Abstract
Motivated by the work of Ghoussoub and Moradifam, we prove some improved Hardy inequalities on the Heisenberg group via Bessel function.
Mathematics Subject Classification (2000):
Primary 26D10
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1 Introduction
Hardy inequality in ℝ N reads, for all and N ≥ 3,
and is the best constant in (1.1) and is never achieved. A similar inequality with the same best constant holds in ℝ N is replaced by an arbitrary domain Ω ⊂ ℝ N and Ω contains the origin. Moreover, in case Ω ⊂ ℝ N is a bounded domain, Brezis and Vázquez [1] have improved it by establishing that for ,
where ω N and |Ω| denote the volume of the unit ball and Ω, respectively, and z0 = 2.4048... denotes the first zero of the Bessel function J0(z). Inequality (1.2) is optimal in case Ω is a ball centered at zero. Triggered by the work of Brezis and Vázquez (1.2), several Hardy inequalities have been established in recent years. In particular, Adimurthi et al.([2]) proved that, for , there exists a constant C n,k such that
where
log(1)(.) = log(.) and log(k)(.) = log(log(k-1)(.)) for k ≥ 2. Filippas and Tertikas ([3]) proved that, for , there holds
where D ≥ sup x∈Ω|x|,
for k ≥ 2 and is the best constant in (1.4) and is never achieved. More recently, Ghoussoub and Moradifam ([4]) give a necessary and sufficient condition on a radially symmetric potential V(|x|) on Ω that makes it an admissible candidate for an improved Hardy inequality. It states that the following improved Hardy inequality holds for , where B ρ = {x ∈ ℝ n :|x| < ρ},
if and only if the ordinary differential equation
has a positive solution on (0, ρ]. These include inequalities (1.2)-(1.4).
Motivated by the work of Ghoussoub and Moradifam ([4]), in this note, we shall prove similar improved Hardy inequality on the Heisenberg group . Recall that the Heisenberg group is the Carnot group of step two whose group structure is given by
The vector fields
(j = 1,..., n) are left invariant and generate the Lie algebra of . The horizontal gradient on is the (2n) -dimensional vector given by
where is a skew symmetric and orthogonal matrix given by
For more information about , we refer to [5–8]. To this end we have:
Theorem 1.1
Let B R = {x ∈ ℝ2n: |x| < R} and Ω H = B R × ℝ ∈ . Let V(|x|) be a radially symmetric decreasing nonnegative function on B R . If the ordinary differential equation
has a positive solution on (0, R], then the following improved Hardy inequality holds for
and the constant (n - 1)2in (1.6) is sharp in the sense of
Corollary 1.2
There holds, for,
and the constant 1/4 is sharp in the sense of
Corollary 1.3
There holds, for and D ≥ R,
and the constant 1/4 is sharp in the sense of
2 Proof
To prove the main result, we first need the following preliminary result.
Lemma 2.1
Let B R = {x ∈ ℝ2n: |x| < R} and V(|x|) be a radially symmetric decreasing nonnegative function on B R . If the ordinary differential equation
has a positive solution on (0, R], then the following improved Hardy inequality holds for,
where r = |x| andis the radial derivation.
Proof
Observe that if f is radial, i.e., f(x) = f(r), then |∇ f| = |∂ r f|. By inequality (1.5), inequality (2.1) holds.
Now let . If we extend f as zero outside B R , we may consider . Decomposing f into spherical harmonics we get (see e.g., [9])
where ϕ k (σ) are the orthonormal eigenfunctions of the Laplace-Beltrami operator with responding eigenvalues
The functions f k (r) belong to , satisfying f k (r) = O(rk ) and as r → 0. So
and
Note that if f is radial, then inequality (2.1) holds. We have, since ,
Therefore, by (2.2) and (2.3),
This completes the proof of lemma 2.1.
Proof of Theorem 1.1
Recall that the horizontal gradient on is the (2n)-dimensional vector given by
where is a skew symmetric and orthogonal matrix given by
Therefore, for any ,
Here we use the fact ⟨x, Λx⟩ = 0 since Λ is a skew symmetric matrix.
Since , for every t ∈ ℝ,. By Lemma 2.1,
Integrating both sides of the inequality (2.5) with respect to t, we have,
By (2.4) and the pointwise Schwartz inequality, we have
Therefore, we obtain, by (2.6)
To see the constant (n - 1)2 is sharp, we choose u(x, t) = ϕ(|x|)w(t) with and . Since ϕ is radial, we have
Here we use the fact |Λx| = |x| since Λ is a orthogonal matrix. Therefore,
Since
we obtain
The proof of Theorem 1.1 is completed.
Proof of Corollary 1.2
By Theorem 1.1 and following [4], it is enough to show the constant 1/4 is sharp. Choose u(x, t) = ϕ(|x|)w(t) with and . By the proof of Theorem 1.1,
Therefore,
Since
we have
Here we use the fact that the sharp constant in inequality (1.3) is 1/4 (see [4]). This completes the proof of Corollary 1.2.
Proof of Corollary 1.3
The proof is similar to that of Corollary 1.2 and it is enough to show the constant 1/4 is sharp. Choose u(x, t) = ϕ(|x|)w(t) with and . Then
Therefore,
Thus
since
This completes the proof of Corollary 1.3.
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The authors thanks the referee for his/her careful reading and very useful comments which improved the final version of this paper.
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YX designed and performed all the steps of proof in this research and also wrote the paper. All authors read and approved the final manuscript.
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Xiao, YX. An improved Hardy type inequality on Heisenberg group. J Inequal Appl 2011, 38 (2011). https://doi.org/10.1186/1029-242X-2011-38
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DOI: https://doi.org/10.1186/1029-242X-2011-38