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Sharp affine weighted L 2 Sobolev inequalities on the upper half space

  • Jingbo Dou EMAIL logo , Yunyun Hu and Caihui Yue
Published/Copyright: March 14, 2024

Abstract

We establish some sharp affine weighted L 2 Sobolev inequalities on the upper half space, which involves a divergent operator with degeneracy on the boundary. Moreover, for some certain exponents cases, we also characterize the extremal functions and best constants. Our approach only relies on the L 2 structure of gradient norm, affine invariance and a class of weighted L 2 Sobolev inequality on the upper half space. This is a simple approach which does not depend on the geometric structure of Euclidean space such as Brunn–Minkowski theory on convex geometry.

Mathematics Subject Classification (2010): 26D10; 46E35; 52A40

1 Introduction

The classical L p Sobolev inequality on R n states that: for any 1 ≤ p < n, there exists a positive constant S n,p such that for any function u C 0 ( R n ) ,

(1.1) R n | u ( x ) | n p n p d x n p n p S n , p R n | u ( x ) | p d x 1 p

holds. It is a basic integral inequality in partial differential equations and harmonic analysis theory, see e.g., [1], [2]. For 1 < p < n, Sobolev [3] established inequality (1.1) via Hardy–Littlewood–Sobolev inequality. For p = 1, Federer, Fleming [4] and Maz’ya [5] established inequality (1.1). See [6] and [7] for general case 1 ≤ p < n. It is well known that the sharp L 1 Sobolev inequality is equivalent to the classical isoperimetric inequality in R n . Later, Aubin [8], [9] employed an alternative method to reprove inequality (1.1), see also [1], [10]. Using Bliss’s lemma and classical calculus of variations, they proved that the equality holds in (1.1) if and only if

u ( x ) = 1 a + b | x x 0 | p p 1 n p p

for 1 < p < n, a, b > 0, x 0 R n . The sharp constant S n,p (1 < p < n) is given by

S n , p = π 1 2 n 1 p p 1 n p 1 1 p Γ ( 1 + n / 2 ) Γ ( n ) Γ ( n / p ) Γ ( 1 + n n / p ) 1 n .

Years later, an affine L p Sobolev inequality

(1.2) R n | u ( x ) | n p n p d x n p n p S n , p E ( u )

was established by Zhang [11] for p = 1, Lutwak, Yang and Zhang [12] for any 1 < p < n and u W 1 , p ( R n ) . Here E ( u ) is called L p affine energy defined as

E ( u ) = n w n w p 1 2 w n + p 2 1 p 1 n w n S n 1 R n | u ( x ) , ξ | p d x n p d ξ 1 n ,

and w n = π n 2 Γ ( n 2 + 1 ) denotes the volume of the unit ball B in R n . Inequality (1.2) is rather surprising since it was not at all expected that affine versions of the classical L p Sobolev inequalities are independent of any Euclidean structure, and it is stronger than classical L p Sobolev inequalities. Recently, for p = 2, Schindler and Tintarev [13] showed the relation between L 2 integral of classical gradient and affine gradient. Moreover, they studied the compactness properties of inequality (1.2), existence and regularity of extremal function for p = 2.

The affine Sobolev inequality can be applied to information theory, see [12]. It is also highly relevant to Minkowski-type problem, see e.g. [14], [15], [16], [17], and references therein.

In [11], the proof of inequality (1.2) bases on the Petty projection inequality in compact sets and L 1 affine isoperimetric inequality. In the case 1 < p < n, Lutwak, Yang and Zhang [12] proved inequality (1.2) by using the solution of L p -Minkowski problem and affine isoperimetric inequality. The authors also pointed out the corresponding geometry of sharp L p Sobolev inequality is the isoperimetric inequality, but the corresponding geometry of sharp affine L p Sobolev inequality is the L p Petty projection inequality. Moreover, they showed that the affine L p Sobolev inequality (1.2) is stronger than classical L p Sobolev inequality (1.1). Later, Haddad, Jiménez and Montenegro [18] provided a new method to prove inequality (1.2), which they only use L p Busemann–Petty centroid inequality. Recently, Nguyen [19] introduced a general affine L p energy to prove affine L p Sobolev inequality.

The affine Sobolev inequality has been widely studied after the seminal works of Lutwak, Yang, Zhang [12]. Let us recall some results for affine weighted Sobolev inequalities on the upper half space in [20], [21]. Let R + n + 1 = ( x , t ) : x = ( x 1 , x 2 , x n ) R n , t > 0 denotes the upper half space. For α R , we define

(1.3) E α ( u ) = c n S n 1 R + n + 1 | x u ( t , x ) , ξ | 2 t α d x d t n 2 d ξ 1 n

as the L 2 weighted affine energy on R + n + 1 , where c n = n ( 2 π n 2 Γ ( n 2 ) ) 1 n .

Recently, De Nápoli et al. [20] established a sharp affine L 2 Sobolev trace inequality

(1.4) R n | u ( 0 , x ) | 2 n n 1 d x n 1 n 2 S n E 0 ( u ) R + n + 1 u ( t , x ) t 2 d x d t 1 2

for any u ( t , x ) W 1,2 R + n + 1 , where E 0 ( u ) is defined as (1.3) for α = 0, S n is the best constant. A central tool used in [20] is L p Busemann–Petty centroid inequality. Nguyen [21] proved a stronger form of inequality (1.4) by using L p Busemann–Petty centroid inequality and sharp Gagliardo–Nirenberg trace inequality. More precisely, it states that

R n | u ( 0 , x ) | 2 n n 1 d x n 1 n 2 S n E 0 ( u ) inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 d x d t 1 2

for any u ( t , x ) C 0 R + n + 1 , where a = ( a 1 , a 2 , , a n ) R n .

Let α R and 1 ≤ p < . In the sequel we define

E α , p ( u ) = n ω n ω p 1 2 ω n + p 2 1 p 1 n ω n S n 1 R + n + 1 | x u ( t , x ) , ξ | p t α d x d t n p d ξ 1 n

as the L p weighted affine energy on R + n + 1 . Obviously, E α ( u ) = E α , 2 ( u ) . For α ≥ 0 and 1 ≤ p < n + 1 + α, Haddad, Jiménez and Montenegro [22] established the following sharp affine weighted L p Sobolev inequality on R + n + 1

(1.5) R + n + 1 | u ( x , t ) | p ( n + 1 + α ) n + 1 + α p t α d x d t n + 1 + α p p ( n + 1 + α ) S n + 1 , p , α E α , p ( u ) n n + 1 + α R + n + 1 u ( t , x ) t p t α d x d t 1 + α p ( n + 1 + α )

for all u C 0 ( R n + 1 ) , where S n+1,p,α is the best constant. The proof bases on L p Busemann–Petty centroid inequality and weighted L p Sobolev inequality on R + n + 1 established by Nguyen [23]. In the case α ∈ (−1, 1) and n + α > 1, this inequality was then improved in [24] by proving the following sharp affine fractional L 2 Sobolev trace inequality on R + n + 1 ,

R n | u ( 0 , x ) | 2 n n + α 1 d x n + α 1 n S n , α 1 α 2 1 α 2 1 + α 2 1 + α 2 E α ( u ) 1 α inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t 1 + α 2

for u D α 1,2 R + n + 1 , where S n,α is the best constant. Here d α 1,2 R + n + 1 denotes the completion of the space C 0 R + n + 1 ̄ under the norm

u D α 1,2 R + n + 1 = R + n + 1 | u ( t , x ) | 2 t α d x d t 1 2 .

The author also proved the following sharp affine fractional L 2 Sobolev inequality in the fractional α homogeneous Sobolev space

R n | f ( x ) | 2 n n + α 1 d x n + α 1 n S n , α 1 α 2 1 α 2 1 + α 2 1 + α 2 E ̃ α ( f ) 2 ,

where

E ̃ α ( f ) = inf E α ( u ) 1 α 2 inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t 1 + α 4 : u D α 1,2 R + n + 1 , u ( 0 , x ) = f ( x )

is an affine fractional energy. Nguyen [24] gave a simpler approach by using quantity of the L 2 structure of the gradient norm and fractional L 2 Sobolev trace inequality on R + n + 1 . Recently, Haddad and Ludwig [25], [26] also established the affine fractional L p Sobolev inequality by introducing s-fractional L p polar projection body and the affine fractional Pólya–Szegö inequalities.

In a recent paper, Dou, Sun, Wang and Zhu [27] established the following sharp weighted Sobolev inequality involving a divergent operator with degeneracy on the boundary

(1.6) R + n + 1 | u ( t , x ) | p * t β d x d t 2 / p * C n + 1 , α , β R + n + 1 | u ( t , x ) | 2 t α d x d t

for all u D α 1,2 R + n + 1 , where C n+1,α,β is best constant, p * = p α , β * 2 n + 2 β + 2 n + α 1 is critical exponent and α, β satisfy

(1.7) α > 0 , β > 1 , n 1 n + 1 β α β + 2 .

Moreover, the authors classified the nonnegative extremal functions of (1.6) as follows.

Theorem A.

Let n ≥ 1, α > 0 , β > 1 , n 1 n + 1 β < α < β + 2 . Assume that u ( t , x ) D α 1,2 R + n + 1 is the positive extremal function of (1.6), we have, up to the multiple of some constant,

(1.8) u ( t , x ) = 1 ( t + μ ) 2 + | x x 0 | 2 n + α 1 2 ψ ( t + μ , x x 0 ) | t + μ | 2 + | x x 0 | 2 1 2 μ , 0 ,

for some x 0 R n , μ > 0, ψ(r) > 0, ψ C 2 [ 0 , 1 2 μ ) C 0 [ 0 , 1 2 μ ] satisfying an ordinary differential equation

(1.9) ψ + n r 2 α r 1 4 μ 2 r 2 ψ α ( n + α 1 ) 1 4 μ 2 r 2 ψ = C 1 4 μ 2 r 2 β α ψ n + 2 β α + 3 n + α 1 , 0 < r < 1 2 μ , ψ 1 2 μ = μ n + α 1 2 , ψ ( 0 ) = 0 , lim r 1 2 μ 1 4 μ 2 r 2 α ψ ( r ) = 0 .

Moreover, the explicit extremal functions of two specific exponential cases are given as follows:

  1. For β = α − 1, if α > 0 for n ≥ 2 or α ( 0 , 1 2 ] [ 1 + 17 4 , ) for n = 1, then up to the multiple of some constant, u(t, x) must be in the form of

    u ( t , x ) = μ ( μ + t ) 2 + | x x 0 | 2 n + α 1 2 ,

    where x 0 R n . Moreover,

    C n + 1 , α , α 1 = α ( n + α 1 ) π n 2 Γ ( α ) Γ n 2 + α Γ ( n + 2 α ) 1 n + α .

  2. For β = α, if α > 0 for n ≥ 2 or α 2 for n = 1, then up to the multiple of some constant, u(t, x) must be in the form of

    u ( t , x ) = μ μ 2 + t 2 + | x x 0 | 2 n + α 1 2 ,

    where x 0 R n . Moreover,

    C n + 1 , α , α = ( n + α 1 ) ( n + α + 1 ) π n 2 2 Γ α + 1 2 Γ n + α + 1 2 Γ ( n + α + 1 ) 2 n + α + 1 .

An interesting question: can we establish the sharp affine weighted L 2 Sobolev inequality on R + n + 1 corresponding to inequality (1.6) (βα)?

This paper is to establish sharp affine weighted L 2 Sobolev inequalities on R + n + 1 involving a divergent operator with degeneracy on the boundary, which is significantly stronger than weighted inequality (1.6).

In order to state our results, we introduce the following notations.

For smooth functions u(t, x) with compact support on R + n + 1 with n ≥ 1, we denote u t as partial derivative of function u(t, x) with respect to variable t. We also define ∇ x u(t, x), ∇u(t, x) as gradient with respect to variable x and variable (t, x), respectively.

We define the weighted space L α 2 R + n + 1 endowed with the norm

u L α 2 R + n + 1 = R + n + 1 | u ( t , x ) | 2 t α d x d t 1 2 .

Let GL n+1,+ be (n + 1) × (n + 1) matrices defined by

(1.10) λ 0 0 0 0 λ 1 + β n B ,

where λ > 0, BGL n the set of all invertible real n × n-matrices. Let ξ be any point on the unit sphere S n 1 R n .

Our first result of this paper is as follows.

Theorem 1.1.

Let n ≥ 1 and α, β satisfy

(1.11) α > 0 , β > 1 , n 1 n + 1 β α < β + 2 .

Then for all u D α 1,2 R + n + 1 , it holds

(1.12) R + n + 1 | u ( t , x ) | p * t β d x d t 1 / p * H n + 1 , α , β E α ( u ) n ( 2 + β α ) 2 ( n + β + 1 ) u t L α 2 R + n + 1 n ( α β ) + 2 β + 2 2 ( n + β + 1 ) ,

where H n + 1 , α , β = ( C n + 1 , α , β d n , α , β ) 1 2 , d n , α , β = 2 ( n + β + 1 ) n ( 2 + β α ) n ( α β ) + 2 β + 2 n ( 2 + β α ) n ( β α ) 2 β 2 2 ( n + β + 1 ) , C n+1,α,β is constant defined in inequality (1.6). Moreover, for n 1 n + 1 β < α < β + 2 , equality in (1.12) holds if and only if

(1.13) u ( t , x ) = 1 ( λ t + μ ) 2 + | λ 1 + β n B ( x x ) | 2 n + α 1 2 ψ λ t + μ , λ 1 + β n B ( x x ) λ t + μ | 2 + | λ 1 + β n B ( x x ) 2 1 2 μ , 0

for some x R n , m > 0, BGL n , where ψ satisfies (1.9).

Furthermore, in following two cases, we write the explicit form of the extremal functions.

  1. If β = α − 1, with α > 0 for n ≥ 2 or α ( 0 , 1 2 ] [ 1 + 17 4 , ) for n = 1, up to the multiple of some constant, u(t, x) must be in the following form

    (1.14) u ( t , x ) = 1 ( 1 + λ t ) 2 + | λ α n B ( x x ) | 2 n + α 1 2 ,

    where λ > 0, x R n , BGL n and

    H n + 1 , α , α 1 = [ ( n + α ) ( n + 2 α 1 ) ] n + 2 α 4 ( n + α ) 2 α 1 2 n + 2 α + 1 n π ( n + 2 α ) n 4 ( n + α ) Γ ( n + 2 α ) Γ ( α ) Γ 2 n + α 1 2 ( n + α ) .

  2. If β = α, with α > 0 for n ≥ 2 or α 2 for n = 1, up to the multiple of some constant, u(t, x) must be in the form of

    (1.15) u ( t , x ) = 1 1 + ( λ t ) 2 + | λ 1 + α n B ( x x ) | 2 n + α 1 2 ,

    where λ > 0, x R n , BGL n and

    H n + 1 , α , α = n + 2 α n π ( n + α 1 ) 2 n 2 ( n + α + 1 ) 1 ( 1 + α ) ( n + α 1 ) 1 + α 2 ( n + α + 1 ) Γ ( n + α + 1 ) Γ α + 1 2 Γ n + α + 1 2 1 n + α + 1 .

Remark 1.2.

If β = α > 0, the inequality (1.12) is equivalent to inequality (1.5) with p = 2 established early by Haddad, Jiménez and Montenegro [22].

Remark 1.3.

The inequality (1.12) is stronger than (1.6), that is,

E α ( u ) R + n + 1 | x u ( t , x ) | 2 t α d x d t 1 2 .

We can directly check this inequality by using Hölder’s inequality (see Lemma 4.1 for details). Furthermore, it is remarkable that the inequality (1.12) is invariant under the action of the elements in GL n+1,+ (see Lemma 2.4).

As an immediate consequence of Theorem 1.1, using Young’s inequality we have

Corollary 1.4.

Let n ≥ 1 and α, β satisfy (1.11). Then for all u D α 1,2 R + n + 1 , it holds

(1.16) R + n + 1 | u ( t , x ) | p * t β d x d t 1 / p * C n + 1 , α , β 1 2 E α ( u ) 2 + u t L α 2 R + n + 1 2 1 2 .

Moreover, for n 1 n + 1 β < α < β + 2 , equality in (1.16) is attained if and only if

u ( t , x ) = 1 ( λ B t + μ ) 2 + | λ B 1 + β n B ( x x ) | 2 n + α 1 2 ψ t + μ , λ B 1 + β n B ( x x ) | λ B t + μ | 2 + | λ B 1 + β n B ( x x ) | 2 1 2 μ , 0 ,

where λ B = | det ( B ) | 1 n + β + 1 , BGL n , C n+1,α,β is constant defined in inequality (1.6).

Define SGL n+1,+ as the set of all (n + 1) × (n + 1)-matrices of the form

(1.17) λ 0 0 a 1 a n λ 1 + β n B ,

where λ > 0, a = ( a 1 , a 2 , , a n ) R n , BGL n . Notice that GL n+1,+ is a subset of the set SGL n+1,+.

We also improve inequality (1.12) to a more general form, which is SGL n+1,+ invariant. We have the following result.

Theorem 1.5.

Let n ≥ 1 and α, β satisfy (1.11). Then for all u D α 1,2 R + n + 1 , it holds

(1.18) R + n + 1 | u ( t , x ) | p * t β d x d t 1 / p * H n + 1 , α , β E α ( u ) n ( 2 + β α ) 2 ( n + β + 1 ) inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t n ( α β ) + 2 β + 2 4 ( n + β + 1 ) .

Moreover, for n 1 n + 1 β < α < β + 2 , the equality in (1.18) holds if and only if

u ( t , x ) = c u * ( A ( t , x ) )

for some c R , ASGL n+1,+, where u *(t, x) satisfies (1.8).

Remark 1.6.

Obviously, if the equality holds in (1.12) and (1.18), then

inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t = R + n + 1 u ( t , x ) t 2 t α d x d t .

It implies that it must a = ( 0,0 , , 0 ) R n when the equality holds in (1.18).

Remark 1.7.

By Young’s inequality, we show that inequality (1.18) is stronger than inequality (1.6). According to Lemma 3.3 (see below), it is easy to verify that inequality (1.18) is invariant under the action of the elements in SGL n+1,+.

Our approach mainly exploits the L 2 structure of the gradient norm of inequalities (1.12) and (1.18) and weighted Sobolev inequality (1.6), which improves the idea of Nguyen in [24]. It only relies on the vector space structure, affine invariance and weighted Sobolev inequality (1.6), This is a simple proof which is different from geometric method. We do not use neither the sharp L p Busemann–Petty centroid inequality and the sharp L p Petty projection inequality, nor the solution of the even L p Minkowski problem.

When α = β = 0, inequality (1.12) becomes affine L 2 Sobolev inequality. This is a subclass of the class of affine L p Sobolev inequality established by Lutwak, Yang and Zhang [12]. Notice also that all inequalities established in above theorems are stronger than their euclidean counterparts or existing affine version. Recalling the previous statement, after the works of Zhang [11] and Lutwak, Yang and Zhang [28], the proof of the affine Sobolev type inequality mainly employed L p Brunn–Minkowski theory on convex geometry and Euclidean versions Sobolev type inequalities. The reader can see [17], [18], [19], [20], [25], [26], [29], [30], [31], [32] and references therein.

The rest of this paper is organized as follows. In Section 2, we show Theorem 1.1 and Corollary 1.4. Section 3 is devoted to proving Theorem 1.5. Section 4 is the Appendix, in which shows the inequality (1.12) is stronger than (1.6), and best constants are calculated.

2 Proof of Theorem 1.1

In this section, we first employ weighted inequality (1.6) and analysis method to prove inequality (1.12). Next we prove the existence of extremal functions for inequality (1.12). Moreover, in the following two cases β = α, β = α − 1, the solutions can be explicitly written out. Finally, we give the proof of Corollary 1.4.

We first state the following lemma which is an extension that of Lemma 2.1 in [24]. It plays a significant role in the proof of Theorem 1.1.

Lemma 2.1.

Let n ≥ 1, α R . Then for all u D α 1,2 R + n + 1 , the following equation holds

(2.1) E α ( u ) 2 = inf R + n + 1 | x ( u ( t , B x ) ) | 2 t α d x d t : B G L n , det ( B ) = ± 1 .

The proof process of Lemma 2.1 is full same that of Lemma 2.1 in [24], we omit it here. We only need to point out that the infimum of (2.1) is attained if and only if

B = [ det ( M [ u ] ) ] 1 2 n M [ u ] 1 2 O ,

(see the proof process of Lemma 2.1 in [24]), where O is n × n orthogonal matrix and M [ u ] = ( M i , j [ u ] ) i , j = 1 n with

M i , j [ u ] = R + n + 1 u ( t , x ) x i u ( t , x ) x j t α d x d t , i , j = 1,2 , , n .

Now we define a general GL n+1,m,l,+ as (n + 1) × (n + 1)-matrices with the form

(2.2) λ m 0 0 0 0 λ l B ,

where λ > 0, m , l R , BGL n the set of all invertible real n × n-matrices, det(B) = ±1. Obviously, G L n + 1 , + = G L n + 1,1 , 1 + β n , + is a subset of GL n+1,m,l,+.

For simplify, we write the assumption of m, l as

(2.3) ( α 1 ) m + n l < 0 , ( α + 1 ) m + ( n 2 ) l > 0 .

Lemma 2.2.

Let n ≥ 1, α, β satisfy (1.11), m, l satisfy (2.3) and ml. Then for all u D α 1,2 R + n + 1 , it holds

(2.4) d n , α , m , l E α ( u ) m m α ln m l R + n + 1 u ( t , x ) t 2 t α d x d t m + m α + ln 2 l 2 ( m l ) = inf R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t : A G L n + 1 , m , l , + ,

where d n , α , m , l = 2 ( l m ) ln + m α m ( 2 l ln m α m ln + m α m ) 2 l ln m α m 2 ( m l ) .

Proof.

The proof is similar to that of Lemma 2.2 in [24]. If u ≡ 0, then (2.4) is trivial. Hence, we only consider the case u ≢ 0. For any AGL n+1,m,l,+ of the form (2.2), we have

R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t = R + n + 1 | ( u ( λ m t , λ l B x ) ) | 2 t α d x d t = R + n + 1 | x ( u ( λ m t , λ l B x ) ) | 2 t α d x d t + R + n + 1 | λ m u ( λ m t , λ l B x ) ( λ m t ) | 2 t α d x d t = λ 2 l ln m α m R + n + 1 | x ( u ( t , B x ) ) | 2 t α d x d t + λ m m α ln R + n + 1 u ( t , x ) t 2 t α d x d t .

Using Lemma 2.1, one can see that

(2.5) inf R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t : A G L n + 1 , m , l , + = inf λ > 0 , det ( B ) = ± 1 λ 2 l ln m α m R + n + 1 | x ( u ( t , B x ) ) | 2 t α d x d t + λ m m α ln R + n + 1 u ( t , x ) t 2 t α d x d t = inf λ > 0 λ 2 l ln m α m inf d e t B = ± 1 R + n + 1 | x ( u ( t , B x ) ) | 2 t α d x d t + λ m m α ln R + n + 1 u ( t , x ) t 2 t α d x d t = inf λ > 0 λ 2 l ln m α m E α ( u ) 2 + λ m m α ln R + n + 1 u ( t , x ) t 2 t α d x d t = d n , α , m , l E α ( u ) m m α ln m l R + n + 1 u ( t , x ) t 2 t α d x d t m + m α + ln 2 l 2 ( m l ) .

Moreover, the infimum of the last equality in (2.5) is attained if and only if

λ = 2 l ln m α m ln + m α m E α ( u ) 2 R + n + 1 u ( t , x ) t 2 t α d x d t 1 2 ( m l ) .

The proof of Lemma 2.2 is completed.□

In particular, we take m = 1, l = 1 + β n in (2.3). A direct computation gives

n ( β α ) 2 β 2 n ( n 2 ) β 2 n n 1 n + 1 β n = 2 ( β + 1 + n ) n ( n + 1 ) < 0 ,

β + 2 α > 0 .

Obviously, taking m = 1, l = 1 + β n , the following corollary is an immediate consequence of Lemma 2.2.

Corollary 2.3.

Let n ≥ 1, α, β satisfy (1.11). Then, for all u D α 1,2 R + n + 1 , it holds

d n , α , β E α ( u ) n ( 2 + β α ) n + β + 1 R + n + 1 u ( t , x ) t 2 t α d x d t n ( α β ) + 2 β + 2 2 ( n + β + 1 ) = inf R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t : A G L n + 1 , + .

Furthermore, there exists AGL n+1,+ such that the infimum is attained.

Lemma 2.4.

Assume n ≥ 1, α, β satisfy (1.11) and u D α 1,2 R + n + 1 is given in Theorem 1.1, then inequality (1.12) is invariant under the action of the elements in GL n+1,+.

Proof.

Let u(t, x) be the extremal function given in Theorem 1.1, and denote u A (t, x) = u(A(t, x)) with AGL n+1,+ of the form (1.10). Using the change of variable, one has

(2.6) R + n + 1 | u A ( t , x ) | p * t β d x d t 1 / p * = R + n + 1 | u λ t , λ 1 + β n B x | p * t β d x d t 1 / p * = ( det ( B ) ) 1 / p * R + n + 1 | u ( t , x ) | p * t β d x d t 1 / p * .

On the one hand, let

B = b 11 b 12 b 1 n b 21 b 22 b 2 n b n 1 b n 2 b n n , x = x 1 x 2 x n

and

y = y 1 y 2 y n = B x = b 11 x 1 + b 12 x 2 + + b 1 n x n b 21 x 1 + b 22 x 2 + + b 2 n x n b n 1 x 1 + b n 2 x 2 + + b n n x n .

It is easy to verify that

x ( u ( t , B x ) ) = ( u x 1 , , u x n ) = ( u y 1 b 11 + + u y n b n 1 , , u y 1 b 1 n + + u y n b n n ) = B T y u ( t , y ) = B T B x u ( t , B x ) .

Using the change of variable again, one can deduce that

(2.7) E α ( u A ) = c n S n 1 R + n + 1 | x u λ t , λ 1 + β n B x , ξ | 2 t α d x d t n 2 d ξ 1 n = c n S n 1 R + n + 1 | λ 1 + β n B T λ 1 + β n B x u λ t , λ 1 + β n B x , ξ | 2 t α d x d t n 2 d ξ 1 n = c n S n 1 R + n + 1 | x u ( t , x ) , ξ | 2 t α λ α 1 | λ 1 + β n B | 1 d x d t n 2 | λ 1 + β n B | 1 d ξ 1 n = λ n ( β α ) 2 β 2 2 n | det ( B ) | 1 n 1 2 E α ( u ) ,

and

(2.8) u A t L α 2 R + n + 1 = R + n + 1 | u λ t , λ 1 + β n B x ( λ t ) λ | 2 t β d x d t 1 2 = λ β + 2 α 2 det ( B ) 1 2 u t L α 2 R + n + 1 .

Therefore,

(2.9) E α ( u A ) n ( 2 + β α ) 2 ( n + β + 1 ) u A t L α 2 R + n + 1 n ( α β ) + 2 β + 2 2 ( n + β + 1 ) = det ( B ) n + α 1 2 ( n + β + 1 ) E α ( u ) n ( 2 + β α ) 2 ( n + β + 1 ) u t L α 2 R + n + 1 n ( α β ) + 2 β + 2 2 ( n + β + 1 ) .

It follows from (2.6), (2.7), (2.8) and (2.9) that inequality (1.12) is invariant under the action of the elements in GL n+1,+. The proof is completed.□

We are now ready to prove Theorem 1.1 and Corollary 1.4.

Proof of Theorem 1.1.

For u D α 1,2 R + n + 1 and AGL n+1,+, it follows from (1.6) that

(2.10) C n + 1 , α , β R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t R + n + 1 | u ( A ( t , x ) ) | p * t β d x d t 2 / p * = R + n + 1 | u ( t , x ) | p * t β d x d t 2 / p * .

By Corollary 2.3 and (2.10), one has

C n + 1 , α , β d n , α , β E α ( u ) n ( 2 + β α ) n + β + 1 R + n + 1 u ( t , x ) t 2 t α d x d t n ( α β ) + 2 β + 2 2 ( n + β + 1 ) = C n + 1 , α , β inf R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t : A G L n + 1 , + R + n + 1 | u ( t , x ) | p * t β d x d t 2 / p * .

That is,

(2.11) R + n + 1 | u ( t , x ) | p * t β d x d t 1 / p * ( C n + 1 , α , β d n , α , β ) 1 2 E α ( u ) n ( 2 + β α ) 2 ( n + β + 1 ) R + n + 1 u ( t , x ) t 2 t α d x d t n ( α β ) + 2 β + 2 4 ( n + β + 1 ) .

Inequality (1.12) is proved.

Next, we check the equality condition in (1.12). Let

u * ( t , x ) = 1 ( t + μ ) 2 + | x x | 2 n + α 1 2 ψ ( t + μ , x x ) | t + μ | 2 + | x x | 2 1 2 μ , 0 ,

where ψ satisfies (1.9). It follows from Lemma 2.4 that E α ( u ) is GL n+1,+ invariant. By (1.6) and (1.12), we show that if u(t, x) = cu *(A(t, x)) for some c R , A G L n + 1 , + , then equality holds in (1.12).

Conversely, suppose that equality holds in (1.12) for u ≢ 0. From Lemma 2.1 and Corollary 2.3, we know that there exists AGL n+1,+ such that

d n , α , β E α ( u ) n ( 2 + β α ) n + β + 1 u t L α 2 R + n + 1 n ( α β ) + 2 β + 2 n + β + 1 = R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t .

Therefore, we have

C n + 1 , α , β R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t = R + n + 1 | u ( A ( t , x ) ) | p * t β d x d t 2 / p * .

This implies that uA is an extremal function of the weighted Sobolev inequality (1.6). By the uniqueness of extremal function, uA must be the form of (1.8). That is,

u ( A ( t , x ) ) = 1 ( t + μ ) 2 + | x x | 2 n + α 1 2 ψ ( t + μ , x x ) | t + μ | 2 + | x x | 2 1 2 μ , 0 = u * ( t , x ) .

Thus, we have

u ( t , x ) = u * ( A 1 ( t , x ) ) .

In particular, we observe that in following two cases, the extremal function can be explicitly written out together with Theorem A (in introduction). The computation of best constants in Theorem 1.1 can be found in Appendix.

  1. For β = α − 1, it follows from Theorem A that (1.14) is the extremal function of the inequality (1.12). Moreover, a straightforward calculation yields the best constant

    H n + 1 , α , α 1 = [ ( n + α ) ( n + 2 α 1 ) ] n + 2 α 4 ( n + α ) 2 α 1 2 n + 2 α + 1 n π ( n + 2 α ) n 4 ( n + α ) Γ ( n + 2 α ) Γ ( α ) Γ 2 n + α 1 2 ( n + α ) .

  2. For β = α, from Theorem A, the extremal function of the inequality (1.12) has the form of (1.15). Moreover, the best constant is

    H n + 1 , α , α = n + 2 α n π ( n + α 1 ) 2 n 2 ( n + α + 1 ) 1 ( 1 + α ) ( n + α 1 ) 1 + α 2 ( n + α + 1 ) Γ ( n + α + 1 ) Γ α + 1 2 Γ n + α + 1 2 1 n + α + 1 .

    This finishes the proof of Theorem 1.1.□

Proof of Corollary 1.4.

Using Young’s inequality for the right hand side of (2.11), we obtain

R + n + 1 | u ( t , x ) | p * t β d x d t 1 / p * ( C n + 1 , α , β d n , α , β ) 1 2 [ n ( 2 + β α ) ] n ( 2 + β α ) 2 ( n + β + 1 ) × [ n ( α β ) + 2 β + 1 ] n ( α β ) + 2 β + 2 2 ( n + β + 1 ) E α ( u ) 2 + u t L α 2 R + n + 1 2 2 ( n + β + 1 ) 1 2 = C n + 1 , α , β 1 2 E α ( u ) 2 + u t L α 2 R + n + 1 2 1 2 .

The inequality (1.16) is established.

It suffices to find a suitable λ so that the terms E α ( u ) and u t L α 2 R + n + 1 have the same factor for the extremal functions (1.13). Argument as Lemma 2.4, it is easy to verify that E α ( u ) and u t L α 2 R + n + 1 have the same factor in inequality (1.16) when λ = | det ( B ) | 1 n + β + 1 . Hence, we have

E α ( u A ) 2 + u A t L α 2 R + n + 1 2 = λ β α + 2 | det ( B ) | 1 E α ( u ) 2 + u t L α 2 R + n + 1 2 .

Hence we know that equality holds in (1.16) if and only if λ = | det ( B ) | 1 n + β + 1 .□

3 Proof of Theorem 1.5

In this section, we are devoted to proving Theorem 1.5. To this end, we first give the following two lemmas.

We first define a general SGL n+1,m,l,+ as (n + 1) × (n + 1)-matrices with the form

(3.1) λ m 0 0 a 1 a n λ l B ,

where λ > 0, m , l R , a = ( a 1 , a 2 , , a n ) R n , BGL n , det(B) = ±1. Obviously, S G L n + 1 , + = S G L n + 1,1 , 1 + β n , + is a subset of the set SGL n+1,m,l,+, and GL n+1,m,l,+ is a subset of the set SGL n+1,m,l,+.

Lemma 3.1.

Let n ≥ 1, α, β satisfy (1.11), m, l satisfy (2.3) and ml. Then for all u D α 1,2 R + n + 1 , it holds

(3.2) d n , α , m , l E α ( u ) m m α ln m l inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t m + m α + ln 2 l 2 ( m l ) = inf R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t : A S G L n + 1 , m , l , + .

Moreover, there exists ASGL n+1,m,l,+ of the form (3.1) such that infimum of (3.2) can be attained.

Proof.

The proof is similar to Lemma 2.2. We include details for the completion of the paper.

If u ≡ 0, (3.2) is trivial. So we only need to consider u ≢ 0. It is easy to see that there exists a 0 R n such that

R + n + 1 | u ( t , x ) , ( 1 , a 0 ) | 2 t α d x d t = inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t .

For any ASGL n+1,m,l,+, we have

R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t = R + n + 1 | x ( u ( λ m t , λ l B x + t a ) ) | 2 t α d x d t + R + n + 1 λ m u ( λ m t , λ l B x + t a ) ( λ m t ) + a , λ l B x + t a u ( λ m t , λ l B x + t a ) 2 t α d x d t = λ 2 l ln m α m R + n + 1 | x ( u ( t , B x ) ) | 2 t α d x d t + λ m m α ln R + n + 1 | u ( t , x ) , ( 1 , λ m a ) | 2 t α d x d t .

It follows from Lemma 2.1 that

(3.3) inf R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t : A S G L n + 1 , + = inf λ > 0 , a R n , det ( B ) = ± 1 λ 2 l ln m α m R + n + 1 | x ( u ( t , B x ) ) | 2 t α d x d t + λ m m α ln R + n + 1 | u ( t , x ) , ( 1 , λ m a ) | 2 t α d x d t = inf λ > 0 , a R n λ 2 l ln m α m inf det ( B ) = ± 1 R + n + 1 | x ( u ( t , B x ) ) | 2 t α d x d t + λ m m α ln R + n + 1 | u ( t , x ) , ( 1 , λ m a ) | 2 t α d x d t = inf λ > 0 λ 2 l ln m α m E α ( u ) 2 + λ m m α ln inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t = d n , α , β E α ( u ) m m α ln m l inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t m + m α + ln 2 l 2 ( m l ) .

Moreover, the infimum of the last equality in (3.3) can be arrived if and only if

λ = 2 l ln m α m ln + m α m E α ( u ) 2 inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t 1 2 ( m l ) .

The lemma is completed.□

Suppose m = 1, l = 1 + β n , the following corollary is an immediate consequence of Lemma 3.1.

Corollary 3.2.

Let n ≥ 1 and α, β satisfy (1.11). Then for all u D α 1,2 R + n + 1 , it holds

(3.4) d n , α , β E α ( u ) n ( 2 + β α ) n + β + 1 inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t n ( α β ) + 2 β + 2 2 ( n + β + 1 ) = inf R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t : A S G L n + 1 , + .

Moreover, there exists ASGL n+1,+ such that infimum of (3.4) can be attained.

Lemma 3.3.

Let n ≥ 1, α, β satisfy (1.11) and v D α 1,2 R + n + 1 be the extremal function given in Theorem 1.5. Then inequality (1.18) is invariant under the action of the elements in SL n+1,+.

Proof.

Let v be an extremal function of inequality (1.18). Denote v A (t, x) = v(A(t, x)) with ASGL n+1,+. An immediate computation has

R + n + 1 | v A ( t , x ) | p * t β d x d t 1 / p * = ( det ( B ) ) 1 / p * R + n + 1 | v ( t , x ) | p * t β d x d t 1 / p * ,

E α ( v A ) = λ n ( β α ) 2 β 2 2 n | det ( B ) | 1 n 1 2 E α ( v )

and

R + n + 1 | v , ( 1 , a ) | 2 t α d x d t = λ β α + 2 | det ( B ) | 1 R + n + 1 | u , ( 1 , a ) | 2 t α d x d t ,

which implies the invariance of (1.18).□

Proof of Theorem 1.5.

Let u D α 1,2 R + n + 1 and ASGL n+1,+ of the form (1.17). From (1.6), we have

C n + 1 , α , β R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t R + n + 1 | u ( A ( t , x ) ) | p * t β d x d t 2 / p * = R + n + 1 | u ( t , x ) | p * t β d x d t 2 / p * .

By Corollary 3.2, one can deduce that

C n + 1 , α , β d n , α , β E α ( u ) n ( 2 + β α ) n + β + 1 inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t n ( α β ) + 2 β + 2 2 ( n + β + 1 ) = C n + 1 , α , β inf R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t : A S G L n + 1 , + R + n + 1 | u ( t , x ) | p * t β d x d t 2 / p * .

Therefore, we have

R + n + 1 | u ( t , x ) | p * t β d x d t 1 / p * ( C n + 1 , α , β d n , α , β ) 1 2 E α ( u ) n ( 2 + β α ) 2 ( n + β + 1 ) inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t n ( α β ) + 2 β + 2 4 ( n + β + 1 ) .

The inequality (1.18) is completed.

Now we show the equality in (1.18) holds. Let

u * ( t , x ) = 1 ( t + μ ) 2 + | x x | 2 n + α 1 2 ψ ( t + μ , x x ) | t + μ | 2 + | x x | 2 1 2 μ , 0 ,

and ψ satisfies (1.9). By Lemma 3.3, one can see that (1.18) and E α ( u ) are invariant in SGL n+1,+. Then by the equality condition in (1.6), we obtain

u ( t , x ) = c u * ( A ( t , x ) ) ,

where c R , A S G L n + 1 , + . This implies that equality holds in (1.18).

Conversely, assume equality holds in (1.18) for all u ≢ 0. It follows from Lemma 2.1 and Corollary 3.2 that there exists ASGL n+1,+ such that

d n , α , β E α ( u ) n ( 2 + β α ) n + β + 1 inf a R n R + n + 1 | u ( t , x ) , ( 1 , a ) | 2 t α d x d t n ( α β ) + 2 β + 2 2 ( n + β + 1 ) = R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t .

Thus,

C n + 1 , α , β R + n + 1 | u ( A ( t , x ) ) | 2 t α d x d t = R + n + 1 | u ( A ( t , x ) ) | p * t β d x d t 2 / p * .

That is, uA is an extremal function of the weighted Sobolev inequality (1.6). By the uniqueness of extremal function, uA must be the form of

u ( A ( t , x ) ) = 1 ( t + μ ) 2 + | x x | 2 n + α 1 2 ψ ( t + μ , x x ) | t + μ | 2 + | x x | 2 1 2 μ , 0 = u * ( t , x ) .

That is,

u ( t , x ) = u * ( A 1 ( t , x ) ) .

This completes the proof of Theorem 1.5.□


Corresponding author: Jingbo Dou, School of Mathematics and Statistics, Shaanxi Normal University, Xi’an, Shaanxi, 710119, China, E-mail:

Funding source: National Natural Science Foundation of China

Award Identifier / Grant number: (Grant Nos. 12071269, 12101380)

Funding source: Youth Innovation Team of Shaanxi Universities

Funding source: Fundamental Research Funds for the Central Universities

Award Identifier / Grant number: (Grant No. GK202307001)

  1. Research ethics: This research did not involve the use of animals.

  2. Author contributions: Jingbo Dou raised the questions, designed the study and provided critical feedback on the manuscript. Yunyun Hu and Caihui Yue analyed and solved the questions, and wrote the manuscript. All authors read and approved the final manuscript.

  3. Competing interests: The authors declare that they have no competing interests.

  4. Research funding: This study was supported by the National Natural Science Foundation of China (Grant Nos. 12071269, 12101380), Youth Innovation Team of Shaanxi Universities and the Fundamental Research Funds for the Central Universities (Grant No. GK202307001).

  5. Data availability: All data generated or analysed during this study are included in this published article.

Appendix

In this section, we show the sharp affine weighted L 2 Sobolev inequality (1.12) which is stronger than the weighted L 2 Sobolev inequality (1.6). Moreover, we compute the best constants of (1.12) for two cases β = α − 1 and β = α.

We first prove the following lemma.

Lemma 4.1.

Let n ≥ 1 and α, β satisfy (1.11). Then, for all u D α 1,2 R + n + 1 , it holds

(4.1) E α ( u ) R + n + 1 | x u ( t , x ) | 2 t α d x d t 1 2 .

Moreover, inequality (1.12) is stronger than inequality (1.6).

Proof.

By Hölder’s inequality and Fubini’s theorem, we have

E α ( u ) = c n S n 1 R + n + 1 | x u ( t , x ) , ξ | 2 t α d x d t n 2 d ξ 1 n c n S n 1 R + n + 1 | x u ( t , x ) , ξ | 2 t α d x d t d ξ 1 n n 2 S n 1 1 d ξ 1 n n + 2 2 = ( n w n ) n + 2 2 n c n S n 1 | u 0 ξ | 2 d ξ R + n + 1 | x u ( t , x ) , ξ | 2 t α d x d t 1 2 = c n ( n w n ) 1 n n 1 2 R + n + 1 | x u ( t , x ) | 2 t α d x d t 1 2 = R + n + 1 | x u ( t , x ) | 2 t α d x d t 1 2 .

This completes the proof of inequality (4.1).

Plugging (4.1) into (1.12), and using Young’s inequality we have

R + n + 1 | u ( t , x ) | p * t β d x d t 2 / p * C n + 1 , α , β d n , α , β R + n + 1 | x u ( t , x ) | 2 t α d x d t n ( 2 + β α ) 2 ( n + β + 1 ) R + n + 1 u ( t , x ) t 2 t α d x d t n ( α β ) + 2 β + 2 2 ( n + β + 1 ) C n + 1 , α , β R + n + 1 | x u ( t , x ) | 2 t α d x d t + R + n + 1 u ( t , x ) t 2 t α d x d t = C n + 1 , α , β R + n + 1 | u ( t , x ) | 2 t α d x d t ,

which implies (1.12) is stronger than (1.6).□

Next, together with Theorem A, we compute the best constants of (1.12) for two cases β = α − 1 and β = α. From (1.12), we have

(4.2) H n + 1 , α , β = sup u D α 1,2 R + n + 1 \ { 0 } R + n + 1 | u ( t , x ) | p * t β d x d t 1 / p * E α ( u ) n ( 2 + β α ) 2 ( n + β + 1 ) u t L α 2 R + n + 1 n ( α β ) + 2 β + 2 2 ( n + β + 1 ) .

Lemma 4.2

  1. If β = α − 1, with α > 0 for n ≥ 2 or α ( 0 , 1 2 ] [ 1 + 17 4 , ) for n = 1, then

    (4.3) H n + 1 , α , α 1 = [ ( n + α ) ( n + 2 α 1 ) ] n + 2 α 4 ( n + α ) 2 α 1 2 n + 2 α + 1 n π ( n + 2 α ) n 4 ( n + α ) Γ ( n + 2 α ) Γ ( α ) Γ 2 n + α 1 2 ( n + α ) .

  2. If β = α with α > 0 for n ≥ 2 or α 2 for n = 1, then

    (4.4) H n + 1 , α , α = n + 2 α n π ( n + α 1 ) 2 n 2 ( n + α + 1 ) 1 ( 1 + α ) ( n + α 1 ) 1 + α 2 ( n + α + 1 ) Γ ( n + α + 1 ) Γ α + 1 2 Γ n + α + 1 2 1 n + α + 1 .

Proof.

For β = α − 1, we know that the extremal function of inequality (1.12) has the following form

u ( t , x ) = 1 ( 1 + λ t ) 2 + | λ α n B ( x x ) | 2 n + α 1 2 .

By the spherical polar coordinate transformation, we obtain

(4.5) R + n + 1 | u ( t , x ) | p * t α 1 d x d t = R + n + 1 1 ( 1 + λ t ) 2 + | λ α n B ( x x ) | 2 n + α t α 1 d x d t = det λ α n B 1 0 t α 1 d t R n 1 1 + | x | 2 ( 1 + λ t ) 2 n + α 1 1 + λ t 2 ( n + α ) d x = det ( B ) 1 ( n w n ) 0 1 1 + t n + 2 α t α 1 d t 0 1 1 + r 2 n + α r n 1 d r = n w n 2 det ( B ) 1 Γ ( α ) Γ n 2 Γ n 2 + α Γ ( n + 2 α ) .

Noting that

u ( t , x ) t = k n , α λ ( 1 + λ t ) 1 ( 1 + λ t ) 2 + | λ α n B ( x x ) | 2 n + α + 1 2

with k n,α = (1 − nα), then using the change of polar coordinates again we have

(4.6) R + n + 1 u ( t , x ) t 2 t α d x d t = k n , α 2 R + n + 1 λ 2 ( 1 + λ t ) 2 1 ( 1 + λ t ) 2 + | λ α n B ( x x ) | 2 n + α + 1 t α d x d t = k n , α 2 n w n λ 2 ( n + α ) ( n + α 1 ) det ( B ) 1 Γ ( α + 1 ) Γ n 2 Γ n 2 + α + 1 Γ ( n + 2 α ) .

A direct computation has

u ( t , x ) x i = λ α n k n , α 1 ( 1 + λ t ) 2 + | λ α n B ( x x ) | 2 n + α 2 b i , λ α n B ( x x ) ,

where b i is the ith column vector of the matrix B. It follows

R + n + 1 u ( t , x ) x i u ( t , x ) x j t α d x d t = λ 2 α n k n , α 2 R + n + 1 1 ( 1 + λ t ) 2 + | λ α n B ( x x ) | 2 n + α + 1 b i , λ α n B ( x x ) b j , λ α n B ( x x ) t α d x d t = λ 2 α n k n , α 2 d e t λ α n B 1 0 t α d t R n 1 ( 1 + λ t ) 2 + | x | 2 n + α + 1 b i , x b j , x d x d t = λ ( n 2 ) α n k n , α 2 d e t ( B ) 1 0 t α d t R n 1 ( 1 + λ t ) 2 + | x | 2 n + α + 1 k = 1 n b k i x k s = 1 n b s j x s d x = λ ( n 2 ) α n k n , α 2 d e t ( B ) 1 k = 1 n b k i b k j 0 t α d t R n 1 ( 1 + λ t ) 2 + | x | 2 n + α + 1 x k 2 d x = k = 1 n b k i b k j λ ( n 2 ) α n k n , α 2 n det ( B ) 1 0 t α d t R n 1 ( 1 + λ t ) 2 + | x | 2 n + α + 1 | x | 2 d x = k = 1 n b k i b k j λ n + 2 α n k n , α 2 n w n 2 n det ( B ) 1 Γ ( α + 1 ) Γ n 2 + 1 Γ n 2 + α Γ ( n + 2 α + 2 ) .

And then

det ( M [ u ] ) = λ n + 2 α n k n , α 2 n w n 2 n det ( B ) 1 Γ ( α + 1 ) Γ n 2 + 1 Γ n 2 + α Γ ( n + 2 α + 2 ) n d e t ( B B T ) .

From Lemma 2.1 in [24], we know

E α ( u ) 2 = n ( det ( M [ u ] ) ) 1 n .

Therefore,

(4.7) E α ( u ) n ( 2 + β α ) 2 ( n + β + 1 ) = n ( det ( M [ u ] ) ) 1 n n 4 ( n + α ) = λ n + 2 α n k n , α 2 n w n 2 n det ( B ) 2 n 1 Γ ( α + 1 ) Γ n 2 + 1 Γ n 2 + α Γ ( n + 2 α + 2 ) n 4 ( n + α ) .

Substituting (4.5), (4.6) and (4.7) into (4.2), we arrive at (4.3).

For β = α, the extremal function of inequality (1.12) has the following form

u ( t , x ) = 1 1 + ( λ t ) 2 + | λ 1 + α n B ( x x ) | 2 n + α 1 2 .

Similar to previous, we can obtain (4.4), so we omit the details.□

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Received: 2022-12-29
Accepted: 2023-11-14
Published Online: 2024-03-14

© 2024 the author(s), published by De Gruyter, Berlin/Boston

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

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