Journal of Function Spaces and Applications

Volume 2013, Article ID 628250, 8 pages

http://dx.doi.org/10.1155/2013/628250

## A More Accurate Half-Discrete Hilbert Inequality with a Nonhomogeneous Kernel

Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong 510303, China

Received 8 May 2013; Accepted 3 June 2013

Academic Editor: Kehe Zhu

Copyright © 2013 Qiliang Huang and Bicheng Yang. 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.

#### Abstract

By using the way of weight functions and the idea of introducing parameters, a more accurate half-discrete Hilbert inequality with a nonhomogeneous kernel and a best constant factor is presented. We also consider its best extension with the parameters, the equivalent forms, and the operator expressions as well as some reverses.

#### 1. Introduction

If,, , , and and , then we have the following discrete Hilbert inequality (cf. [1]): where the constant factoris the best possible. Moreover, for, , , and , we have the following Hilbert integral inequality (cf. [2]): with the same best constant factor. Inequalities (1)-(2) are important in the analysis and its applications (cf. [3]). There are lots of improvements, generalizations, and applications of inequalities (1)-(2) for more details, refer to the literatures [4–15].

We find a result on the half-discrete Hilbert-type inequality with the nonhomogeneous kernel, which were published early as follows (cf. [16, Theorem 351]): where, , , and is the beta function. But the constant factors were not considered whether the best possible. Recently, Yang gave a half-discrete Hilbert inequality with the nonhomogeneous kernel as follows (cf. [17]): where the constant factor is the best possible. And some half-discrete Hilbert-type inequalities with the nonhomogeneous kernel were given (cf. [18]).

In this paper, by using the way of weight functions and the idea of introducing parameters and by means of Hadamard’s inequality, we give the more accurate inequalities of (3) and (4) as follows: with the best constant factors and . We also consider their best extensions with the parameters, the equivalent forms, and the operator expressions as well as some reverses.

#### 2. Some Lemmas

In the following lemmas, we assumed that, , , , , , and .

Lemma 1. *Define the weight functions as follows:
**
Setting , one has the following inequalities:
**
where and .*

*Proof. *Putting in (7), we have
For fixed , setting
in view of the conditions, we find that and . By the following Hadamard inequality (cf. [15]):
and letting , it follows that
where
Hence, we prove that (9) is valid.

Lemma 2. *Suppose that , and , and is a real measurable function in . Then the following are considered. (i) For, one has the following:
**
where and are indicated by (7) and (8).**
(ii) For , one has the reverses of (15) and (16).*

*Proof. *(i) By (7)–(9) and Hölder’s inequality (cf. [15]), we find that
Hence, (15) is valid. Using Hölder’s inequality again, we have
Hence, (16) is valid.

(ii) For or , using the reverse Hölder inequality and in the same way, we have the reverses of (15) and (16).

#### 3. Main Results

Some functions and spaces are introduced as follow:

*Note 3. *If , then and are normal spaces; if or , then both and are not normal spaces, but we still use the formal symbols in the following.

Theorem 3. *Suppose that , , , , , , , and , such that, , , and , then we have the following equivalent inequalities:
**
where the constant factor is the best possible. *

* Proof. *By the Lebesgue term-by-term integration theorem, we find that there are two expressions of in (21). In view of (15), , and , we have (22). By Hölder’s inequality, we find that
Hence, (21) is valid by (22). On the other hand, assuming that (21) is valid and setting
then we have
By (15), it follows that . If , then (22) is trivially valid. If , then . By (21), we have
Hence, (22) is valid, which is equivalent to (21).

In view of (16) and , we obtain (23). By Hölder’s inequality again, we have
Then (21) is valid by using (23). Supposing that (21) is valid and setting
then it follows that
In view of (16) and (9), we find . If , then (23) is trivially valid; if , that is,, then, by (21), we have
Hence (23) is valid, which is equivalent to (21). Then (21), (22), and (23) are equivalent. We prove that the constant factor in (21) is the best possible. For , setting and as follows :
if there exists a positive number , such that (21) is still valid as we replace by , then, substitution of and , we have

For , , letting , we find by Lemma 1 that
By simple calculation, we have
In view of (33), (34), and , it follows that
Then we have for . Hence, is the best value of (21). By the equivalency, the constant factor in (22) ((23)) is the best possible. Otherwise, we would reach a contradiction by (24) ((28)) that the constant factor in (21) is not the best possible.

*Remark 4. *(i) Define a half-discrete Hilbert operator as follows. For , we define , satisfying
Then by (22), it follows that ; that is, is the bounded operator with . Since the constant factorin (22) is the best possible, then we have .

(ii) Define a half-discrete Hilbert operator in the following way. For , we define , satisfying
Then by (23), it follows that ; that is, is the bounded operator with . Since the constant factor in (23) is the best possible, then we have .

Theorem 5. *Suppose that , , , , , , , and , such that , , , and . Then one has the following equivalent inequalities:
**
where the constant factor is the best possible. *

*Proof. *In view of (9), the reverse of (15), and , we have (41). Using the reverse Hölder inequality, we obtain the reverse form of (24) as follows:
Then by (41), (40) is valid.

On the other hand, if (40) is valid, setting as in (25), then (26) still holds with . By (40), it follows that . If , then (41) is trivially valid; if , then , and we have
Then (41) is valid, which is equivalent to (40).

By the reverse of (16) and , and by , we have
Hence, (42) is valid. By the reverse Hölder inequality again, we obtain
Then (40) is valid by (42). On the other hand, if (40) is valid, setting
then . By the reverse of (16), we have . If , then (42) is trivially valid; if , then, by (40), we obtain
Hence, (42) is valid, which is equivalent to (40). It follows that (40), (41), and (42) are equivalent.

If there exists a positive number , such that (40) is still valid as we replace by , then, in particular, we have
where and are taken as in (32) (). Since, by (35), we find in
then, by (35), (49), and the above results, we obtain (notice that)
For , we have . Hence, is the best value of (40). By the equivalency, the constant factor in (41) ((42)) is the best possible. Otherwise, we would reach a contradiction by (43) ((46)) that the constant factor in (40) is not the best possible.

In the same way, for , we also have the following result.

Theorem 6. *If the assumption of in Theorem 3 is replaced by , then the reverses of (21), (22), and (23) are valid and equivalent. Moreover, the same constant factor is the best possible. *

*Remark 7. *(i) If , , and , then and (23) reduces to (3). In particular, for , (21) reduces to (4).

(ii) For, , and in (21) and (23), it follows
where the constant factors both and are the best possibles. In particular, for , (53) comes to (5); for and , in (52), we obtain (6). Hence, inequality (21) is the best extension of (4) and (6), and inequality (23) is the best extension of (3) and (5) with the parameters.

#### Acknowledgment

This paper is supported by the 2012 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (no. 2012KJCX0079).

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