‖ L p ( ∂ R + n ) q + C q ‖ P ( f i 1 ) ∣ x ∣ − β + P ( f i 2 ) ∣ x ∣ − β ‖ L q ( R + n ) ‖ f i 3 ∣ y ∣ α ‖ L p ( ∂ R + n ) q − 1 + C q ‖ P ( f i 1 ) ∣ x ∣ − β + P ( f i 2 ) ∣ x ∣ − β ‖ L q ( R + n ) q − 1 ‖ f i 3 ∣ y ∣ α ‖ L p ( ∂ R + n ) ) = lim i → ∞ ‖ P ( f i 1 ) ∣ x ∣ − β + P ( f i 2 ) ∣ x ∣ − β ‖ L q ( R + n ) q .

By inequality (4.4) again, we have

lim i → ∞ ‖ P ( f i 1 ) ∣ x ∣ − β + P ( f i 2 ) ∣ x ∣ − β ‖ L q ( R + n ) q ≤ lim i → ∞ ∫ R + n ( P ( f i 1 ) ∣ x ∣ − β ) q + ( P ( f i 2 ) ∣ x ∣ − β ) q + 2 q − 1 q ( P ( f i 1 ) ∣ x ∣ − β ) q − 1 ( P ( f i 2 ) ∣ x ∣ − β ) + 2 q − 1 q ( P ( f i 1 ) ∣ x ∣ − β ) ( P ( f i 2 ) ∣ x ∣ − β ) q − 1 d x .

Once we prove that

lim i → ∞ ∫ R + n 2 q − 1 q ( P ( f i 1 ) ∣ x ∣ − β ) q − 1 ( P ( f i 2 ) ∣ x ∣ − β ) + 2 q − 1 q ( P ( f i 1 ) ∣ x ∣ − β ) ( P ( f i 2 ) ∣ x ∣ − β ) q − 1 d x = 0 ,

we can obtain the claim (4.3). Without loss of generality, we just prove that the first term ∫ R + n ( P ( f i 1 ) ∣ x ∣ − β ) q − 1 ( P ( f i 2 ) ∣ x ∣ − β ) d x → 0 as i → ∞ , since the second term can be estimated by the same way. We divide the integral into three parts:

lim i → ∞ ∫ R + n ( P ( f i 1 ) ∣ x ∣ − β ) q − 1 ( P ( f i 2 ) ∣ x ∣ − β ) d x = lim i → ∞ ∫ B R + ( P ( f i 1 ) ∣ x ∣ − β ) q − 1 ( P ( f i 2 ) ∣ x ∣ − β ) d x + lim i → ∞ ∫ B ( R i + R ) ∕ 2 + \ B R + ( P ( f i 1 ) ∣ x ∣ − β ) q − 1 ( P ( f i 2 ) ∣ x ∣ − β ) d x + lim i → ∞ ∫ R + n \ B ( R i + R ) ∕ 2 + ( P ( f i 1 ) ∣ x ∣ − β ) q − 1 ( P ( f i 2 ) ∣ x ∣ − β ) d x = I + I I + I I I .

We first estimate I . Choosing small enough ε , by Hölder’s inequality and inequality (1.10), we have

I ≤ lim i → ∞ ‖ P ( f i 1 ) ∣ x ∣ − β ‖ L q ( B R + ) q − 1 ‖ P ( f i 2 ) ∣ x ∣ − β ‖ L q ( B R + ) ≤ lim i → ∞ C n , α , β , p , q ′ , λ q − 1 ∫ B R + ∫ ∂ R + n \ B R i n − 1 f i x n θ ∣ x − y ∣ λ d y q ∣ x ∣ − β q d x 1 q ≤ lim i → ∞ C n , α , β , p , q ′ , λ q − 1 R ε ∣ R i ∣ ε ∫ B R + ∫ ∂ R + n \ B R i n − 1 f i x n θ ∣ x − y ∣ λ − ε d y q ∣ x ∣ − ( β + ε ) q d x 1 q ≤ lim i → ∞ C n , α , β , p , q ′ , λ q − 1 R ε ∣ R i ∣ ε ‖ f i ∣ y ∣ α ‖ L p ( ∂ R + n ) = 0 .

For the estimate of I I I , we also can derive

I I I ≤ lim i → ∞ ‖ P ( f i 1 ) ∣ x ∣ − β ‖ L q ( R + n \ B ( R i + R ) ∕ 2 + ) q − 1 ‖ P ( f i 2 ) ∣ x ∣ − β ‖ L q ( R + n \ B ( R i + R ) ∕ 2 + ) ≤ lim i → ∞ C n , α , β , p , q ′ , λ ∫ R + n \ B ( R i + R ) ∕ 2 + ∫ B R n − 1 f i x n θ ∣ x − y ∣ λ d y q ∣ x ∣ − β q d x q − 1 q ≤ lim i → ∞ C n , α , β , p , q ′ , λ 1 R i + R 2 ε ( q − 1 ) ∫ R + n \ B ( R i + R ) ∕ 2 + ∫ B R n − 1 f i x n θ ∣ x − y ∣ λ − ε d y q ∣ x ∣ − β q d x q − 1 q ≤ lim i → ∞ C n , α , β , p , q ′ , λ 1 R i + R 2 ε ( q − 1 ) ‖ f i χ B R n − 1 ∣ y ∣ α + ε ‖ L p ( ∂ R + n ) q − 1 = 0 .

Now we turn to estimate I I . For small enough ε , we can find r > q satisfying the following identities:

1 r = n − 1 n 1 p + α + β + λ − ε − θ + 1 n − 1 ,

1 ( q − 1 ) r ′ = n − 1 n 1 p + α + ε + β + λ − θ + 1 n − 1 .

Then

I I ≤ lim i → ∞ ‖ P ( f i 1 ) ∣ x ∣ − β ‖ L ( q − 1 ) r ′ ( B ( R i + R ) ∕ 2 + \ B R + ) q − 1 ‖ P ( f i 2 ) ∣ x ∣ − β ‖ L r ( B ( R i + R ) ∕ 2 + \ B R + ) ≤ lim i → ∞ C n , α , β , p , q ′ , λ ‖ f i χ B R n − 1 ∣ y ∣ α + ε ‖ L p ( ∂ R + n ) q − 1 ∫ B ( R i + R ) ∕ 2 + \ B R + ∫ ∂ R + n \ B R i n − 1 f i x n θ ∣ x − y ∣ λ d y r ∣ x ∣ − β r d x 1 r ≤ lim i → ∞ C n , α , β , p , q ′ , λ , ε , R 1 ∣ R i ∣ ε ∫ B ( R i + R ) ∕ 2 + \ B R + ∫ ∂ R + n \ B R i n − 1 f i x n θ ∣ x − y ∣ λ − ε d y r ∣ x ∣ − β r d x 1 r ≤ lim i → ∞ C n , α , β , p , q ′ , λ , ε , R 1 ∣ R i ∣ ε ‖ f i ∣ y ∣ α ‖ L p ( ∂ R + n ) = 0 .

Until now, we exclude vanishing and dichotomy for sequence { f i } . Therefore, compactness (a) hold for it. We claim that { y i } is bounded. If { y i } is not bounded, for small enough ε , we can find large enough y i such that B 1 n − 1 ⊂ ∂ R + n \ B R ε n − 1 ( y i ) . Then there holds

∫ B 1 n − 1 ∣ f i ( y ) ∣ p ∣ y ∣ p α d y ≤ ∫ ∂ R + n \ B R ε n − 1 ( y i ) ∣ f i ( y ) ∣ p ∣ y ∣ p α d y ≤ ε ,

which is contradict to (4.1). Therefore, { ∣ f i ( y ) ∣ p ∣ y ∣ p α } is tight.

Step 2: Assume that f i ( y ) ∣ y ∣ α → f ( y ) ∣ y ∣ α weakly in L p ( ∂ R + n ) . Next we prove that { ∣ P ( f i ) ∣ q ∣ x ∣ − β q } is also tight and P ( f i ) ( x ) ∣ x ∣ − β → P ( f ) ( x ) ∣ x ∣ − β a.e. on R + n .

For R < M < + ∞ , we have

‖ P ( f i ) ∣ x ∣ − β ‖ L q ( R + n \ B M + ) ≤ ‖ P ( f i χ ∂ R + n \ B R n − 1 ) ∣ x ∣ − β ‖ L q ( R + n \ B M + ) + ‖ P ( f i χ B R n − 1 ) ∣ x ∣ − β ‖ L q ( R + n \ B M + ) ≤ C n , α , β , p , q ′ , λ ‖ f i χ ∂ R + n \ B R n − 1 ∣ y ∣ α ‖ L p ( ∂ R + n ) + ‖ f i χ B R n − 1 ‖ L 1 ( ∂ R + n ) ⋅ ∫ R + n \ B M + ∣ x ∣ ( θ − β ) q ( ∣ x ∣ − R ) λ q d x 1 q ≤ ε ( R ) + ‖ f i ( y ) ∣ y ∣ α ‖ L p ( ∂ R + n ) ⋅ ‖ ∣ y ∣ − α χ B R n − 1 ‖ L p ′ ( ∂ R + n ) ⋅ ∫ R + n \ B M + ∣ x ∣ ( θ − β ) q ( ∣ x ∣ − R ) λ q d x 1 q ≤ ε ( R ) + C n , α , p ∫ R + n \ B M + ∣ x ∣ ( θ − β ) q ( ∣ x ∣ − R ) λ q d x 1 q ≤ ε ( R ) + δ R ( M ) ,

where ε ( R ) → 0 as R → + ∞ and δ R ( M ) → 0 as M → + ∞ for any fixed R < + ∞ . This show the tightness of { ∣ P ( f i ) ∣ q ∣ x ∣ − β q } .

To prove the pointwise convergence of P ( f i ) ∣ x ∣ − β q , we just need to show that P ( f i ) ∣ x ∣ − β q → P ( f ) ∣ x ∣ − β q in measure. For any k > 0 ,

∣ { ∣ P ( f i ) ∣ x ∣ − β − P ( f ) ∣ x ∣ − β ∣ ≥ 15 k } ∣ ≤ ∣ { ∣ P ( f i ) ∣ x ∣ − β − P ( f i ) ∣ x ∣ − β χ B M + ( x ) ∣ ≥ 5 k } ∣ + ∣ { ∣ P ( f i ) ∣ x ∣ − β χ B M + ( x ) − P ( f ) ∣ x ∣ − β χ B M + ( x ) ∣ ≥ 5 k } ∣ + ∣ { ∣ P ( f ) ∣ x ∣ − β χ B M + ( x ) − P ( f ) ∣ x ∣ − β ∣ ≥ 5 k } ∣ = K 1 + K 2 + K 3 .

According to the tightness of { ∣ P ( f i ) ∣ q ∣ x ∣ − β q } , we derive that

‖ P ( f i ) ∣ x ∣ − β − P ( f i ) ∣ x ∣ − β χ B M + ( x ) ‖ L q ( R + n ) ≤ ε ( M ) .

Observing that

‖ P ( f ) ∣ x ∣ − β ‖ L q ( R + n ) ≤ C n , α , β , p , q ′ , λ ‖ f ∣ y ∣ α ‖ L p ( ∂ R + n ) ≤ C n , α , β , p , q ′ , λ liminf i → ∞ ‖ f i ∣ y ∣ α ‖ L p ( ∂ R + n ) = C n , α , β , p , q ′ , λ ,

we also have the following inequality:

‖ P ( f ) ∣ x ∣ − β − P ( f ) ∣ x ∣ − β χ B M + ( x ) ‖ L q ( R + n ) ≤ ε ( M ) .

Therefore, K 1 , K 3 → 0 as M → + ∞ . We just need to estimate K 2 .

Denoting by x = ( x ′ , x n ) and

P η ( f ) ( x ) = ∫ ∂ R + n \ B η n − 1 ( x ′ ) f ( y ) x n θ ∣ x − y ∣ λ d y ,

we have

K 2 ≤ ∣ { ∣ P ( f i ) ∣ x ∣ − β χ B M + ( x ) − P ( f i χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) ∣ ≥ k } ∣ + ∣ { ∣ P ( f i χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) − P η ( f i χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) ∣ ≥ k } ∣ + ∣ { ∣ P η ( f i χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) − P η ( f χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) ∣ ≥ k } ∣ + ∣ { ∣ P η ( f χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) − P ( f χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) ∣ ≥ k } ∣ + ∣ { ∣ P ( f χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) − P ( f ) ∣ x ∣ − β χ B M + ( x ) ∣ ≥ k } ∣ = K 2 1 + K 2 2 + K 2 3 + K 2 4 + K 2 5 .

Since

‖ P ( f i ) ∣ x ∣ − β − P ( f i χ B R n − 1 ( y ) ) ∣ x ∣ − β ‖ L q ( R + n ) ≤ C n , α , β , p , q ′ , λ ‖ f i χ ∂ R + n \ B R n − 1 ( y ) ∣ y ∣ α ‖ L p ( ∂ R + n ) ≤ ε ( R )

and

‖ P ( f χ B R n − 1 ( y ) ) ∣ x ∣ − β − P ( f ) ∣ x ∣ − β ‖ L q ( R + n ) ≤ C n , α , β , p , q ′ , λ ‖ f χ ∂ R + n \ B R n − 1 ( y ) ∣ y ∣ α ‖ L p ( ∂ R + n ) ≤ ε ( R ) ,

so K 2 1 , K 2 5 → 0 as R → ∞ . Note that

∫ ( ∂ R + n \ B η n − 1 ( x ′ ) ) ∩ B R n − 1 ∣ y ∣ − α ∣ x − y ∣ λ p ′ d y < ∞

for fixed x ∈ R + n . This yields P η ( f i χ B R n − 1 ( y ) ) ∣ x ∣ − β → P η ( f χ B R n − 1 ( y ) ) ∣ x ∣ − β as i → ∞ . Furthermore, we have P η ( f i χ B R n − 1 ( y ) ) ∣ x ∣ − β → P η ( f χ B R n − 1 ( y ) ) ∣ x ∣ − β as i → ∞ locally in measure. Thus, K 2 3 → 0 as i → ∞ .

As for the K 2 2 , we have that

‖ P ( f i χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) − P η ( f i χ B R n − 1 ( y ) ) ∣ x ∣ − β χ B M + ( x ) ‖ L 1 ( R + n ) = ∫ B M + ∫ B η n − 1 ( x ′ ) f i χ B R n − 1 ( y ) x n θ ∣ x − y ∣ λ ∣ x ∣ β d y d x ≤ ∫ 0 M ∫ ∂ R + n ∫ B η n − 1 ( y ) f i χ B R n − 1 ( y ) x n θ ∣ x − y ∣ λ ∣ x ∣ β d x ′ d y d x n = Λ ( η ) .

For the case α ≥ 0 , choosing small enough ε , we have λ − ε + β < λ + α + β ≤ n − 1 + θ . Then we can derive the following estimate:

∫ B η n − 1 ( y ) x n θ ∣ x − y ∣ λ − ε ∣ x ∣ β d x ′ = ∫ B η n − 1 ( y ) ∩ { ∣ x ′ − y ∣ > ∣ x ′ ∣ } x n θ ∣ x − y ∣ λ − ε ∣ x ∣ β d x ′ + ∫ B η n − 1 ( y ) ∩ { ∣ x ′ − y ∣ < ∣ x ′ ∣ } x n θ ∣ x − y ∣ λ − ε ∣ x ∣ β d x ′ ≤ ∫ B η n − 1 ( y ) { ∩ ∣ x ′ − y ∣ > ∣ x ′ ∣ } ∣ x ∣ θ ∣ x ∣ λ − ε + β d x ′ + ∫ B η n − 1 ( y ) ∩ { ∣ x ′ − y ∣ < ∣ x ′ ∣ } ( ∣ x ′ − y ∣ 2 + ∣ x n ∣ 2 ) θ 2 ∣ x − y ∣ λ − ε + β d x ′ ≤ ∫ B η n − 1 ( y ) 1 ∣ x ′ ∣ λ − ε + β − θ d x ′ + ∫ B η n − 1 ( y ) 1 ∣ x − y ∣ λ − ε + β − θ d x ′ ≤ C n η n − 1 − ( λ − ε + β − θ ) .

By applying the aforementioned estimate, we can derive that

Λ ( η ) = ∫ 0 M ∫ ∂ R + n ∫ B η n − 1 ( y ) f i χ B R n − 1 ( y ) x n θ ∣ x − y ∣ λ ∣ x ∣ β d x ′ d y d x n ≤ ∫ 0 M ∫ ∂ R + n ∫ B η n − 1 ( y ) f i χ B R n − 1 ( y ) x n θ ∣ x − y ∣ λ − ε ∣ x ∣ β x n ε d x ′ d y d x n ≤ ∫ 0 M ‖ f i χ B R n − 1 ‖ L 1 ( ∂ R + n ) ∫ B η n − 1 ( y ) x n θ ∣ x − y ∣ λ − ε ∣ x ∣ β d x ′ L ∞ ( ∂ R + n ) ⋅ 1 x n ε d x n ≤ M 1 − ε C n , α , p , ε , R ⋅ η n − 1 − ( λ − ε + β − θ ) → 0

as η → 0 for fixed M and R .

For the case α < 0 and α + β > 0 , according to the condition of Theorem 1.3, we can obtain that α + λ > n − 1 p ′ > 0 . For sufficiently small ε > 0 , obviously α < n − 1 ( 1 + ε ) ′ . Choose t ′ > 1 such that n − 1 n ( 1 + ε ) + 1 t ′ + λ + α + β + 1 n = 2 . This together with α + β > 0 yields that λ < n t if ε is sufficiently small. Then it follows from Stein-Weiss inequality with the extended Poisson kernel ( θ = 0 ) that

(4.5) Λ ( η ) = ∫ 0 M ∫ ∂ R + n ∫ B η n − 1 ( y ) f i χ B R n − 1 ( y ) x n θ ∣ y ∣ α ∣ y ∣ α ∣ x − y ∣ λ ∣ x ∣ β d x ′ d y d x n ≤ M θ ∫ ∂ R + n ∫ 0 M ∫ B η n − 1 ( 0 ) f i χ B R n − 1 ( y ) ∣ y ∣ α ∣ y ∣ α ∣ x ∣ λ ∣ y − x ∣ β d x ′ d x n d y ≤ M θ ‖ f i χ B R n − 1 ∣ y ∣ α ‖ L 1 + ε ( ∂ R + n ) ‖ χ B η n − 1 ( 0 ) × [ 0 , M ] ( x ) ‖ L t ′ ( R + n ) ≤ M θ C n , α , p , θ , ε η n − 1 t ′ → 0

as η → 0 for fixed M and R .

For the case α < 0 and α + β = 0 , we just need do a perturbation for α in (4.5). Assume τ is small enough such that α ′ = α + τ < 0 . It is easy to check that Λ ( η ) → 0 from the process of (4.5). Therefore, K 2 2 → 0 as η → 0 . By the same argument as K 2 2 , we can obtain K 2 4 → 0 as η → 0 . By combining the estimate of K 2 1 , K 2 2 , K 2 3 , K 2 4 and K 2 5 , we deduce that P ( f i ) ( x ) ∣ x ∣ − β → P ( f ) ( x ) ∣ x ∣ − β a.e. on R + n .

Step 3: Finally, we prove that f is the maximum of problem (1.12). We follow Lions’ second concentration compactness lemma [33,41,42].

Lemma 4.2

Let f i ( y ) ∣ y ∣ α → f ( y ) ∣ y ∣ α weakly in L p ( ∂ R + n ) and assume { ∣ f i ∣ p ∣ y ∣ p α } is tight. We further assume { ∣ P ( f i ) ∣ q ∣ x ∣ − q β } , { ∣ f i ∣ p ∣ y ∣ p α } converge weakly in the sense of measure to some bounded nonnegative measures ν , μ , respectively, on R + n , ∂ R + n , then there exists some at most countable set of points { y i } i ∈ I ⊂ ∂ R + n such that

ν = ∣ P ( f ) ∣ q ∣ x ∣ − q β + ∑ i ∈ I v i δ y i

and

μ ≥ ∣ f ∣ p ∣ y ∣ p α + ∑ i ∈ I u i δ y i ,

where v i = ν ( y i ) , u i = μ ( y i ) . Furthermore, we also have v i ≤ C n , α , β , p , q ′ , λ q ( u i ) q p .

We claim that μ ( ∂ R + n ) = 1 . By the lower semi-continuity of the measure, there holds

∫ ∂ R + n d μ ≤ ∫ ∂ R + n ∣ f i ∣ p ∣ y ∣ p α d y = 1 .

On the other hand, for any ε > 0 , there exist R ε > 0 and ϕ ∈ C c ∞ ( ∂ R + n ) with ϕ ∣ B R ε n − 1 = 1 , 0 ≤ ϕ ≤ 1 such that

(4.6) ∫ ∂ R + n d μ ≥ ∫ ∂ R + n ϕ ( y ) d μ = lim i → ∞ ∫ ∂ R + n ∣ f i ∣ p ∣ y ∣ p α ϕ ( y ) d y ≥ lim i → ∞ ∫ B R ε n − 1 ∣ f i ∣ p ∣ y ∣ p α d y ≥ 1 − ε ,

where we have used the tightness of { ∣ f i ∣ p ∣ y ∣ p α } in the last step. Let ε → 0 , we have μ ( ∂ R + n ) ≥ 1 . Since { ∣ P ( f i ) ∣ q ∣ x ∣ − q β } is also tight, we can similarly derive that ν ( R + n ) = C n , α , β , p , q ′ , λ as the aforementioned arguments.

Now we are prepared to prove that ∫ ∂ R + n ∣ f ∣ p ∣ y ∣ p α d y = 1 . By Lemma 4.2, we derive that

(4.7) C n , α , β , p , q ′ , λ = ν ( R + n ) = ‖ P ( f ) ∣ x ∣ − β ‖ L q ( R + n ) q + ∑ i ∈ I v i ≤ C n , α , β , p , q ′ , λ q ‖ f ∣ y ∣ α ‖ L p ( ∂ R + n ) q + ∑ i ∈ I C n , α , β , p , q ′ q ( u i ) q p ≤ C n , α , β , p , q ′ , λ q k q p + C n , α , β , p , q ′ , λ q ∑ i ∈ I u i q p ≤ C n , α , β , p , q ′ , λ q k q p + C n , α , β , p , q ′ , λ q ( 1 − k ) q p , ≤ C n , α , β , p , q ′ , λ q .

Then the aforementioned inequalities must be equalities since p is strictly smaller than q , which is implied by α + β + λ ≤ n − 1 + θ . If ∫ ∂ R + n ∣ f ∣ p ∣ y ∣ p α d y = k < 1 , we must have

(4.8) f = 0 , μ = δ y 0 , ν = C n , α , β , p , q ′ , λ q δ y 0 .

If (4.8) happen, for the case y 0 = 0 , take

(4.9) ϕ ( y ) ∈ C c ∞ ( B 1 n − 1 ) , 0 ≤ ϕ ≤ 1 , ϕ ≡ 1 in B 1 2 n − 1 ,

then

1 = ∫ B 1 2 n − 1 d μ ≤ ∫ B 1 n − 1 ϕ ( y ) d μ = lim i → ∞ ∫ B 1 n − 1 ∣ f i ∣ p ∣ y ∣ p α ϕ ( y ) d y ≤ lim i → ∞ ∫ B 1 n − 1 ∣ f i ∣ p ∣ y ∣ p α d y ,

which is a contradiction with ∫ B 1 n − 1 ∣ f i ∣ p ∣ y ∣ p α d y = 1 2 .

For the case y 0 ≠ 0 , we claim that

(4.10) lim i → ∞ ∫ B δ + ( y 0 ) ∣ P ( f i ) ∣ q ∣ x ∣ − q β d x = 0 .

Let us discuss this claim in two cases. If β > 0 , define 1 t = n − 1 n 1 p − n − 1 n + α + λ − θ n , then t > q . Since y 0 ≠ 0 , we can find some δ > 0 such that 0 ∉ B δ + ( y 0 ) ¯ . Then it follows from Theorem 1.2 ( β = 0 ) that

∫ B δ + ( y 0 ) ∣ P ( f i ) ∣ t ∣ x ∣ − t β d x ≤ C n , t , β ∫ B δ + ( y 0 ) ∣ P ( f i ) ∣ t d x ≤ C n , t , β , δ , α , p ∫ ∂ R + n ∣ f i ∣ p ∣ y ∣ p α d y t p , ≤ C n , t , β , δ , α , p .

This together with Vitali convergence theorem gives

(4.11) lim i → ∞ ∫ B δ + ( y 0 ) ∣ P ( f i ) ∣ q ∣ x ∣ − q β d x = 0 .

If β = 0 , choosing sufficiently small ζ > 0 , we have

lim i → ∞ ∫ B δ + ( y 0 ) ∣ P ( f i χ B ζ n − 1 ) ∣ q d x ≤ lim i → ∞ C n , α , p , q ′ , λ q ∫ B ζ n − 1 ∣ f i ∣ p ∣ y ∣ α d y q p = 0 .

On the other hand, choosing suitable ε > 0 , since

∫ B δ + ( y 0 ) ∣ P ( f i χ ∂ R + n \ B ζ n − 1 ) ∣ q + ε d x ≤ C n , p , q ′ , λ , ε ∫ ∂ R + n \ B ζ n − 1 ∣ f i ∣ p d y q + ε p ≤ C n , p , q ′ , λ , ε , ζ ∫ ∂ R + n \ B ζ n − 1 ∣ f i ∣ p ∣ y ∣ p α d y q + ε p ≤ C n , p , q ′ , λ , ε , ζ ,

then

lim i → ∞ ∫ B δ + ( y 0 ) ∣ P ( f i χ ∂ R + n \ B ζ n − 1 ) ∣ q d x = 0 .

By combining the aforementioned estimates, we derive that

lim i → ∞ ∫ B δ + ( y 0 ) ∣ P ( f i ) ∣ q d x = 2 q lim i → ∞ ∫ B δ + ( y 0 ) ∣ P ( f i χ B ζ n − 1 ) ∣ q d x + 2 q lim i → ∞ ∫ B δ + ( y 0 ) ∣ P ( f i χ ∂ R + n \ B ζ n − 1 ) ∣ q d x = 0 .

Take

(4.12) φ ( x ) ∈ C c ∞ ( B δ + ( y 0 ) ) , 0 ≤ φ ≤ 1 , φ ≡ 1 in B δ 2 + ( y 0 ) ,

then we have

0 = lim i → ∞ ∫ B δ + ( y 0 ) ∣ P ( f i ) ∣ q ∣ x ∣ − q β d x ≥ lim i → ∞ ∫ B δ + ( y 0 ) ∣ P ( f i ) ∣ q ∣ x ∣ − q β φ d x ≥ ∫ B δ + ( y 0 ) φ d ν ≥ ∫ B δ 2 + ( y 0 ) d ν = C n , α , β , p , q ′ , λ ,

which is a contradiction with C n , α , β , p , q ′ , λ > 0 . Therefore, we have ∫ ∂ R + n ∣ f ∣ p ∣ y ∣ p α d y = 1 and f is the maximum of Stein-Weiss inequality with the extended Poisson kernel.□

5 The proof of Theorem 1.4

In this section, we provide the regularity estimate for positive solutions of the integral system (1.15). Let V be a topological vector space. Suppose there are two extended norms (i.e., the norm of an element in V might be infinity) defined on V ,

‖ ⋅ ‖ X , ‖ ⋅ ‖ Y : V ⟶ [ 0 , ∞ ] .

Let

X ≔ { f ∈ V : ‖ f ‖ X < ∞ } and Y ≔ { f ∈ V : ‖ f ‖ Y < ∞ } .

The operator T : X → Y is said to be contracting if for any f , g ∈ X , there exists some constant η ∈ ( 0 , 1 ) such that

(5.1) ‖ T ( f ) − T ( g ) ‖ Y ≤ η ‖ f − g ‖ X

and T is said to be shrinking if for any f ∈ X , there exists some constant θ ∈ ( 0 , 1 ) such that

(5.2) ‖ T ( f ) ‖ Y ≤ θ ‖ f ‖ X .

We need the following regularity lifting lemma [9].

Lemma 5.1

Let T be a contraction map from X into itself and from Y into itself. Assume that for any f ∈ X , there exists a function g ∈ Z ≔ X ∩ Y such that f = T f + g ∈ X . Then f ∈ Z .

Now, we start our proof. Denote

u a ( y ) = u ( y ) , ∣ u ( y ) ∣ > a or ∣ y ∣ > a , 0 , otherwise . v a ( x ) = v ( x ) , ∣ v ( x ) ∣ > a or ∣ x ∣ > a , 0 , otherwise .

u b ( y ) = u ( y ) − u a ( y ) and v b ( x ) = v ( x ) − v a ( x ) . Define the linear operator T 1 as follows:

T 1 ( h ) ( y ) = ∫ R + n ∣ y ∣ − α P ( x , y , λ , θ ) v a q 0 − 1 ( x ) h ( x ) ∣ x ∣ − β d x , y ∈ ∂ R + n

and

T 2 ( h ) ( x ) = ∫ ∂ R + n ∣ x ∣ − β P ( x , y , λ , θ ) u a p 0 − 1 ( y ) h ( y ) ∣ y ∣ − α d y , x ∈ R + n .

Since ( u ( y ) , v ( x ) ) ∈ L p 0 + 1 ( ∂ R + n ) × L q 0 + 1 ( R + n ) is a pair of positive solutions of the integral system (1.15), then

u ( y ) = ∫ R + n ∣ y ∣ − α P ( x , y , λ , θ ) v q 0 ( x ) ∣ x ∣ − β d x , = ∫ R + n ∣ y ∣ − α P ( x , y , λ , θ ) ( v a + v b ) q 0 − 1 ( x ) v ( x ) ∣ x ∣ − β d x = ∫ R + n ∣ y ∣ − α P ( x , y , λ , θ ) v a q 0 − 1 ( x ) v ( x ) ∣ x ∣ − β d x + ∫ R + n ∣ y ∣ − α P ( x , y , λ , θ ) v b q 0 ( x ) ∣ x ∣ − β d x = T 1 ( v ) ( y ) + F ( y )

and

v ( x ) = ∫ ∂ R + n ∣ x ∣ − β P ( x , y , λ , θ ) u p 0 ( y ) ∣ y ∣ − α d y , = ∫ ∂ R + n ∣ x ∣ − β P ( x , y , λ , θ ) ( u a + u b ) p 0 − 1 ( y ) u ( y ) ∣ y ∣ − α d y = ∫ ∂ R + n ∣ y ∣ − α P ( x , y , λ , θ ) u a p 0 − 1 ( y ) u ( y ) ∣ x ∣ − β d y + ∫ ∂ R + n ∣ y ∣ − α P ( x , y , λ , θ ) u b p 0 ( y ) ∣ x ∣ − β d y = T 2 ( u ) ( x ) + G ( x ) ,

where

F ( y ) = ∫ R + n ∣ y ∣ − α P ( x , y , λ , θ ) v b q 0 ( x ) ∣ x ∣ − β d x , G ( x ) = ∫ ∂ R + n ∣ y ∣ − α P ( x , y , λ , θ ) u b p 0 ( y ) ∣ x ∣ − β d y .

Define the operator T ( h 1 , h 2 ) = ( T 1 ( h 2 ) , T 2 ( h 1 ) ) , equip the product space L r ( ∂ R + n ) × L s ( R + n ) with the norm ‖ ( h 1 , h 2 ) ‖ r , s = ‖ h 1 ‖ L r ( ∂ R + n ) + ‖ h 2 ‖ L s ( R + n ) . It is easy to see the product space is complete under these norms.

Observe that ( u , v ) solves the equation ( u , v ) = T ( u , v ) + ( F , G ) . We first show that T is shrinking from L p 0 + 1 ( ∂ R + n ) × L q 0 + 1 ( R + n ) to itself. By using the integral inequality (1.10) together with the Hölder inequality, we obtain for ( h 1 , h 2 ) ∈ L p 0 + 1 ( ∂ R + n ) × L q 0 + 1 ( R + n ) ,

‖ T 1 ( h 2 ) ‖ L p 0 + 1 ( ∂ R + n ) ≲ ‖ v a q 0 − 1 ‖ L q 0 + 1 q 0 − 1 ( R + n ) ‖ h 2 ‖ L q 0 + 1 ( R + n ) ≲ ‖ v a ‖ L q 0 + 1 ( R + n ) q 0 − 1 ‖ h 2 ‖ L q 0 + 1 ( R + n )

and

‖ T 2 ( h 1 ) ‖ L q 0 + 1 ( R + n ) ≲ ‖ u a p 0 − 1 ‖ L p 0 + 1 p 0 − 1 ( ∂ R + n ) ‖ h 1 ‖ L p 0 + 1 ( ∂ R + n ) ≲ ‖ u a ‖ L p 0 + 1 ( ∂ R + n ) p 0 − 1 ‖ h 1 ‖ L p 0 + 1 ( ∂ R + n ) .

By choosing sufficient large a , in view of the integrability L p 0 + 1 ( ∂ R + n ) × L q 0 + 1 ( R + n ) , we derive that

‖ T ( h 1 , h 2 ) ‖ p 0 + 1 , q 0 + 1 = ‖ T 1 ( h 2 ) ‖ L p 0 + 1 ( ∂ R + n ) + ‖ T 2 ( h 1 ) ‖ L q 0 + 1 ( R + n ) ≤ 1 2 ‖ ( h 1 , h 2 ) ‖ p 0 + 1 , q 0 + 1 .

This shows that T is shrinking from L p 0 + 1 ( ∂ R + n ) × L q 0 + 1 ( R + n ) to itself.

To apply the regularity lifting lemma by contracting operators (Lemma 5.1), we fix the indices r and s whose range will be determined later satisfying

(5.3) 1 s − n − 1 n 1 r = 1 q 0 + 1 − n − 1 n 1 p 0 + 1 .

To prove that the conclusion that ( u , v ) ∈ L r ( ∂ R + n ) × L s ( R + n ) , we need to verify the following conditions:

  1. T is shrinking from L r ( ∂ R + n ) × L s ( R + n ) to itself.

  2. ( F , G ) ∈ L p 0 + 1 ( ∂ R + n ) × L q 0 + 1 ( R + n ) ∩ L r ( ∂ R + n ) × L s ( R + n ) .

Next, we show that T is shrinking from L r ( ∂ R + n ) × L s ( R + n ) to itself. By using the inequality, we can obtain

‖ T 2 ( h 1 ) ‖ L s ( R + n ) ≤ C ‖ u a p 0 − 1 h 1 ‖ L t ( ∂ R + n ) ≤ C ‖ u a ‖ L p 0 + 1 ( ∂ R + n ) p 0 − 1 ‖ h 1 ‖ L r ( ∂ R + n ) ,

where s , t , r ∈ ( 1 , + ∞ ) satisfying

max { β − θ , 0 } < n s , max { α , 0 } < n − 1 t ′ ,

1 s = n − 1 n 1 t + α + β + λ + 1 − n − θ n , 1 t = p 0 − 1 p 0 + 1 + 1 r ,

which is equivalent to

(5.4) max { β − θ , 0 } n < 1 s < β + λ + α − max { α , 0 } − θ n , 1 s = n − 1 n 1 t + α + β + λ + 1 − n − θ n , 1 r + p 0 − 1 p 0 + 1 = 1 t < 1 − max { α , 0 } n − 1 .

Choosing a sufficiently large a that C ‖ u a ‖ L p 0 + 1 ( ∂ R + n ) p 0 − 1 ≤ 1 2 since u ∈ L p 0 + 1 ( ∂ R + n ) . Thus, ‖ T 2 ( h 1 ) ‖ L s ( R + n ) ≤ 1 2 ‖ h 1 ‖ L r ( ∂ R + n ) for all h 1 ∈ L r ( ∂ R + n ) . Similarly, by using the inequality, we can also obtain

‖ T 1 ( h 2 ) ‖ L r ( ∂ R + n ) ≤ C ‖ v a q 0 − 1 h 2 ‖ L w ( ∂ R + n ) ≤ C ‖ v a ‖ L q 0 + 1 ( R + n ) q 0 − 1 ‖ h 2 ‖ L s ( R + n ) ,

where s , w , r ∈ ( 1 , + ∞ ) satisfying

max { α , 0 } < n − 1 r , max { β − θ , 0 } < n w ′ , 1 r = n n − 1 1 w + α + β + λ − n − θ n − 1 , 1 w = q 0 − 1 q 0 + 1 + 1 s ,

which is equivalent to

(5.5) max { α , 0 } n − 1 < 1 r < α + λ − max { β − θ , 0 } + β − θ n − 1 , 1 r = n n − 1 1 w + α + β + λ − n − θ n − 1 , q 0 − 1 q 0 + 1 + 1 s = 1 w < 1 − max { β − θ , 0 } n .

Choosing a sufficiently large a that C ‖ v a ‖ L q 0 + 1 ( R + n ) q 0 − 1 ≤ 1 2 since v ∈ L q 0 + 1 ( R + n ) . Thus, ‖ T 1 ( h 2 ) ‖ L r ( ∂ R + n ) ≤ 1 2 ‖ h 2 ‖ L s ( R + n ) for all h 2 ∈ L s ( R + n ) . By combining the above estimate, we derive that ‖ T ( h 1 , h 2 ) ‖ r , s ≤ 1 2 ‖ ( h 1 , h 2 ) ‖ r , s . This shows that T is shrinking from L r ( ∂ R + n ) × L s ( R + n ) to itself.

Finally, we show that ( F , G ) ∈ L p 0 + 1 ( ∂ R + n ) × L q 0 + 1 ( R + n ) ∩ L r ( ∂ R + n ) × L s ( R + n ) . It is evident once one notices that u b ( y ) and v b ( x ) are uniformly bounded by a . By applying the regularity lifting Lemma 5.1, we derive that ( u ( y ) , v ( x ) ) ∈ L r ( ∂ R + n ) × L s ( R + n ) . Now, we start to determine the range of 1 r and 1 s .

For any ε > 0 , pick s 0 > 1 and r 0 > 1 such that

1 s 0 > max max { β − θ , 0 } n , 1 q 0 + 1 − n − 1 n 1 p 0 + 1 + max { α , 0 } n + ε ,

1 r 0 > max max { α , 0 } n − 1 , 1 p 0 + 1 − n n − 1 1 q 0 + 1 + max { β − θ , 0 } n − 1 + ε .

It is not easy to check that ( r 0 , s 0 ) satisfying the conditions (5.4) and (5.5). Hence we have ( u ( y ) , v ( x ) ) ∈ L r 0 ( ∂ R + n ) × L s 0 ( R + n ) . Similarly, picking s 1 > 1 and r 1 > 1 such that

(5.6) 1 s 1 < min β + λ + α − max { α , 0 } − θ n , 1 q 0 + 1 − n − 1 n 1 p 0 + 1 + n − 1 n min α + λ − max { β − θ , 0 } + β − θ n − 1 , 2 p 0 + 1 − max { α , 0 } n − 1 , 1 − ε ,

(5.7) 1 r 1 < min α + λ − max { β − θ , 0 } + β − θ n − 1 , 1 p 0 + 1 − n n − 1 1 q 0 + 1 + n n − 1 min β + λ + α − max { α , 0 } − θ n , 2 q 0 + 1 − max { β − θ , 0 } n , 1 − ε ,

we have ( u ( y ) , v ( x ) ) ∈ L r 1 ( ∂ R + n ) × L s 1 ( R + n ) . By interpolation theorem, we obtain that ( u , v ) ∈ L r ( ∂ R + n ) × L s ( R + n ) for all r and s such that

(5.8) 1 r ∈ 1 r 0 , 1 r 1 , 1 s ∈ 1 s 0 , 1 s 1 .

Let ε → 0 , we conclude that

(5.9) 1 r ∈ max max { α , 0 } n − 1 , 1 p 0 + 1 − n n − 1 1 q 0 + 1 + max { β − θ , 0 } n − 1 , min α + λ − max { β − θ , 0 } + β − θ n − 1 , 1 p 0 + 1 − n n − 1 1 q 0 + 1 + n n − 1 min β + λ + α − max { α , 0 } − θ n , 2 q 0 + 1 − max { β − θ , 0 } n , 1

and

(5.10) 1 s ∈ max max { β − θ , 0 } n , 1 q 0 + 1 − n − 1 n 1 p 0 + 1 + max { α , 0 } n , min β + λ + α − max { α , 0 } − θ n , 1 q 0 + 1 − n − 1 n 1 p 0 + 1 + n − 1 n min α + λ − max { β − θ , 0 } + β − θ n − 1 , 2 p 0 + 1 − max { α , 0 } n − 1 , 1

Acknowledgements

The authors would like to thank the referees for having carefully read the article and for many constructive comments, which have improved the exposition of the article.

  1. Funding information: The work was supported by the National Natural Science Foundation of China (No.12201016) and a grant from Beijing Technology and Business University (QNJJ2022-03).

  2. Conflict of interest: The authors state that there is no conflict of interest.

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Received: 2023-07-21
Revised: 2023-08-11
Accepted: 2023-08-12
Published Online: 2023-08-28

© 2023 the author(s), published by De Gruyter

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  8. Multiple solutions of p-fractional Schrödinger-Choquard-Kirchhoff equations with Hardy-Littlewood-Sobolev critical exponents
  9. Improved fractional Trudinger-Moser inequalities on bounded intervals and the existence of their extremals
  10. The existence of infinitely many boundary blow-up solutions to the p-k-Hessian equation
  11. A priori bounds, existence, and uniqueness of smooth solutions to an anisotropic Lp Minkowski problem for log-concave measure
  12. Existence of nonminimal solutions to an inhomogeneous elliptic equation with supercritical nonlinearity
  13. Non-degeneracy of multi-peak solutions for the Schrödinger-Poisson problem
  14. Gagliardo-Nirenberg-type inequalities using fractional Sobolev spaces and Besov spaces
  15. Ground states of Schrödinger systems with the Chern-Simons gauge fields
  16. Quasilinear problems with nonlinear boundary conditions in higher-dimensional thin domains with corrugated boundaries
  17. A system of equations involving the fractional p-Laplacian and doubly critical nonlinearities
  18. A modified Picone-type identity and the uniqueness of positive symmetric solutions for a prescribed mean curvature problem
  19. On a version of hybrid existence result for a system of nonlinear equations
  20. Special Issue: Geometric PDEs and applications
  21. Preface for the special issue on “Geometric Partial Differential Equations and Applications”
  22. Convex hypersurfaces with prescribed Musielak-Orlicz-Gauss image measure
  23. Total mean curvatures of Riemannian hypersurfaces
  24. On degenerate case of prescribed curvature measure problems
  25. A curvature flow to the Lp Minkowski-type problem of q-capacity
  26. Aleksandrov reflection for extrinsic geometric flows of Euclidean hypersurfaces
  27. A note on second derivative estimates for Monge-Ampère-type equations
  28. The Lp chord Minkowski problem
  29. Widths of balls and free boundary minimal submanifolds
  30. Smooth approximation of twisted Kähler-Einstein metrics
  31. The exterior Dirichlet problem for the homogeneous complex k-Hessian equation
  32. A Carleman inequality on product manifolds and applications to rigidity problems
  33. Asymptotic behavior of solutions to the Monge-Ampère equations with slow convergence rate at infinity
  34. Pinched hypersurfaces are compact
  35. The spinorial energy for asymptotically Euclidean Ricci flow
  36. Geometry of CMC surfaces of finite index
  37. Capillary Schwarz symmetrization in the half-space
  38. Regularity of optimal mapping between hypercubes
  39. Special Issue: In honor of David Jerison
  40. Preface for the special issue in honor of David Jerison
  41. Homogenization of oblique boundary value problems
  42. A proof of a trace formula by Richard Melrose
  43. Compactness estimates for minimizers of the Alt-Phillips functional of negative exponents
  44. Regularity properties of monotone measure-preserving maps
  45. Examples of non-Dini domains with large singular sets
  46. Sharp inequalities for coherent states and their optimizers
  47. Gradient estimates and the fundamental solution for higher-order elliptic systems with lower-order terms
  48. Propagation of symmetries for Ricci shrinkers
  49. Linear extension operators for Sobolev spaces on radially symmetric binary trees
  50. The Neumann problem on the domain in 𝕊3 bounded by the Clifford torus
  51. On an effective equation of the reduced Hartree-Fock theory
  52. Polynomial sequences in discrete nilpotent groups of step 2
  53. Integral inequalities with an extended Poisson kernel and the existence of the extremals
  54. On singular solutions of Lane-Emden equation on the Heisenberg group
https://doi.org/10.1515/ans-2023-0104
Keywords for this article
Hardy-Littlewood-Sobolev inequality; Stein-Weiss inequality; Poisson kernel; concentration compactness principle
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Articles in the same Issue

  1. Research Articles
  2. Asymptotic properties of critical points for subcritical Trudinger-Moser functional
  3. The existence of positive solution for an elliptic problem with critical growth and logarithmic perturbation
  4. On some dense sets in the space of dynamical systems
  5. Sharp profiles for diffusive logistic equation with spatial heterogeneity
  6. Generic properties of the Rabinowitz unbounded continuum
  7. Global bifurcation of coexistence states for a prey-predator model with prey-taxis/predator-taxis
  8. Multiple solutions of p-fractional Schrödinger-Choquard-Kirchhoff equations with Hardy-Littlewood-Sobolev critical exponents
  9. Improved fractional Trudinger-Moser inequalities on bounded intervals and the existence of their extremals
  10. The existence of infinitely many boundary blow-up solutions to the p-k-Hessian equation
  11. A priori bounds, existence, and uniqueness of smooth solutions to an anisotropic Lp Minkowski problem for log-concave measure
  12. Existence of nonminimal solutions to an inhomogeneous elliptic equation with supercritical nonlinearity
  13. Non-degeneracy of multi-peak solutions for the Schrödinger-Poisson problem
  14. Gagliardo-Nirenberg-type inequalities using fractional Sobolev spaces and Besov spaces
  15. Ground states of Schrödinger systems with the Chern-Simons gauge fields
  16. Quasilinear problems with nonlinear boundary conditions in higher-dimensional thin domains with corrugated boundaries
  17. A system of equations involving the fractional p-Laplacian and doubly critical nonlinearities
  18. A modified Picone-type identity and the uniqueness of positive symmetric solutions for a prescribed mean curvature problem
  19. On a version of hybrid existence result for a system of nonlinear equations
  20. Special Issue: Geometric PDEs and applications
  21. Preface for the special issue on “Geometric Partial Differential Equations and Applications”
  22. Convex hypersurfaces with prescribed Musielak-Orlicz-Gauss image measure
  23. Total mean curvatures of Riemannian hypersurfaces
  24. On degenerate case of prescribed curvature measure problems
  25. A curvature flow to the Lp Minkowski-type problem of q-capacity
  26. Aleksandrov reflection for extrinsic geometric flows of Euclidean hypersurfaces
  27. A note on second derivative estimates for Monge-Ampère-type equations
  28. The Lp chord Minkowski problem
  29. Widths of balls and free boundary minimal submanifolds
  30. Smooth approximation of twisted Kähler-Einstein metrics
  31. The exterior Dirichlet problem for the homogeneous complex k-Hessian equation
  32. A Carleman inequality on product manifolds and applications to rigidity problems
  33. Asymptotic behavior of solutions to the Monge-Ampère equations with slow convergence rate at infinity
  34. Pinched hypersurfaces are compact
  35. The spinorial energy for asymptotically Euclidean Ricci flow
  36. Geometry of CMC surfaces of finite index
  37. Capillary Schwarz symmetrization in the half-space
  38. Regularity of optimal mapping between hypercubes
  39. Special Issue: In honor of David Jerison
  40. Preface for the special issue in honor of David Jerison
  41. Homogenization of oblique boundary value problems
  42. A proof of a trace formula by Richard Melrose
  43. Compactness estimates for minimizers of the Alt-Phillips functional of negative exponents
  44. Regularity properties of monotone measure-preserving maps
  45. Examples of non-Dini domains with large singular sets
  46. Sharp inequalities for coherent states and their optimizers
  47. Gradient estimates and the fundamental solution for higher-order elliptic systems with lower-order terms
  48. Propagation of symmetries for Ricci shrinkers
  49. Linear extension operators for Sobolev spaces on radially symmetric binary trees
  50. The Neumann problem on the domain in 𝕊3 bounded by the Clifford torus
  51. On an effective equation of the reduced Hartree-Fock theory
  52. Polynomial sequences in discrete nilpotent groups of step 2
  53. Integral inequalities with an extended Poisson kernel and the existence of the extremals
  54. On singular solutions of Lane-Emden equation on the Heisenberg group
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