Open Access

Hardy-type inequalities on a half-space in the Heisenberg group

Journal of Inequalities and Applications20132013:291

https://doi.org/10.1186/1029-242X-2013-291

Received: 30 April 2012

Accepted: 23 May 2013

Published: 10 June 2013

Abstract

We prove some Hardy-type inequalities on half-spaces for Kohn’s sub-Laplacian in the Heisenberg group. Furthermore, the constants we obtained are sharp.

MSC:26D10, 35H20.

Keywords

Hardy inequalityHeisenberg groupsharp constant

1 Introduction

The Hardy inequality in R N reads that for all u C 0 ( R N ) and N 3 ,
R N | u | 2 d x ( N 2 ) 2 4 R N u 2 | x | 2 d x
(1.1)
and the constant ( N 2 ) 2 4 in (1.1) is sharp. Recently, it has been proved by Nazarov ([1], Proposition 4.1, see also [2]) that the following Hardy inequality is valid for f C 0 ( R + N ) :
R + N | u ( x ) | 2 d x N 2 4 R + N u ( x ) 2 | x | 2 d x ,
(1.2)

where R + N = { ( x 1 , , x n ) | x 1 > 0 } , and the constant N 2 4 is sharp. This shows that the Hardy constant jumps from ( N 2 ) 2 4 to N 2 4 when the singularity of the potential reaches the boundary. For more information about this inequality and its applications, we refer to [310] and the references therein.

The aim of this note is to prove an analogous Hardy-type inequality on a half-space for Kohn’s sub-Laplacian in Heisenberg groups H n . It has been proved by D’Ambrosio ([11], Theorem 3.3) that for u C 0 ( H n ) , the following holds:
H n | H u | 2 d x d t ( n 1 ) 2 H n u 2 | x | 2 d x d t ,
(1.3)

where H is the horizontal gradient associated with Kohn’s sub-Laplacian on H n (for details, see Section 2). Furthermore, the constant ( n 1 ) 2 in (1.3) is sharp (see [12], Theorem 3.13). In this note we shall show that when the singularity is on the boundary, the Hardy constant also jumps. In fact, we have the following.

Theorem 1.1 For all u C 0 ( H + n ) , the following holds:
H + n | H u | 2 d x d t n 2 H + n u 2 | x | 2 d x d t ,
(1.4)

where H + n = { ( x , t ) H n : x 1 > 0 } , and the constant n 2 in (1.4) is sharp.

In order to prove Theorem 1.1, we use a new technique which is different from that in [1, 2]. In fact, it seems that the method used in [1, 2] cannot be applied to Kohn’s sub-Laplacian.

With the same technique, we obtain the following sharp Hardy inequality on H k + n = { ( x , t ) H n : x 1 > 0 , , x k > 0 } .

Theorem 1.2 Let 1 k 2 n . For all u C 0 ( H k + n ) , the following holds:
H k + n | H u | 2 d x d t ( n + k 1 ) 2 H k + n u 2 | x | 2 d x d t .
(1.5)

Furthermore, the constant ( n + k 1 ) 2 in (1.5) is sharp.

2 Proofs

Let H n = ( R 2 n × R , ) be the ( 2 n + 1 ) -dimensional Heisenberg group whose group structure is given by
( x , t ) ( x , t ) = ( x + x , t + t + 2 j = 1 n ( x 2 j x 2 j 1 x 2 j 1 x 2 j ) ) .
The vector fields
X 2 j 1 = x 2 j 1 + 2 x 2 j t , X 2 j = x 2 j 2 x 2 j 1 t
( j = 1 , , n ) are left invariant and generate the Lie algebra of H n . Kohn’s sub-Laplace on H n is
Δ H = j = 1 2 n X j 2 = j = 1 2 n 2 x j 2 + 4 | x | 2 2 t 2 + 4 j = 1 n ( x 2 j x 2 j 1 x 2 j 1 x 2 j ) t
and the horizontal gradient is the ( 2 n ) -dimensional vector given by
H = ( X 1 , , X 2 n ) = x + 2 Λ x t ,
where x = ( x 1 , , x 2 n ) , Λ is a skew symmetric and orthogonal matrix given by
Λ = diag ( J 1 , , J n ) , J 1 = = J n = ( 0 1 1 0 ) .
By the definition of H , we have, for α R and | x | 0 ,
Δ H ( x 1 | x | α ) = j = 1 2 n 2 x j 2 ( x 1 | x | α ) = j = 2 2 n 2 x j 2 ( x 1 | x | α ) + 2 x 1 2 ( x 1 | x | α ) = x 1 j = 2 2 n 2 x j 2 | x | α + x 1 2 x 1 2 | x | α + 2 | x | α x 1 = x 1 j = 1 2 n 2 x j 2 | x | α + 2 | x | α x 1 = α ( 2 n + α ) x 1 | x | α 2 .
(2.1)
Similarly,
Δ H ( | x | α i = 1 k x i ) = j = 1 2 n 2 x j 2 ( | x | α i = 1 k x i ) = j = k + 1 2 n 2 x j 2 ( | x | α i = 1 k x i ) + l = 1 k 2 x l 2 ( | x | α i = 1 k x i ) = i = 1 k x i j = 1 2 n 2 x j 2 | x | α + 2 k | x | α 2 i = 1 k x i = α ( 2 n + 2 k + α 2 ) | x | α 2 i = 1 k x i .
(2.2)
Proof of Theorem 1.1 Using the substitution u = x 1 | x | n f , we get
H + n | H u | 2 = H + n [ | H ( x 1 | x | n ) | 2 f 2 + | H f | 2 x 1 2 | x | 2 n + H ( x 1 2 | x | 2 n ) , H f 2 2 ] H + n ( | H ( x 1 | x | n ) | 2 f 2 + H ( x 1 2 | x | 2 n ) , H f 2 2 ) = H + n f 2 ( | H ( x 1 | x | n ) | 2 1 2 Δ H ( x 1 2 | x | 2 n ) ) .
Using the following identity, for g C 2 ( H n ) ,
1 2 Δ H g 2 = 1 2 j = 1 2 n X j 2 g 2 = g j = 1 2 n X j 2 g + j = 1 m | X j g | 2 = g Δ H g + | H g | 2 ,
(2.3)
we have, by (2.1),
| H ( x 1 | x | n ) | 2 1 2 Δ H ( x 1 2 | x | 2 n ) = x 1 | x | n Δ H ( x 1 | x | n ) = x 1 | x | n ( n ) n x 1 | x | n 2 = n 2 x 1 2 | x | 2 n 2 .
Therefore,
H + n | H u | 2 d x d t H + n f 2 ( | H ( x 1 | x | n ) | 2 1 2 Δ H ( x 1 2 | x | 2 n ) ) d x d t = n 2 H + n f 2 x 1 2 | x | 2 n 2 d x d t = n 2 H + n u 2 | x | 2 d x d t .
(2.4)
Now we show the constant n 2 in (1.4) is sharp. Choosing
g ( x , t ) = ϕ ( x ) ω ( t ) ,
where ϕ C 0 ( R + 2 n ) and ω C 0 ( R ) , we have
H g ( x , t ) = x g ( x , t ) + 2 Λ x t g ( x , t ) = ω ( t ) x ϕ ( x ) + 2 ϕ ( x ) ω ( t ) Λ x .
Therefore,
| H g ( x , t ) | 2 = ω ( t ) x ϕ ( x ) + 2 ϕ ( x ) ω ( t ) Λ x , ω ( t ) x ϕ ( x ) + 2 ϕ ( x ) ω ( t ) Λ x = ω 2 ( t ) | x ϕ | 2 + 4 ϕ 2 | x | 2 | ω ( t ) | 2 + 4 ω ( t ) ω ( t ) ϕ Λ x , x ϕ .
(2.5)

To get the last equation, we use the fact | Λ x | 2 = | x | 2 .

Since
+ ω ( t ) ω ( t ) d t = 1 2 + d ω 2 ( t ) d t d t = 0 ,
we have, by (2.5),
H + n | H g ( x , t ) | 2 d x d t H + n g 2 | x | 2 d x d t = H + n ω 2 ( t ) | x ϕ | 2 d x d t + 4 H + n ϕ 2 | x | 2 | ω ( t ) | 2 d x d t R + 2 n ϕ 2 | x | 2 d x R ω 2 d t = R + 2 n | x ϕ | 2 d x R + 2 n ϕ 2 | x | 2 d x + 4 R | ω ( t ) | 2 d t R ω 2 d t R + 2 n ϕ 2 | x | 2 d x R + 2 n ϕ 2 | x | 2 d x .
Notice that
inf ω C 0 ( R ) { 0 } R | ω ( t ) | 2 d t R ω 2 d t = 0 ,
(2.6)
we have
inf u C 0 ( H + n ) { 0 } H + n | H u | 2 d x d t H + n u 2 | x | 2 d x d t inf ϕ C 0 ( R 2 n ) { 0 } R + 2 n | x ϕ | 2 d x R + 2 n ϕ 2 | x | 2 d x = n 2 .

Here we use the sharp Hardy inequality (1.2). This completes the proof of Theorem 1.1. □

Proof of Theorem 1.2 The proof is similar to that of Theorem 1.1. Using the substitution u = f | x | n k + 1 i = 1 k x i , we get
H k + n | H u | 2 = H k + n [ | H ( | x | n k + 1 i = 1 k x i ) | 2 f 2 + | H f | 2 i = 1 k x i 2 | x | 2 n + 1 2 H ( | x | 2 n 2 k + 2 i = 1 k x i 2 ) , H f 2 ] H k + n ( | H ( | x | n k + 1 i = 1 k x i ) | 2 f 2 + 1 2 H ( | x | 2 n 2 k + 2 i = 1 k x i 2 ) , H f 2 ) = H k + n f 2 ( | H ( | x | n k + 1 i = 1 k x i ) | 2 1 2 Δ H ( | x | 2 n 2 k + 2 i = 1 k x i 2 ) ) .
Using the identities (2.3) and (2.2), we have
| H ( | x | n k + 1 i = 1 k x i ) | 2 1 2 Δ H ( | x | 2 n 2 k + 2 i = 1 k x i 2 ) = | x | n k + 1 i = 1 k x i Δ H ( | x | n k + 1 i = 1 k x i ) = ( n + k 1 ) 2 | x | 2 n 2 k i = 1 k x i 2 .
Therefore,
H k + n | H u | 2 H k + n f 2 ( | H ( | x | n k + 1 i = 1 k x i ) | 2 1 2 Δ H ( | x | 2 n 2 k + 2 i = 1 k x i 2 ) ) = ( n + k 1 ) 2 H k + n f 2 | x | 2 n 2 k i = 1 k x i 2 = ( n + k 1 ) 2 H k + n u 2 | x | 2 .
To see the constant ( n + k 1 ) 2 in (1.5) is sharp, we consider the function
h ( x , t ) = ψ ( x ) ω ( t ) ,
where ψ C 0 ( R k + 2 n ) and ω C 0 ( R ) . Here we denote by R k + 2 n = { x R 2 n : x 1 > 0 , , x k > 0 } . Then
| H h ( x , t ) | 2 = ω ( t ) x ψ ( x ) + 2 ψ ( x ) ω ( t ) Λ x , ω ( t ) x ψ ( x ) + 2 ψ ( x ) ω ( t ) Λ x = ω 2 ( t ) | x ψ | 2 + 4 ψ 2 | x | 2 | ω ( t ) | 2 + 4 ω ( t ) ω ( t ) ψ Λ x , x ψ
and
H k + n | H h ( x , t ) | 2 d x d t = H k + n ( ω 2 ( t ) | x ψ | 2 + 4 ψ 2 | x | 2 | ω ( t ) | 2 ) d x d t + 4 R k + 2 n ψ Λ x , x ψ d x R ω ( t ) ω ( t ) d t = H k + n ( ω 2 ( t ) | x ψ | 2 + 4 ψ 2 | x | 2 | ω ( t ) | 2 ) d x d t + 4 R k + 2 n ψ Λ x , x ψ d x 1 2 R d ω 2 ( t ) = H k + n ( ω 2 ( t ) | x ψ | 2 + 4 ψ 2 | x | 2 | ω ( t ) | 2 ) d x d t .
Therefore,
H k + n | H h ( x , t ) | 2 d x d t H k + n h 2 | x | 2 d x d t = H k + n ω 2 ( t ) | x ψ | 2 d x d t + 4 H k + n ψ 2 | x | 2 | ω ( t ) | 2 d x d t R k + 2 n ψ 2 | x | 2 d x R ω 2 d t = R + 2 n | x ψ | 2 d x R k + 2 n ψ 2 | x | 2 d x + 4 R | ω ( t ) | 2 d t R ω 2 d t R k + 2 n ψ 2 | x | 2 d x R k + 2 n ψ 2 | x | 2 d x .
Thus, by (2.6),
inf u C 0 ( H k + n ) { 0 } H k + n | H u | 2 d x d t H k + n u 2 | x | 2 d x d t inf ψ C 0 ( R 2 n ) { 0 } R + 2 n | x ψ | 2 d x R k + 2 n ψ 2 | x | 2 d x = ( n + k 1 ) 2 .
Here we use the sharp Hardy inequality ([9], Theorem 1.1)
R + 2 n | f | 2 d x ( n + k 1 ) 2 R + 2 n f 2 | x | 2 d x .

The proof of Theorem 1.2 is therefore completed. □

Declarations

Acknowledgements

The first author is supported by the National Natural Science Foundation of China (No. 11171259) and the second author is supported by the National Natural Science Foundation of China (No. 11201346).

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Wuhan University, Wuhan, People’s Republic of China

References

  1. Nazarov AI: Hardy-Sobolev inequalities in a cone. J. Math. Sci. 2006, 132(4):419–427. 10.1007/s10958-005-0508-1MathSciNetView ArticleGoogle Scholar
  2. Filippas S, Tertikas A, Tidblom J: On the structure of Hardy-Sobolev-Maz’ya inequalities. J. Eur. Math. Soc. 2009, 11(6):1165–1185.MathSciNetView ArticleGoogle Scholar
  3. Caldiroli P, Musina R: Stationary states for a two-dimensional singular Schrödinger equation. Boll. Unione Mat. Ital, B 2001, 4(3):609–633.MathSciNetGoogle Scholar
  4. Caldiroli P, Musina R: Rellich inequalities with weights. Calc. Var. Partial Differ. Equ. 2012, 45(1–2):147–164. doi:10.1007/s00526–011–0454–3 10.1007/s00526-011-0454-3MathSciNetView ArticleGoogle Scholar
  5. Cazacu C: On Hardy inequalities with singularities on the boundary. C. R. Math. Acad. Sci. Paris, Sér. I 2011, 349: 273–277. 10.1016/j.crma.2011.02.005MathSciNetView ArticleGoogle Scholar
  6. Cazacu, C: Hardy inequalities with boundary singularities. arXiv:1009.0931Google Scholar
  7. Fall, MM: On the Hardy-Poincaré inequality with boundary singularities. arXiv:1008.4785Google Scholar
  8. Fall MM, Musina R: Hardy-Poincaré inequalities with boundary singularities. Proc. R. Soc. Edinb., Sect. A, Math. 2012, 142(4):769–786. 10.1017/S0308210510000740MathSciNetView ArticleGoogle Scholar
  9. Su D, Yang Q-H:On the best constants of Hardy inequality in R n k × ( R + ) k and related improvements. J. Math. Anal. Appl. 2012, 389: 48–53. 10.1016/j.jmaa.2011.11.033MathSciNetView ArticleGoogle Scholar
  10. Yang Q: Hardy type inequalities related to Carnot-Carathéodory distance on the Heisenberg group. Proc. Am. Math. Soc. 2013, 141(1):351–362.View ArticleGoogle Scholar
  11. D’Ambrosio L: Some Hardy inequalities on the Heisenberg group. Differ. Equ. 2004, 40: 552–564.MathSciNetView ArticleGoogle Scholar
  12. D’Ambrosio L: Hardy-type inequalities related to degenerate elliptic differential operators. Ann. Sc. Norm. Super. Pisa, Cl. Sci. 2005, IV: 451–486.MathSciNetGoogle Scholar

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© Liu and Luan; licensee Springer 2013

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