Open Access

An Improved Hardy-Rellich Inequality with Optimal Constant

Journal of Inequalities and Applications20092009:610530

DOI: 10.1155/2009/610530

Received: 25 May 2009

Accepted: 11 September 2009

Published: 27 September 2009

Abstract

We show that a Hardy-Rellich inequality with optimal constants on a bounded domain can be refined by adding remainder terms. The procedure is based on decomposition into spherical harmonics.

1. Introduction

Hardy inequality in reads, for all and ,
(1.1)
and is the best constant in (1.1) and is never achieved. A similar inequality with the same best constant holds if is replaced by an arbitrary domain and contains the origin. Moreover, Brezis and Vázquez [1] have improved it by establishing that for ,
(1.2)

where and denote the volume of the unit ball and , respectively, and is the first eigenvalue of the Dirichlet Laplacian of the unit disc in . In case is a ball centered at zero, the constant in (1.2) is sharp.

Similar improved inequalities have been recently proved if instead of (1.1) one considers the corresponding Hardy inequalities. In all these cases a correction term is added on the right-hand side (see, e.g., [24]).

On the other hand, the classical Rellich inequality states that, for ,
(1.3)
and is the best constant in (1.3) and is never achieved (see [5]). And, more recently, Tertikas and Zographopoulos [6] obtained a stronger version of Rellich's inequality. That is, for all ,
(1.4)
Both inequalities are valid when is replaced by a bounded domain containing the origin and the corresponding constants are known to be optimal. Recently, Gazzola et al. [4] have improved (1.3) by establishing that for and ,
(1.5)
where
(1.6)

and is the unit ball in . Our main concern in this note is to improve (1.4). In fact we have the following theorem.

Theorem 1.1.

There holds, for and ,
(1.7)

Inequality (1.7) is optimal in case is a ball centered at zero.

Combining Theorem 1.1 with (1.2), we have the following.

Corollary 1.2.

There holds, for and ,
(1.8)

Next we consider analogous inequality (1.5). The main result is the following theorem.

Theorem 1.3.

Let and let be such that . Then for every one has
(1.9)

Remark 1.4.

Since
(1.10)

inequality (1.5) is implied by (1.9) in case of .

2. The Proofs

To prove the main results, we first need the following preliminary result.

Lemma 2.1.

Let and . Set . If is a radial function, that is, , then
(2.1)

Proof.

Observe that if , then
(2.2)
Therefore, we have
(2.3)
Though integration by parts, when ,
(2.4)
and hence
(2.5)
By Lemma 2.1 and inequality (1.2), we have, when restricted to radial functions,
(2.6)

Our next step is to prove the following. If is not a radial function, inequality (2.6) also holds.

Let . If we extend as zero outside , we may consider . Decomposing into spherical harmonics we get
(2.7)
where are the orthonormal eigenfunctions of the Laplace-Beltrami operator with responding eigenvalues
(2.8)
The functions belong to , satisfying and as . In particular, and , for any . Then, for any , we have
(2.9)

So

(2.10)

In addition,

(2.11)

Using equality (2.10), we have that (see, e.g., [6, page 452])

(2.12)
Therefore, we have that, by (2.12),
(2.13)

Lemma 2.2.

There holds, for and ,
(2.14)

Proof.

Set . Then satisfies and as . Moreover, since belong to , we have that
(2.15)
Here we use the fact when and ,
(2.16)
Using inequalities (1.2) and (2.15), we have that, for and ,
(2.17)

An immediate consequence of the inequalities (2.13) and Lemma 2.2 is the following result. For ,

(2.18)

Using inequalities (2.18) and Lemma 2.1, we have that, since , for ,

(2.19)
Inequality (2.19) implies that, if is not a radial function, then
(2.20)

Proof of Theorem 1.1.

Using inequality (2.6) and (2.20), we have that, for and ,
(2.21)
In case is a ball centered at zero, a simple scaling allows to consider the case . Set
(2.22)

Using Lemma 2.1 and inequality (1.2), we have that . On the other hand, we have, by inequality (2.21), . Thus . The proof is complete.

Proof of Theorem 1.3.

A scaling argument shows that we may assume and .

Step 1.

Assume is radial, and , then (see [6, Lemma ])
(2.23)
and (see [6, (6.4)])
(2.24)
Therefore
(2.25)
Since is radial,
(2.26)
where denote the surface area of the unit sphere in , is the unit ball in , and
(2.27)

is the radial Laplacian in .

Therefore, for ,
(2.28)

Step 2.

For , set
(2.29)
We get, by (2.18),
(2.30)

In getting the last equality, we used Lemma 2.1.

Using inequality (1.9) for radial functions from step 1,
(2.31)
one obtains, by (2.11),
(2.32)

which demonstrates inequality (1.9).

Declarations

Acknowledgment

This work was supported by National Science Foundation of China under Grant no. 10571044.

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Xiaogan University
(2)
School of Mathematics and Statistics, Wuhan University

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Copyright

© Y.-X. Xiao and Q.-H. Yang 2009

This article is published under license to BioMed Central Ltd. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.