- Research Article
- Open Access

# A Hilbert-Type Integral Inequality in the Whole Plane with the Homogeneous Kernel of Degree −2

- Dongmei Xin
^{1}Email author and - Bicheng Yang
^{1}

**2011**:401428

https://doi.org/10.1155/2011/401428

© D. Xin and B. Yang 2011

**Received:**20 December 2010**Accepted:**29 January 2011**Published:**22 February 2011

## Abstract

By applying the way of real and complex analysis and estimating the weight functions, we build a new Hilbert-type integral inequality in the whole plane with the homogeneous kernel of degree −2 involving some parameters and the best constant factor. We also consider its reverse. The equivalent forms and some particular cases are obtained.

## Keywords

- Weight Function
- Constant Factor
- Equivalent Form
- Integral Formula
- Integral Inequality

## 1. Introduction

where the constant factor is the best possible. Zeng and Xie[11] also give a new inequality in the whole plane.

By applying the method of [10, 11] and using the way of real and complex analysis, the main objective of this paper is to give a new Hilbert-type integral inequality in the whole plane with the homogeneous kernel of degree −2 involving some parameters and a best constant factor. The reverse form is considered. As applications, we also obtain the equivalent forms and some particular cases.

## 2. Some Lemmas

Lemma 2.1.

Proof.

Hence we find .

By the same way, we still can find that . The lemma is proved.

Lemma 2.2.

Proof.

The lemma is proved.

## 3. Main Results and Applications

Theorem 3.1.

where the constant factor and are the best possible and is defined by Lemma 2.1. Inequality (3.1) and (3.2) are equivalent.

Proof.

If (2.11) takes the form of equality for a , then there exist constants and , such that they are not all zero, and a.e. in . Hence, there exists a constant , such that a.e. in . We suppose (otherwise ). Then a. e. in , which contradicts the fact that . Hence (2.11) takes the form of strict inequality, so does (2.10), and we have (3.2).

Hence we have (3.2), which is equivalent to (3.1).

which contradicts the fact that . Hence the constant factor in (3.1) is the best possible.

If the constant factor in (3.2) is not the best possible, then by (3.3), we may get a contradiction that the constant factor in (3.1) is not the best possible. Thus the theorem is proved.

In view of Note (2) and Theorem 3.1, we still have the following theorem.

Theorem 3.2.

where the constant factors and are the best possible. Inequality (3.12) is equivalent.

Theorem 3.3.

As the assumptions of Theorem 3.1, replacing by , we have the equivalent reverses of (3.1) and (3.2) with the best constant factors.

Proof.

By the reverse Hölder's inequality [13], we have the reverse of (2.10) and (3.3). It is easy to obtain the reverse of (3.2). In view of the reverses of (3.2) and (3.3), we obtain the reverse of (3.1). On the other hand, suppose that the reverse of (3.1) is valid. Setting the same as Theorem 3.1, by the reverse of (2.10), we have . If , then the reverse of (3.2) is obvious value; if , then by the reverse of (3.1), we obtain the reverses of (3.5). Hence we have the reverse of (3.2), which is equivalent to the reverse of (3.1).

By (3.14), (3.15), and (3.17), for , we have , which contradicts the fact that . Hence the constant factor in the reverse of (3.1) is the best possible.

If the constant factor in reverse of (3.2) is not the best possible, then by the reverse of (3.3), we may get a contradiction that the constant factor in the reverse of (3.1) is not the best possible. Thus the theorem is proved.

By the same way of Theorem 3.3, we still have the following theorem.

Theorem 3.4.

By the assumptions of Theorem 3.2, replacing by , we have the equivalent reverses of (3.12) with the best constant factors.

## Authors’ Affiliations

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