Journal of Inequalities in Pure and
Applied Mathematics
SOME HARDY TYPE INEQUALITIES IN THE HEISENBERG GROUP
YAZHOU HAN AND PENGCHENG NIU
Department of Applied Mathematics,
Northwestern Polytechnical University,
Xi’an, Shaanxi, 710072,
volume 4, issue 5, article 103,
2003.
Received 21 January, 2003;
accepted 10 June, 2003.
Communicated by: L. Pick
P.R. China.
EMail: taoer558@163.com
EMail: pengchengniu@yahoo.com.cn
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
Some Hardy type inequalities on the domain in the Heisenberg group are established by using the Picone type identity and constructing suitable auxiliary
functions.
2000 Mathematics Subject Classification: 35H20.
Key words: Hardy inequality, Picone identity, Heisenberg group.
Some Hardy Type Inequalities in
the Heisenberg Group
The research supported in part by National Nature Science Foundation and Natural
Science Foundation of Shaanxi Province, P.R. China.
Yazhou Han and Pengcheng Niu
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Hardy Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
Title Page
3
4
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J. Ineq. Pure and Appl. Math. 4(5) Art. 103, 2003
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1.
Introduction
The Hardy inequality in the Euclidean space (see [3], [4], [7]) has been established using many methods. In [1], Allegretto and Huang found a Picone’s identity for the p-Laplacian and pointed out that one can prove the Hardy inequality
via the identity. Niu, Zhang and Wang in [6] obtained a Picone type identity for
the p-sub-Laplacian in the Heisenberg group and then established a Hardy type
inequality. When p = 2, the result of [6] coincides with the inequality in [2]. As
stated in [1], the Picone type identity allows us to avoid postulating regularity
conditions on the boundary of the domain under consideration. Since there is a
presence of characteristic points in the sub-Laplacian Dirichlet problem in the
Heisenberg group (see [2]), we understand that such an identity is especially
useful.
We recall that the Heisenberg group Hn of real dimension N = 2n + 1, n ∈
N , is the nilpotent Lie group of step two whose underlying manifold is R2n+1 .
A basis for the Lie algebra of left invariant vector fields on Hn is given by
∂
∂
∂
∂
Xj =
+ 2yj , Yj =
− 2xj ,
j = 1, 2, . . . , n.
∂xj
∂t
∂yj
∂t
The number Q = 2n + 2 is the homogeneous dimension of Hn . There exists a
Heisenberg distance
n
o1
2 4
0 0
0 2
0 2 2
0
0
0
d ((z, t), (z , t )) = (x − x ) + (y − y ) + [t − t − 2(x · y − x · y)]
between (z, t) and (z 0 , t0 ). We denote the Heisenberg gradient by
∇Hn = (X1 , . . . , Xn , Y1 , . . . , Yn ).
In this note we give some Hardy type inequalities on the domain in the
Heisenberg group by considering different auxiliary functions.
Some Hardy Type Inequalities in
the Heisenberg Group
Yazhou Han and Pengcheng Niu
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2.
Hardy Inequalities
First we state two lemmas given in [6] which will be needed in the sequel.
Lemma 2.1. Let Ω be a domain in Hn , v > 0, u ≥ 0 be differentiable in Ω.
Then
(2.1)
L(u, v) = R(u, v) ≥ 0,
where
up−1
up
L(u, v) = |∇Hn u|p + (p − 1) p |∇Hn v|p − p p−2 ∇Hn · |∇Hn v|p−2 ∇Hn v,
v
vp u
R(u, v) = |∇Hn u|p − ∇Hn
· |∇Hn v|p−2 ∇Hn v.
p−1
v
Denote the p-sub-Laplacian by ∆Hn ,p v = ∇Hn · (|∇Hn v|p−2 ∇Hn v).
Lemma 2.2. Assume that the differentiable function v > 0 satisfies the condition −∆Hn ,p v ≥ λgv p−1 , for some λ > 0 and nonnegative function g. Then for
every u ∈ C0∞ (Ω), u ≥ 0,
Z
Z
p
(2.2)
|∇Hn u| ≥ λ g|u|p .
Ω
Ω
Let BR = { (z, t) ∈ Hn | d ((z, t), (0, 0)) < R} be the Heisenberg group and
δ(z, t) = dist ((z, t), ∂BR ) , (z, t) ∈ BR , in the sense of distance functions on
the Heisenberg group.
Some Hardy Type Inequalities in
the Heisenberg Group
Yazhou Han and Pengcheng Niu
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Theorem 2.3. Let Ω = BR \{(0, 0}, p > 1. Then for every u ∈ C0∞ (Ω),
p Z
Z
|z|p |u|p
p−1
p
,
(2.3)
|∇Hn u| ≥
p δp
p
Ω d
Ω
p
where |z| = x2 + y 2 , d = d ((z, t), (0, 0)) .
Proof. We first consider u ≥ 0. The following equations are evident:

Xj d = d−3 (|z|2 xj + yj t) , Yj d = d−3 (|z|2 yj − xj t) ,





 2
2
Xj d = −3d−7 (|z|2 xj + yj t) + d−3 |z|2 + 2x2j + 2yj2 ,
(2.4)


2
2
−7
2
−3
2
2
2

Y
d
=
−3d
(|z|
y
−
x
t)
+
d
|z|
+
2x
+
2y

j
j
j
j ,

 j
j = 1, . . . , n
and
Some Hardy Type Inequalities in
the Heisenberg Group
Yazhou Han and Pengcheng Niu
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(2.5)
|∇Hn d| = |z|d−1 , ∆Hn d = (Q − 1)d−3 |z|2 .
Choose v(z, t) = δ(z, t)β = (R − d)β , in which β =
p−1
,
p
one has
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Xj v = −βδ β−1 Xj d, Yj v = −βδ β−1 Yj d, j = 1, . . . , n,
∇Hn v = −βδ β−1 ∇Hn d, |∇Hn v| = |β|δ β−1 |z|d−1 ,
and
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−∆Hn v = −∇Hn · |∇Hn v|p−2 ∇Hn v
= −∇Hn · −β|β|p−2 δ (β−1)(p−1) |z|p−2 d2−p ∇Hn d
J. Ineq. Pure and Appl. Math. 4(5) Art. 103, 2003
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p−2
= β|β|
− (β − 1)(p − 1)δ (β−1)(p−1)−1 |z|p−2 d2−p |∇Hn d|2
+ δ (β−1)(p−1) d2−p ∇Hn |z|p−2 · ∇Hn d
+ (2 − p)δ (β−1)(p−1) |z|p−2 d1−p |∇Hn d|2
(β−1)(p−1)
p−2 2−p
+δ
|z| d 4Hn d .
From the fact ∇Hn (|z|p−2 ) · ∇Hn d = (p − 2)|z|p−4 d−3 |z|4 = (p − 2)|z|p d−3
and (2.5), it follows that
p−2
−∆Hn v = β|β|
− (β − 1)(p − 1)δ (β−1)(p−1)−1 |z|p d−p
+ (p − 2)δ (β−1)(p−1) |z|p d−1−p
− (p − 2)δ
(β−1)(p−1)
Some Hardy Type Inequalities in
the Heisenberg Group
Yazhou Han and Pengcheng Niu
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p −1−p
|z| d
Contents
+ (Q − 1)δ (β−1)(p−1) |z|p d−1−p
δ |z|p v p−1
p−2
= β|β|
−(β − 1)(p − 1) + (Q − 1)
d dp δ p
p−1 p p−1
p−1
p−1
δ |z| v
=
+ (Q − 1)
p
p
d dp δ p
p p p−1
p−1
|z| v
≥
.
p
dp δ p
The desired inequality (2.3) is obtained by Lemma 2.2. For general u, by letting
u = u+ − u− , we directly obtain (2.3).
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Theorem 2.4. Let Ω = Hn \{BHn ,R }, Q > p > 1. Then for every u ∈ C0∞ (Ω),
there exists a constant C Z> 0, such that Z
|z|p |u|p
(2.6)
|∇Hn u|p ≥ C
.
dp d2p
Ω
Ω α
Proof. Suppose that u ≥ 0. Take v = log Rd , R < d = d ((z, t), (0, 0)) <
+∞, α < 0. Using (2.4) and (2.5) show that
α α−1
d
1
R
α
∇Hn v =
α
∇Hn d = ∇Hn d,
d
R
R
d
−2
|∇Hn v| = |α||z|d ,
−4Hn v = −∇Hn · |∇Hn v|p−2 ∇Hn v
= −α|α|p−2 ∇Hn · |z|p−2 d−2(p−2)−1 ∇Hn d
p−2
= −α|α|
(p − 2)|z|p−3 d2(2−p)−1 ∇Hn (|z|) · ∇Hn d
+ (2(2 − p) − 1) |z|p−2 d2(1−p) |∇Hn d|2
p−2 2(2−p)−1
+ |z| d
4Hn d .
Some Hardy Type Inequalities in
the Heisenberg Group
Yazhou Han and Pengcheng Niu
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Since ∇Hn (|z|) · ∇Hn d = |z|3 d−3 , the last equation above becomes
p−2
−4Hn v = −α|α|
(p − 2)|z|p−3 d2(2−p)−1 |z|3 d−3
+ (3 − 2p)|z|p−2 d2(1−p) |z|2 d−2
p−2 3−2p
2 −3
+ (Q − 1)|z| d
|z| d
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(2.7)
= −α|α|p−2 |z|p d−2p (p − 2 + 3 − 2p + Q − 1)
= −α|α|p−2 (Q − p)|z|p d−2p .
Noting
v p−1
= 0,
d→+∞ dp
p−1
there exists a positive number M ≥ R, such that v dp < 1, for d > M . Since
v p−1
is continuous on the interval [R, M ], we find a constant C 0 > 0, such that
dp
v p−1
< C 0 . Pick out C 00 = max{C 0 , 1} and one has v p−1 < C 00 dp in Ω. This
dp
leads to the following
|z|p v p−1
−∆Hn v ≥ C 2p p ,
d
d
p−2
where C = −α|α| C 00 (Q−p) , and to (2.6) by Lemma 2.2. A similar treatment for
general u completes the proof.
lim
In particular, α = p − Q (1 < p < Q) satisfies the assumption in the proof
above.
Theorem 2.5. Let Ω be as defined in Theorem 2.4 and p ≥ Q. Then there exists
a constant C > 0, such that for every u ∈ C0∞ (Ω),
Z
Z
|z|p
|u|p
p
.
(2.8)
|∇Hn u| ≥ C
p
p log d
dp
Ω
Ω d
R
Proof. It is sufficient to show that (2.8) holds for u ≥ 0. Choose v = φα , φ =
log Rd , where R < d < +∞, 0 < α < 1. We know that from (2.4) and (2.5),
−1
Some Hardy Type Inequalities in
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−2
∇Hn φ = d ∇Hn d, |∇Hn φ| = d |z|,
∆Hn φ = d−1 4Hn d − d−2 |∇Hn d|2 = (Q − 2)|z|2 d−4 .
J. Ineq. Pure and Appl. Math. 4(5) Art. 103, 2003
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This allows us to obtain
−∆Hn v = −∇Hn · |∇Hn v|p−2 ∇Hn v
= −∇Hn · |αφα−1 ∇Hn φ|p−2 αφα−1 ∇Hn φ
= −α|α|p−2 ∇Hn · φ(α−1)(p−1) |z|p−2 d2(2−p) ∇Hn φ
p−2
= −α|α|
(α − 1)(p − 1)φ(α−1)(p−1)−1 |z|p−2 d2(2−p) |∇Hn φ|2
+ (p − 2)φ(α−1)(p−1) |z|p−3 d2(2−p) ∇Hn (|z|) · ∇Hn φ
+ 2(2 − p)φ(α−1)(p−1) |z|p−2 d2(2−p)−1 ∇Hn d · ∇Hn φ
(α−1)(p−1)
p−2 2(2−p)
+φ
|z| d
4Hn φ
p−2
= −α|α|
(α − 1)(p − 1)φ(α−1)(p−1)−1 |z|p−2 d2(2−p) |z|2 d−4
+ (p − 2)φ(α−1)(p−1) |z|p−3 d2(2−p) |z|3 d−4
(α−1)(p−1)
+ 2(2 − p)φ
p−2 2(2−p)−1
|z|
2 −3
|z| d
(α−1)(p−1)
p−2 2(2−p)
2 −4
+φ
|z| d
(Q − 2)|z| d
d
v p−1 |z|p
{(α − 1)(p − 1) + (p − 2)φ
φp d2p
+2(2 − p)φ + (Q − 2)φ}
p−1
|z|p
v
= −α|α|p−2 p 2p {(α − 1)(p − 1) + (Q − p)φ} ,
φ d
= −α|α|p−2
Some Hardy Type Inequalities in
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Taking into account that 0 < α < 1 and p ≥ Q, we have
−α|α|p−2 (Q − p)φ ≥ 0,
and therefore
−4Hn v ≥ −α|α|p−2 (α − 1)(p − 1)
v p−1 |z|p
v p−1 |z|p
=
C
,
φp d2p
φp d2p
where C = −α|α|p−2 (α − 1)(p − 1). An application of Lemma 2.2 completes
the proof of Theorem 2.5.
Some Hardy Type Inequalities in
the Heisenberg Group
Yazhou Han and Pengcheng Niu
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References
[1] W. ALLEGRETTO AND Y.X. HUANG, A Picone’s identity for the pLaplacian and applications, Nonlinear Anal., 32 (1998), 819–830.
[2] N. GAFOFALO AND E. LANCONELLI, Frequency functions on the
Heisenberg group, the uncertainty priciple and unique continuation, Ann.
Inst. Fourier (Grenoble), 40 (1990), 313–356.
[3] G.H. HARDY, Note on a Theorem of Hilbert, Math. Zeit., 6 (1920), 314–
317.
[4] G.H. HARDY, An inequality betweeen integrals, Messenger of Mathematics, 54 (1925), 150–156.
[5] D. JERISON, The Dirichlet problem for the Laplacian on the Heisenberg
group, I, II, Journal of Functional Analysis, 43 (1981), 97–141; 43 (1981),
224–257.
[6] P. NIU, H. ZHANG AND Y. WANG, Hardy type and Rellich type inequalities on the Heisenberg group, Proc. A.M.S., 129 (2001), 3623–3630.
[7] A. WANNEBO, Hardy inequalities, Proc. A.M.S., 109 (1990), 85–95.
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Yazhou Han and Pengcheng Niu
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