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A more accurate half-discrete reverse Hilbert-type inequality with a non-homogeneous kernel
Journal of Inequalities and Applications volume 2014, Article number: 96 (2014)
Abstract
By means of weight functions and the improved Euler-Maclaurin summation formula, a more accurate half-discrete reverse Hilbert-type inequality with the kernel and a best constant factor is given. Some equivalent forms, the dual forms as well as some related homogeneous cases, are also considered.
MSC:26D15.
1 Introduction
Assuming that , , , we have the following Hilbert integral inequality (cf. [1]):
where the constant factor π is best possible. If , , , , then we have the following analogous discrete Hilbert inequality:
with the same best constant factor π. Inequalities (1) and (2) are important in analysis and its applications (cf. [2–4]).
In 1998, by introducing an independent parameter , Yang [5] gave an extension of (1). For generalizing the results from [5], Yang [6] gave some best extensions of (1) and (2): If , , , is a non-negative homogeneous function of degree −λ satisfying , , , , , and , then
where the constant factor is best possible. Moreover if the value of is finite and () is decreasing for (), then for , , and , , we have
where the constant is still best value. Clearly, for , , , , (3) reduces to (1), while (4) reduces to (2). The reverses of (3) and (4) as well as the equivalent forms are also considered by [6].
Some other results about integral and discrete Hilbert-type inequalities can be found in [7–15]. On half-discrete Hilbert-type inequalities with the general non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors are best possible. In 2005, Yang [16] gave a result with the kernel by introducing a variable and proved that the constant factor is best possible. Very recently, Yang [17] and [18] gave the following half-discrete reverse Hilbert inequality with best constant factor: For , , , , , , ,
In this paper, by means of weight functions and the improved Euler-Maclaurin summation formula, a more accurate half-discrete reverse Hilbert-type inequality with the kernel similar to (5) and a best constant factor is given. Moreover, some equivalent forms, the dual forms as well as some relating homogeneous cases are also considered.
2 Some lemmas
Lemma 1 If , , (), () are continuous decreasing functions satisfying , , define a function as follows:
Then there exists , such that
where is the Bernoulli function of the first order. In particular, for , , we have and
for , , if , then it follows and
Proof Define a continuous decreasing function as follows:
Then it follows that
Since , is a non-constant continuous decreasing function with , by the improved Euler-Maclaurin summation formula (cf. [6], Theorem 2.2.2), it follows that
and then in view of the above results and by simple calculation, we have (7). □
Lemma 2 If , , , and and are weight functions given by
then we have
Proof Substituting in (10), and by simple calculation, we have
For fixed , we find
By the Euler-Maclaurin summation formula (cf. [6]), it follows that
-
(i)
For , we obtain , and
Setting , wherefrom , and
then by (7), we find
In view of (11) and the above results, since for , namely , it follows that
-
(ii)
For , we obtain , and
Since for , , by the improved Euler-Maclaurin summation formula (cf. [6]), it follows that
In view of (13) and the above results, for , we find
Hence for , we have , and then .
On the other-hand, since is decreasing with respect to , we find
where .
-
(i)
For , we obtain
-
(ii)
For , it follows that
Hence we have (11) and (12). □
Lemma 3 Let the assumptions of Lemma 2 be fulfilled and additionally, let or , , , , be a non-negative measurable function in . Then we have the following inequalities:
Proof For , setting , by the reverse Hölder inequality (cf. [19]) and (11), it follows that
Then by the Lebesgue term by term integration theorem (cf. [20]), we have
and then (14) follows. By the reverse Hölder inequality, for , we have
By the Lebesgue term by term integration theorem, we have
and in view of (11), inequality (15) follows. For , by the same way we still have (14) and (15). □
Lemma 4 Let the assumptions of Lemma 2 be fulfilled and additionally, let , , . Setting , ; , , and , , then we have
Proof We find
and then (16) is valid. We obtain
and so (17) is valid. □
3 Main results
We introduce the functions
wherefrom , and .
Theorem 1 If , , , , , , , , and , then we have the following equivalent inequalities:
where the constant is the best possible in the above inequalities.
Proof The two expressions for I in (18) follow from Lebesgue’s term by term integration theorem. By (14) and (11), we have (19). By the reverse Hölder inequality, we have
Then by (19), we have (18). On the other-hand, assume that (18) is valid. Setting
it follows that . By (14), we find . If , then (19) is trivially valid; if , then by (18), we have
therefore , that is, (19) is equivalent to (18). On the other-hand, by (11) we have . Then in view of (15), we have (20). By the Hölder inequality, we find
Then by (20), we have (18). On the other-hand, assume that (18) is valid. Setting
then . By (15), we find . If , then (20) is trivially valid; if , then by (18), we have
therefore , that is, (20) is equivalent to (18). Hence, (18), (19), and (20) are equivalent.
If there exists a positive number k (), such that (18) is valid as we replace with k, then in particular, it follows that . In view of (16) and (17), we have
and (). Hence is the best value of (18).
By the equivalence of the inequalities, the constant factor in (19) and (20) is the best possible. □
For , we have the dual forms of (18), (19), and (20) as follows:
Theorem 2 If , , , , , , , , and , then we have the following equivalent inequalities:
where the constant is the best possible in the above inequalities.
Proof By means of Lemma 3 and the same way, we can prove that (21), (22), and (23) are valid and equivalent. For , setting and as Lemma 4, if there exists a positive number k (), such that (21) is valid as we replace with k, then in particular, by (16), it follows that
and (). Hence is the best value of (21). By the equivalence of the inequalities, the constant factor in (21) and (22) is the best possible. □
Remark (i) Since we find
then for in (18), we have
and the following inequality:
Hence (18) is a more accurate inequality of (24).
-
(ii)
For in (18), we have , , ,
and the following inequality:
for in (18), we have , , ,
and the following inequality:
for in (18), we have , , ,
and the following inequality:
-
(iii)
Setting , , and in (18), by simplification, we obtain the following inequality with the homogeneous kernel:
(28)
It is evident that (28) is equivalent to (18), and then the same constant factor in (28) is still the best possible. In the same way, we can find the following inequalities equivalent to (28) with the same best possible constant factor :
-
(iv)
Applying the same way in Theorem 2, we still can obtain some particular dual forms as (i) and (ii) and some equivalent inequalities similar to (28), (29), and (30).
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Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 61370186), 2012 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2012KJCX0079), and Guangzhou Science and Technology Plan Item (No. 2013J4100009).
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The main idea of this paper was proposed by BY and QC. BY prepared the manuscript initially and performed all the steps of the proofs in this research. All authors read and approved the final manuscript.
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Yang, B., Chen, Q. A more accurate half-discrete reverse Hilbert-type inequality with a non-homogeneous kernel. J Inequal Appl 2014, 96 (2014). https://doi.org/10.1186/1029-242X-2014-96
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DOI: https://doi.org/10.1186/1029-242X-2014-96