Research Article A More Accurate Half-Discrete...
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Hindawi Publishing Corporation Journal of Function Spaces and Applications Volume 2013, Article ID 628250, 8 pages http://dx.doi.org/10.1155/2013/628250 Research Article A More Accurate Half-Discrete Hilbert Inequality with a Nonhomogeneous Kernel Qiliang Huang and Bicheng Yang Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong 510303, China Correspondence should be addressed to Bicheng Yang; [email protected] Received 8 May 2013; Accepted 3 June 2013 Academic Editor: Kehe Zhu Copyright © 2013 Q. Huang and B. Yang. is 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. 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 , ≥0, = { } ∞ =1 ∈ 2 , = { } ∞ =1 ∈ 2 , and ‖‖ = {∑ ∞ =1 2 } 1/2 >0 and ‖‖ = {∑ ∞ =1 2 } 1/2 >0, then we have the following discrete Hilbert inequality (cf. [1]): ∞ ∑ =1 ∞ ∑ =1 + < ‖‖ ‖‖ , (1) where the constant factor is the best possible. Moreover, for , (≥0) ∈ 2 ( + ), ‖‖ = {∫ ∞ 0 2 ()} 1/2 >0, and ‖‖ > 0, we have the following Hilbert integral inequality (cf. [2]): ∬ ∞ 0 () () + < , (2) with the same best constant factor . Inequalities (1)-(2) are important in the analysis and its applications (cf. [3]). ere 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 inequal- ity with the nonhomogeneous kernel, which were published early as follows (cf. [16, eorem 351]): ∫ ∞ 0 −2 ( ∞ ∑ =1 1 + ) < ( 1 , 1 ) ∞ ∑ =1 , (3) where > 1, > 1, 1/ + 1/ = 1, and (, V) = ∫ ∞ 0 ( −1 /(1 + ) +V ) (, V > 0) is the beta function. But the constant factors (1/, 1/) were not considered whether the best possible. Recently, Yang gave a half-discrete Hilbert inequality with the nonhomogeneous kernel as follows (cf. [17]): ∞ ∑ =1 ∫ ∞ 0 () 1 + < ( ∞ ∑ =1 2 ∫ ∞ 0 2 ()) 1/2 , (4) where the constant factor is the best possible. And some half-discrete Hilbert-type inequalities with the nonhomoge- neous 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: ∫ ∞ 0 −2 ( ∞ ∑ =1 1 + − /2 ) < ( 1 , 1 ) ∞ ∑ =1 , (5) ∞ ∑ =1 ∫ ∞ 0 () 1 + − /2 < ( ∞ ∑ =1 2 ∫ ∞ 0 2 ()) 1/2 , (6) with the best constant factors (1/, 1/) and . We also consider their best extensions with the parameters, the
Transcript of Research Article A More Accurate Half-Discrete...
Hindawi Publishing CorporationJournal of Function Spaces and ApplicationsVolume 2013 Article ID 628250 8 pageshttpdxdoiorg1011552013628250
Research ArticleA More Accurate Half-Discrete Hilbert Inequality witha Nonhomogeneous Kernel
Qiliang Huang and Bicheng Yang
Department of Mathematics Guangdong University of Education Guangzhou Guangdong 510303 China
Correspondence should be addressed to Bicheng Yang bcyanggdeieducn
Received 8 May 2013 Accepted 3 June 2013
Academic Editor Kehe Zhu
Copyright copy 2013 Q Huang and B Yang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
By using the way of weight functions and the idea of introducing parameters a more accurate half-discrete Hilbert inequality witha nonhomogeneous kernel and a best constant factor is presented We also consider its best extension with the parameters theequivalent forms and the operator expressions as well as some reverses
1 Introduction
If 119886119899 119887119899ge 0 119886 = 119886
119899infin
119899=1isin 1198972 119887 = 119887
119899infin
119899=1isin 1198972 and 119886 =
suminfin
119899=11198862
119899
12
gt 0 and 119887 = suminfin
119899=11198872
119899
12
gt 0 then we havethe following discrete Hilbert inequality (cf [1])
infin
sum
119899=1
infin
sum
119898=1
119886119898119887119899
119898 + 119899
lt 120587 119886 119887 (1)
where the constant factor 120587 is the best possible Moreoverfor 119891 119892(ge0) isin 119871
2
(119877+) 119891 = int
infin
0
1198912
(119909)119889119909
12
gt 0 and119892 gt 0 we have the following Hilbert integral inequality (cf[2])
∬
infin
0
119891 (119909) 119892 (119910)
119909 + 119910
119889119909 119889119910 lt 12058710038171003817100381710038171198911003817100381710038171003817
10038171003817100381710038171198921003817100381710038171003817 (2)
with the same best constant factor 120587 Inequalities (1)-(2) areimportant in the analysis and its applications (cf [3]) Thereare lots of improvements generalizations and applications ofinequalities (1)-(2) formore details refer to the literatures [4ndash15]
We find a result on the half-discrete Hilbert-type inequal-ity with the nonhomogeneous kernel which were publishedearly as follows (cf [16 Theorem 351])
int
infin
0
119909119902minus2
(
infin
sum
119899=1
119886119899
1 + 119909119899
)
119902
119889119909 lt 119861119902
(
1
119901
1
119902
)
infin
sum
119899=1
119886119902
119899 (3)
where 119901 gt 1 119902 gt 1 1119901 + 1119902 = 1 and 119861(119906 V) =
int
infin
0
(119905119906minus1
119889119905(1 + 119905)119906+V) (119906 V gt 0) is the beta function But
the constant factors119861(1119901 1119902) were not consideredwhetherthe best possible Recently Yang gave a half-discrete Hilbertinequality with the nonhomogeneous kernel as follows (cf[17])
infin
sum
119899=1
119886119899int
infin
0
119891 (119909)
1 + 119909119899
119889119909 lt 120587(
infin
sum
119899=1
1198862
119899int
infin
0
1198912
(119909)119889119909)
12
(4)
where the constant factor 120587 is the best possible And somehalf-discrete Hilbert-type inequalities with the nonhomoge-neous kernel were given (cf [18])
In this paper by using theway of weight functions and theidea of introducing parameters and by means of Hadamardrsquosinequality we give the more accurate inequalities of (3) and(4) as follows
int
infin
0
119909119902minus2
(
infin
sum
119899=1
119886119899
1 + 119909119899 minus 1199092
)
119902
119889119909 lt 119861119902
(
1
119901
1
119902
)
infin
sum
119899=1
119886119902
119899 (5)
infin
sum
119899=1
119886119899int
infin
0
119891 (119909)
1 + 119909119899 minus 1199092
119889119909 lt 120587(
infin
sum
119899=1
1198862
119899int
infin
0
1198912
(119909)119889119909)
12
(6)
with the best constant factors 119861(1119901 1119902) and 120587 We alsoconsider their best extensions with the parameters the
2 Journal of Function Spaces and Applications
equivalent forms and the operator expressions as well assome reverses
2 Some Lemmas
In the following lemmas we assumed that 120572 120573 120582 gt 0 120572 lt
120582120573 120572 le min1 2 minus 120573 1205741isin (minusinfininfin) and 0 le 120574
2le 12
Lemma 1 Define the weight functions as follows
120596120572(119899) = (119899 minus 120574
2)120572
int
infin
1205741
(119909 minus 1205741)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
(119899 isin N)
(7)
120603120572(119909) = (119909 minus 120574
1)120572
infin
sum
119899=1
(119899 minus 1205742)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119909 isin (1205741infin))
(8)
Setting 119896 = 119896(120572 120573 120582) = (1120573)119861(120572120573 120582 minus 120572120573) one has thefollowing inequalities
0 lt 119896 (1 minus 120579 (119909)) lt 120603120572(119909) lt 120596
120572(119899) = 119896 (9)
where 120579(119909) = (1120573119896) int
(119909minus1205741)120573(1minus1205742)120573
0
(119906120572120573minus1
(1 + 119906)120582
)119889119906 isin
(0 1) and 120579(119909) = 119874((119909 minus 1205741)120572
) (119909 isin (1205741infin))
Proof Putting 119906 = (119909 minus 1205741)120573
(119899 minus 1205742)120573 in (7) we have
120596120572(119899) =
1
120573
int
infin
0
119906120572120573minus1
(1 + 119906)120582
119889119906 =
1
120573
119861(
120572
120573
120582 minus
120572
120573
) = 119896 (10)
For fixed 119909 isin (1205741infin) setting
119891 (119905) =
(119909 minus 1205741)120572
(119905 minus 1205742)120572minus1
[1 + (119909 minus 1205741)120573
(119905 minus 1205742)120573
]
120582
(119905 isin (1205742infin)) (11)
in view of the conditions we find that 1198911015840(119905) = minus(1 minus 120572)(119909 minus
1205741)120572
(119905 minus 1205742)120572minus2
[1 + (119909 minus 1205741)120573
(119905 minus 1205742)120573
]120582
minus 120582120573(119909 minus 1205741)120572+120573
(119905 minus
1205742)120572+120573minus2
[1+(119909minus1205741)120573
(119905minus1205742)120573
]120582+1
lt 0 and 11989110158401015840(119905) gt 0 By thefollowing Hadamard inequality (cf [15])
119891 (119899) lt int
119899+12
119899minus12
119891 (119905) 119889119905 (119899 isin N) (12)
and letting 119906 = (119909 minus 1205741)120573
(119905 minus 1205742)120573 it follows that
120603120572(119909) =
infin
sum
119899=1
119891 (119899) lt
infin
sum
119899=1
int
119899+12
119899minus12
119891 (119905) 119889119905 = int
infin
12
119891 (119905) 119889119905
le int
infin
1205742
119891 (119905) 119889119905 =
1
120573
int
infin
0
119906120572120573minus1
(1 + 119906)120582
119889119906
=
1
120573
119861(
120572
120573
120582 minus
120572
120573
) = 119896
120603120572(119909) =
infin
sum
119899=1
119891 (119899) gt int
infin
1
119891 (119905) 119889119905
= int
infin
1205742
119891 (119905) 119889119905 minus int
1
1205742
119891 (119905) 119889119905
= 119896 minus
1
120573
int
(119909minus1205741)120573(1minus1205742)120573
0
119906120572120573minus1
(1 + 119906)120582
119889119906
= 119896 (1 minus 120579 (119909)) gt 0
(13)
where
0 lt 120579 (119909) =
1
120573119896
int
(119909minus1205741)120573(1minus1205742)120573
0
119906120572120573minus1
(1 + 119906)120582
119889119906
lt
1
120573119896
int
(119909minus1205741)120573(1minus1205742)120573
0
119906120572120573minus1
119889119906 =
(1 minus 1205742)120572
(119909 minus 1205741)120572
120572119896
(14)
Hence we prove that (9) is valid
Lemma 2 Suppose that 1119901 + 1119902 = 1 (119901 = 0 1) and 119886119899ge 0
and119891(119909) ge 0 is a realmeasurable function in (1205741infin)Then the
following are considered (i) For 119901 gt 1 one has the following
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1
times[
[
[
int
infin
1205741
119891(119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
119901
1119901
le 1198961119902
int
infin
1205741
120603120572(119909)(119909 minus 120574
1)119901(1minus120572)minus1
119891119901
(119909)119889119909
1119901
(15)
Journal of Function Spaces and Applications 3
1198711=
int
infin
1205741
(119909 minus 1205741)119902120572minus1
120603119902minus1
120572 (119909)
times[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902
119889119909
1119901
le 119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
(16)
where 120603120572(119909) and 120596
120572(119899) are indicated by (7) and (8)
(ii) For 119901 lt 1 (119901 = 0) one has the reverses of (15) and (16)
Proof (i) By (7)ndash(9) andHolderrsquos inequality (cf [15]) we findthat
[
[
[
int
infin
1205741
119891(119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
119901
=
int
infin
1205741
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times [
(119909 minus 1205741)(1minus120572)119902
(119899 minus 1205742)(1minus120572)119901
119891 (119909)]
times[
(119899 minus 1205742)(1minus120572)119901
(119909 minus 1205741)(1minus120572)119902
]119889119909
119901
le int
infin
1205741
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119909 minus 1205741)(1minus120572)(119901minus1)
(119899 minus 1205742)1minus120572
119891119901
(119909) 119889119909
times
[
[
[
[
[
int
infin
1205741
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119899 minus 1205742)(1minus120572)(119902minus1)
(119909 minus 1205741)1minus120572
119889119909
]
]
]
]
]
119901minus1
= int
infin
1205741
119891119901
(119909) (119909 minus 1205741)(1minus120572)(119901minus1)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
(119899 minus 1205742)1minus120572
times [(119899 minus 1205742)119902(1minus120572)minus1
120596120572(119899)]
119901minus1
= (119899 minus 1205742)1minus119901120572
119896119901minus1
times int
infin
1205741
119891119901
(119909) (119909 minus 1205741)(1minus120572)(119901minus1)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
1
(119899 minus 1205742)1minus120572
119889119909
(17)
119869119901
le 119896119901minus1
infin
sum
119899=1
int
infin
1205741
119891119901
(119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119909 minus 1205741)(1minus120572)(119901minus1)
(119899 minus 1205742)1minus120572
119889119909
= 119896119901minus1
int
infin
1205741
infin
sum
119899=1
(119899 minus 1205742)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times (119909 minus 1205741)120572+119901(1minus120572)minus1
119891119901
(119909) 119889119909
= 119896119901minus1
int
infin
1205741
120603120572(119909) (119909 minus 120574
1)119901(1minus120572)minus1
119891119901
(119909) 119889119909
(18)
Hence (15) is valid Using Holderrsquos inequality again we have
[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902
=
infin
sum
119899=1
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times[
(119909 minus 1205741)(1minus120572)119902
(119899 minus 1205742)(1minus120572)119901
][
(119899 minus 1205742)(1minus120572)119901
(119909 minus 1205741)(1minus120572)119902
119886119899]
119902
le [120603120572(119909) (119909 minus 120574
1)119901(1minus120572)minus1
]
119902minus1
times
infin
sum
119899=1
119886119902
119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119899 minus 1205742)(1minus120572)119902119901
(119909 minus 1205741)1minus120572
= 120603119902minus1
120572(119909) (119909 minus 120574
1)1minus119902120572
times
infin
sum
119899=1
(119899 minus 1205742)(1minus120572)(119902minus1)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
1
(119909 minus 1205741)1minus120572
119886119902
119899
4 Journal of Function Spaces and Applications
119871119902
1le int
infin
1205741
infin
sum
119899=1
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119899 minus 1205742)(1minus120572)(119902minus1)
(119909 minus 1205741)1minus120572
119886119902
119899119889119909
=
infin
sum
119899=1
[
[
[
(119899 minus 1205742)
120572
int
infin
1205741
(119909 minus 1205741)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
times (119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
=
infin
sum
119899=1
120596120572(119899) (119899 minus 120574
2)119902(1minus120572)minus1
119886119902
119899
= 119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
(19)
Hence (16) is valid(ii) For 0 lt 119901 lt 1 (119902 lt 0) or 119901 lt 0 (0 lt 119902 lt 1) using the
reverse Holder inequality and in the same way we have thereverses of (15) and (16)
3 Main Results
Some functions and spaces are introduced as follow
120601 (119909) = (119909 minus 1205741)119901(1minus120572)minus1
120601 (119909) = (1 minus 120579 (119909)) 120601 (119909)
120595 (119899) = (119899 minus 1205742)119902(1minus120572)minus1
(119909 isin (1205741infin) 119899 isin N 120579 (119909) is indicated as Lemma 1)
119871119901120601(1205741infin) =119891
10038171003817100381710038171198911003817100381710038171003817119901120601
=int
infin
1205741
120601 (119909)1003816100381610038161003816119891 (119909)
1003816100381610038161003816
119901
119889119909
1119901
ltinfin
119897119902120595
= 119886=119886119899infin
119899=1 119886119902120595=
infin
sum
119899=1
120595 (119899)1003816100381610038161003816119886119899
1003816100381610038161003816
119902
1119902
ltinfin
(20)
Note If 119901 gt 1 then 119871119901120601(1205741infin) and 119897
119902120595are normal spaces
if 0 lt 119901 lt 1 or 119901 lt 0 then both 119871119901120601(1205741infin) and 119897
119902120595are
not normal spaces but we still use the formal symbols in thefollowing
Theorem 3 Suppose that 119901 gt 1 1119901 + 1119902 = 1 120572 120573 120582 gt
0 120572 lt 120582120573 120572 le min1 2 minus 120573 1205741
isin (minusinfininfin) 0 le
1205742le 12 and 119891(119909) 119886
119899ge 0 such that 119891 isin 119871
119901120601(1205741infin)
119886 = 119886119899infin
119899=1isin 119897119902120595
119891119901120601
gt 0 and 119886119902120595
gt 0 then we havethe following equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (x)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
(21)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891 (119909) 119889119909
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
]
]
]
119901
1119901
lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
(22)
119871 =
int
infin
1205741
(119909 minus 1205741)119902120572minus1[
[
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
]
]
]
119902
119889119909
1119902
lt 119896119886119902120595
(23)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof By the Lebesgue term-by-term integration theoremwe find that there are two expressions of 119868 in (21) In viewof (15) 120603
120572(119909) lt 119896 and 0 lt 119891
119901120601lt infin we have (22) By
Holderrsquos inequality we find that
119868 =
infin
sum
119899=1
[
[
[
(119899 minus 1205742)120572minus1119901
int
infin
1205741
119891 (119909) 119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(119899 minus 1205742)1119901minus120572
119886119899]
le 119869
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119869119886119902120595
(24)
Hence (21) is valid by (22) On the other hand assuming that(21) is valid and setting
119886119899= (119899 minus 120574
2)119901120572minus1[
[
[
int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
119901minus1
(119899 isin N) (25)
then we have
119886119902
119902120595=
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899= 119869119901
= 119868 (26)
Journal of Function Spaces and Applications 5
By (15) it follows that 119869 lt infin If 119869 = 0 then (22) is triviallyvalid If 119869 gt 0 then 0 lt 119886
119902120595= 119869119901minus1
lt infin By (21) we have
119886119902
119902120595= 119869119901
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595lt 119896
10038171003817100381710038171198911003817100381710038171003817119901120601
(27)
Hence (22) is valid which is equivalent to (21)In view of (16) and120603
120572(119909) lt 119896 we obtain (23) By Holderrsquos
inequality again we have
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
le 119871int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909
1119901
= 11987110038171003817100381710038171198911003817100381710038171003817119901120601
(28)
Then (21) is valid by using (23) Supposing that (21) is validand setting
119891 (119909) = (119909 minus 1205741)119902120572minus1[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(29)
then it follows that
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909) 119889119909 = 119871119902
= 119868 (30)
In view of (16) and (9) we find 119871 lt infin If 119871 = 0 then (23) istrivially valid if 119871 gt 0 that is 0 lt 119891
119901120601lt infin then by (21)
we have
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= 119871119902
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119871 =10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601lt 119896119886
119902120595
(31)
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 0 lt 120576 lt 119902120572 setting 119886 = 119886
119899infin
119899=1
and 119891(119909) as follows (119901 119902 gt 1)
119886119899= (119899 minus 120574
2)120572minus120576119902minus1
119891 (119909) =
(119909 minus 1205741)120572+120576119901minus1
119909 isin (1205741 1 + 120574
1)
0 119909 isin [1 + 1205741infin)
(32)
if there exists a positive number 1198961015840 le 119896 such that (21) is stillvalid as we replace 119896 by 1198961015840 then substitution of 119886 and
119891(119909)we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
(33)
For 119901 119902 gt 1 letting = 120572 minus 120576119902 we find by Lemma 1 that
119868 = int
1205741+1
1205741
(119909 minus 1205741)120576minus1
(119909 minus 1205741)120572minus120576119902
times
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
120603(119909) 119889119909
gt 119896 ( 120573 120582) int
1205741+1
1205741
(119909 minus 1205741)120576minus1
[1 minus 119874((119909 minus 1205741)
)] 119889119909
= 119896(120572 minus
120576
119902
120573 120582) (
1
120576
minus 119874 (1))
(34)
By simple calculation we have
119886119902
119902120595=
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899=
infin
sum
119899=1
(119899 minus 1205742)minus1minus120576
lt (1 minus 1205742)minus1minus120576
+ int
infin
1
(119910 minus 1205742)minus1minus120576
119889119910 =
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1
(35)
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
)
1119901
(36)
In view of (33) (34) and 119902 gt 1 it follows that
119896(120572 minus
120576
119902
120573 120582) [1 minus 120576119874 (1)]
lt 120576119868 lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
lt 1198961015840
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]
1119902
(37)
Thenwe have 119896(120572 120573 120582) = 119896 le 1198961015840 for 120576 rarr 0+ Hence 1198961015840 = 119896 is
the best value of (21) By the equivalency the constant factor kin (22) ((23)) is the best possible Otherwise we would reacha contradiction by (24) ((28)) that the constant factor in (21)is not the best possible
Remark 4 (i) Define a half-discrete Hilbert operator 119879
119871119901120601(1205741infin) rarr 119897
1199011205951minus119901 as follows For 119891 isin 119871
119901120601(1205741infin) we
define 119879119891 isin 1198971199011205951minus119901 satisfying
119879119891 (119899) = int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909 (119899 isin N) (38)
6 Journal of Function Spaces and Applications
Then by (22) it follows that 1198791198911199011205951minus119901 le 119896119891
119901120601 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (22) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)(ii) Define a half-discrete Hilbert operator 119879 119897
119902120595rarr
1198711199021206011minus119902(1205741infin) in the following way For 119886 isin 119897
119902120595 we define
119879119886 isin 119871
1199021206011minus119902 satisfying
119879119886 (119909) =
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119909 isin (1205741infin))
(39)
Then by (23) it follows that 1198791198861199021206011minus119902 le 119896119886
119902120595 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (23) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)
Theorem 5 Suppose that 0 lt 119901 lt 1 1119901 + 1119902 = 1 120572 120573 120582 gt0 120572 lt 120582120573 120572 le min1 2 minus 120573 120574
1isin (minusinfininfin) 0 le 120574
2le 12
and 119891(119909) 119886119899ge 0 such that 119891 isin 119871
119901120601(1205741infin) 119886 = 119886
119899infin
119899=1isin
119897119902120595
119891119901120601gt 0 and 119886
119902120595gt 0 Then one has the following
equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
(40)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891(119909)
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
119889119909]
]
]
119901
1119901
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
(41)
= int
infin
1205741
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]120582
]
119902
119889119909
1119902
gt 119896119886119902120595
(42)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof In view of (9) the reverse of (15) and 0 lt 119891119901120601lt infin
we have (41) Using the reverse Holder inequality we obtainthe reverse form of (24) as follows
119868 ge 119869119886119902120595 (43)
Then by (41) (40) is valid
On the other hand if (40) is valid setting 119886119899as in (25)
then (26) still holds with 0 lt 119901 lt 1 By (40) it follows that119869 gt 0 If 119869 = infin then (41) is trivially valid if 119869 lt infin then0 lt 119886
119902120595= 119869119901minus1
lt infin and we have
119886119902
119902120595= 119869119901
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595gt 119896
10038171003817100381710038171198911003817100381710038171003817119901 120601
(44)
Then (41) is valid which is equivalent to (40)By the reverse of (16) and120603(119909) gt 119896(1minus120579(119909)) and by 119902 lt 0
we have
gt 119896(119902minus1)119902
1198711ge 119896(119902minus1)119902
119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119896119886119902120595
(45)
Hence (42) is valid By the reverse Holder inequality againwe obtain
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
(1 minus 120579 (119909))1119901
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(1 minus 120579 (119909))1119901
(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
ge 10038171003817100381710038171198911003817100381710038171003817119901 120601
(46)
Then (40) is valid by (42) On the other hand if (40) is validsetting
119891 (119909) =
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(47)
then 119891119901119901120601
= int
infin
1205741
[1 minus 120579(119909)](119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909 = 119902
= 119868
By the reverse of (16) we have gt 0 If = infin then (42) istrivially valid if 0 lt lt infin then by (40) we obtain
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601
= 119902
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is = 10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601
gt 119896119886119902120595
(48)
Hence (42) is valid which is equivalent to (40) It follows that(40) (41) and (42) are equivalent
Journal of Function Spaces and Applications 7
If there exists a positive number 1198961015840 ge 119896 such that (40) isstill valid as we replace 119896 by 1198961015840 then in particular we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
gt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
119886119902120595
(49)
where 119886 = 119886119899infin
119899=1and
119891 are taken as in (32) (0 lt 120576 lt 119901(120573120582 minus120572)) Since by (35) we find in
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
= int
1+1205741
1205741
[1 minus 119874 ((119909 minus 1205741)120572
)] (119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
minus 119874(1))
1119901
119868 le
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
int
infin
1205741
(119909 minus 1205741)120572+120576119901minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
=
1
120573
infin
sum
119899=1
(119899 minus 1205742)minus120576minus1
int
infin
0
119906120572120573+120576120573119901minus1
(1 + 119906)120582
119889119906
le
1
120573
[
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
(50)
then by (35) (49) and the above results we obtain (noticethat 119902 lt 0)
1
120573
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
ge 120576119868 gt 1205761198961015840
(
1
120576
minus 119874(1))
1119901
(
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1)
1119902
= 1198961015840
(1 minus 120576119874(1))1119901
(
120576 + 1 minus 1205742
(1 minus 1205742)120576+1)
1119902
(51)
For 120576 rarr 0+ we have 119896 = (1120573)119861(120572120573 120582 minus 120572120573) ge 119896
1015840Hence 119896 = 119896
1015840 is the best value of (40) By the equivalencythe constant factor k in (41) ((42)) is the best possibleOtherwise wewould reach a contradiction by (43) ((46)) thatthe constant factor in (40) is not the best possible
In the same way for 119901 lt 0 we also have the followingresult
Theorem 6 If the assumption of 119901 gt 1 in Theorem 3 isreplaced by 119901 lt 0 then the reverses of (21) (22) and (23)are valid and equivalent Moreover the same constant factor isthe best possible
Remark 7 (i) If 1205741= 1205742= 0 120582 = 120573 = 1 and 120572 = 1119901
then 119896 = 119861(1119901 1119902) and (23) reduces to (3) In particularfor 119901 = 119902 = 2 (21) reduces to (4)
(ii) For 1205741= 0 120574
2= 12 and 120572 = 1119901 in (21) and (23) it
followsinfin
sum
119899=1
119886119899int
infin
0
119891 (119909)
(1 + (119909119899 minus 1199092)120573
)
120582
119889119909
lt 119896int
infin
0
119909119901minus2
119891119901
(119909)119889119909
1119901
infin
sum
119899=1
119886119902
119899
1119902
(52)
int
infin
0
119909119902minus2
(
infin
sum
119899=1
119886119899
(1 + (119909119899 minus 1199092)120573
)
120582
)
119902
119889119909 lt 119896119902
infin
sum
119899=1
119886119902
119899 (53)
where the constant factors both 119896 = (1120573)119861(1119901120573 120582 minus 1119901120573)and 119896119902 are the best possibles In particular for 120582 = 120573 = 1(53) comes to (5) for 120582 = 120573 = 1 and 119901 = 119902 = 2 in (52) weobtain (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 ConstructionSpecial Foundation Item of Guangdong Institution of HigherLearning College and University (no 2012KJCX0079)
References
[1] H Weyl Singulare integral gleichungen mit besonderer be-rucksichtigung des fourierschen integral theorems [Inaugeral-Dissertation] Gottingen 1908
[2] I Schur ldquoBernerkungen sur Theorie der beschrankten Bilin-earformenmit unendlich vielen veranderlichenrdquo Journal fur dieReine und Angewandte Mathematik vol 140 pp 1ndash28 1911
[3] D S Mitrinovic J E Pecaric and A M Fink InequalitiesInvolving Functions and Their Integrals and Derivatives KluwerAcaremic Boston Mass USA 1991
[4] Y Bicheng ldquoOn Hilbertrsquos Integral Inequalityrdquo Journal of Math-ematical Analysis and Applications vol 220 no 2 pp 778ndash7851998
[5] B Yang and L Debnath ldquoOn the extended Hardy-Hilbertrsquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 272 no 1 pp 187ndash199 2002
[6] J Jin and L Debnath ldquoOn a Hilbert-type linear series operatorand its applicationsrdquo Journal of Mathematical Analysis andApplications vol 371 no 2 pp 691ndash704 2010
[7] B Yang and T M Rassias ldquoOn a new extension of Hilbertrsquosinequalityrdquo Mathematical Inequalities and Applications vol 8no 4 pp 575ndash582 2005
[8] M Krnic and J Pecaric ldquoHilbertrsquos inequalities and theirreversesrdquo Publicationes Mathematicae Debrecen vol 67 no 3-4 pp 315ndash331 2005
[9] L E Azar ldquoOn some extensions of hardy-hilbertrsquos inequalityand applicationsrdquo Journal of Inequalities and Applications vol2008 Article ID 546829 2008
8 Journal of Function Spaces and Applications
[10] Q Huang and B Yang ldquoOn a multiple hilbert-type integraloperator and applicationsrdquo Journal of Inequalities and Applica-tions vol 2009 Article ID 192197 2009
[11] Q Huang ldquoOn a Multiple hilbertrsquos inequality with parametersrdquoJournal of Inequalities and Applications vol 2010 Article ID309319 2010
[12] B Yang The Norm of Operator and Hilbert-Type InequalitiesScience Press Beijing China 2009 (Chinese)
[13] B Yang Hilbert-Type Integral Inequalities Bentham ScienceDubai UAE 2009
[14] B Yang Discrete Hilbert-Type Inequalities Bentham ScienceDubai UAE 2011
[15] J Kuang Applied Inequalities Shangdong Science TechnicPress Jinan China 2010 (Chinese)
[16] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1934
[17] B Yang ldquoA mixed Hilbert-type inequality with a best constantfactorrdquo International Journal of Pure and Applied Mathematicsvol 20 no 3 pp 319ndash328 2005
[18] B Yang ldquoOn a half-discrete reverse hilbert-yype inequality witha non-homogeneous kernelrdquo Journal of Inner Mongolia NormalUniversity (Natural Science Edition) vol 40 no 5 pp 433ndash4362011 (Chinese)
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Journal of Function Spaces and Applications
equivalent forms and the operator expressions as well assome reverses
2 Some Lemmas
In the following lemmas we assumed that 120572 120573 120582 gt 0 120572 lt
120582120573 120572 le min1 2 minus 120573 1205741isin (minusinfininfin) and 0 le 120574
2le 12
Lemma 1 Define the weight functions as follows
120596120572(119899) = (119899 minus 120574
2)120572
int
infin
1205741
(119909 minus 1205741)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
(119899 isin N)
(7)
120603120572(119909) = (119909 minus 120574
1)120572
infin
sum
119899=1
(119899 minus 1205742)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119909 isin (1205741infin))
(8)
Setting 119896 = 119896(120572 120573 120582) = (1120573)119861(120572120573 120582 minus 120572120573) one has thefollowing inequalities
0 lt 119896 (1 minus 120579 (119909)) lt 120603120572(119909) lt 120596
120572(119899) = 119896 (9)
where 120579(119909) = (1120573119896) int
(119909minus1205741)120573(1minus1205742)120573
0
(119906120572120573minus1
(1 + 119906)120582
)119889119906 isin
(0 1) and 120579(119909) = 119874((119909 minus 1205741)120572
) (119909 isin (1205741infin))
Proof Putting 119906 = (119909 minus 1205741)120573
(119899 minus 1205742)120573 in (7) we have
120596120572(119899) =
1
120573
int
infin
0
119906120572120573minus1
(1 + 119906)120582
119889119906 =
1
120573
119861(
120572
120573
120582 minus
120572
120573
) = 119896 (10)
For fixed 119909 isin (1205741infin) setting
119891 (119905) =
(119909 minus 1205741)120572
(119905 minus 1205742)120572minus1
[1 + (119909 minus 1205741)120573
(119905 minus 1205742)120573
]
120582
(119905 isin (1205742infin)) (11)
in view of the conditions we find that 1198911015840(119905) = minus(1 minus 120572)(119909 minus
1205741)120572
(119905 minus 1205742)120572minus2
[1 + (119909 minus 1205741)120573
(119905 minus 1205742)120573
]120582
minus 120582120573(119909 minus 1205741)120572+120573
(119905 minus
1205742)120572+120573minus2
[1+(119909minus1205741)120573
(119905minus1205742)120573
]120582+1
lt 0 and 11989110158401015840(119905) gt 0 By thefollowing Hadamard inequality (cf [15])
119891 (119899) lt int
119899+12
119899minus12
119891 (119905) 119889119905 (119899 isin N) (12)
and letting 119906 = (119909 minus 1205741)120573
(119905 minus 1205742)120573 it follows that
120603120572(119909) =
infin
sum
119899=1
119891 (119899) lt
infin
sum
119899=1
int
119899+12
119899minus12
119891 (119905) 119889119905 = int
infin
12
119891 (119905) 119889119905
le int
infin
1205742
119891 (119905) 119889119905 =
1
120573
int
infin
0
119906120572120573minus1
(1 + 119906)120582
119889119906
=
1
120573
119861(
120572
120573
120582 minus
120572
120573
) = 119896
120603120572(119909) =
infin
sum
119899=1
119891 (119899) gt int
infin
1
119891 (119905) 119889119905
= int
infin
1205742
119891 (119905) 119889119905 minus int
1
1205742
119891 (119905) 119889119905
= 119896 minus
1
120573
int
(119909minus1205741)120573(1minus1205742)120573
0
119906120572120573minus1
(1 + 119906)120582
119889119906
= 119896 (1 minus 120579 (119909)) gt 0
(13)
where
0 lt 120579 (119909) =
1
120573119896
int
(119909minus1205741)120573(1minus1205742)120573
0
119906120572120573minus1
(1 + 119906)120582
119889119906
lt
1
120573119896
int
(119909minus1205741)120573(1minus1205742)120573
0
119906120572120573minus1
119889119906 =
(1 minus 1205742)120572
(119909 minus 1205741)120572
120572119896
(14)
Hence we prove that (9) is valid
Lemma 2 Suppose that 1119901 + 1119902 = 1 (119901 = 0 1) and 119886119899ge 0
and119891(119909) ge 0 is a realmeasurable function in (1205741infin)Then the
following are considered (i) For 119901 gt 1 one has the following
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1
times[
[
[
int
infin
1205741
119891(119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
119901
1119901
le 1198961119902
int
infin
1205741
120603120572(119909)(119909 minus 120574
1)119901(1minus120572)minus1
119891119901
(119909)119889119909
1119901
(15)
Journal of Function Spaces and Applications 3
1198711=
int
infin
1205741
(119909 minus 1205741)119902120572minus1
120603119902minus1
120572 (119909)
times[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902
119889119909
1119901
le 119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
(16)
where 120603120572(119909) and 120596
120572(119899) are indicated by (7) and (8)
(ii) For 119901 lt 1 (119901 = 0) one has the reverses of (15) and (16)
Proof (i) By (7)ndash(9) andHolderrsquos inequality (cf [15]) we findthat
[
[
[
int
infin
1205741
119891(119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
119901
=
int
infin
1205741
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times [
(119909 minus 1205741)(1minus120572)119902
(119899 minus 1205742)(1minus120572)119901
119891 (119909)]
times[
(119899 minus 1205742)(1minus120572)119901
(119909 minus 1205741)(1minus120572)119902
]119889119909
119901
le int
infin
1205741
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119909 minus 1205741)(1minus120572)(119901minus1)
(119899 minus 1205742)1minus120572
119891119901
(119909) 119889119909
times
[
[
[
[
[
int
infin
1205741
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119899 minus 1205742)(1minus120572)(119902minus1)
(119909 minus 1205741)1minus120572
119889119909
]
]
]
]
]
119901minus1
= int
infin
1205741
119891119901
(119909) (119909 minus 1205741)(1minus120572)(119901minus1)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
(119899 minus 1205742)1minus120572
times [(119899 minus 1205742)119902(1minus120572)minus1
120596120572(119899)]
119901minus1
= (119899 minus 1205742)1minus119901120572
119896119901minus1
times int
infin
1205741
119891119901
(119909) (119909 minus 1205741)(1minus120572)(119901minus1)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
1
(119899 minus 1205742)1minus120572
119889119909
(17)
119869119901
le 119896119901minus1
infin
sum
119899=1
int
infin
1205741
119891119901
(119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119909 minus 1205741)(1minus120572)(119901minus1)
(119899 minus 1205742)1minus120572
119889119909
= 119896119901minus1
int
infin
1205741
infin
sum
119899=1
(119899 minus 1205742)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times (119909 minus 1205741)120572+119901(1minus120572)minus1
119891119901
(119909) 119889119909
= 119896119901minus1
int
infin
1205741
120603120572(119909) (119909 minus 120574
1)119901(1minus120572)minus1
119891119901
(119909) 119889119909
(18)
Hence (15) is valid Using Holderrsquos inequality again we have
[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902
=
infin
sum
119899=1
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times[
(119909 minus 1205741)(1minus120572)119902
(119899 minus 1205742)(1minus120572)119901
][
(119899 minus 1205742)(1minus120572)119901
(119909 minus 1205741)(1minus120572)119902
119886119899]
119902
le [120603120572(119909) (119909 minus 120574
1)119901(1minus120572)minus1
]
119902minus1
times
infin
sum
119899=1
119886119902
119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119899 minus 1205742)(1minus120572)119902119901
(119909 minus 1205741)1minus120572
= 120603119902minus1
120572(119909) (119909 minus 120574
1)1minus119902120572
times
infin
sum
119899=1
(119899 minus 1205742)(1minus120572)(119902minus1)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
1
(119909 minus 1205741)1minus120572
119886119902
119899
4 Journal of Function Spaces and Applications
119871119902
1le int
infin
1205741
infin
sum
119899=1
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119899 minus 1205742)(1minus120572)(119902minus1)
(119909 minus 1205741)1minus120572
119886119902
119899119889119909
=
infin
sum
119899=1
[
[
[
(119899 minus 1205742)
120572
int
infin
1205741
(119909 minus 1205741)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
times (119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
=
infin
sum
119899=1
120596120572(119899) (119899 minus 120574
2)119902(1minus120572)minus1
119886119902
119899
= 119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
(19)
Hence (16) is valid(ii) For 0 lt 119901 lt 1 (119902 lt 0) or 119901 lt 0 (0 lt 119902 lt 1) using the
reverse Holder inequality and in the same way we have thereverses of (15) and (16)
3 Main Results
Some functions and spaces are introduced as follow
120601 (119909) = (119909 minus 1205741)119901(1minus120572)minus1
120601 (119909) = (1 minus 120579 (119909)) 120601 (119909)
120595 (119899) = (119899 minus 1205742)119902(1minus120572)minus1
(119909 isin (1205741infin) 119899 isin N 120579 (119909) is indicated as Lemma 1)
119871119901120601(1205741infin) =119891
10038171003817100381710038171198911003817100381710038171003817119901120601
=int
infin
1205741
120601 (119909)1003816100381610038161003816119891 (119909)
1003816100381610038161003816
119901
119889119909
1119901
ltinfin
119897119902120595
= 119886=119886119899infin
119899=1 119886119902120595=
infin
sum
119899=1
120595 (119899)1003816100381610038161003816119886119899
1003816100381610038161003816
119902
1119902
ltinfin
(20)
Note If 119901 gt 1 then 119871119901120601(1205741infin) and 119897
119902120595are normal spaces
if 0 lt 119901 lt 1 or 119901 lt 0 then both 119871119901120601(1205741infin) and 119897
119902120595are
not normal spaces but we still use the formal symbols in thefollowing
Theorem 3 Suppose that 119901 gt 1 1119901 + 1119902 = 1 120572 120573 120582 gt
0 120572 lt 120582120573 120572 le min1 2 minus 120573 1205741
isin (minusinfininfin) 0 le
1205742le 12 and 119891(119909) 119886
119899ge 0 such that 119891 isin 119871
119901120601(1205741infin)
119886 = 119886119899infin
119899=1isin 119897119902120595
119891119901120601
gt 0 and 119886119902120595
gt 0 then we havethe following equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (x)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
(21)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891 (119909) 119889119909
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
]
]
]
119901
1119901
lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
(22)
119871 =
int
infin
1205741
(119909 minus 1205741)119902120572minus1[
[
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
]
]
]
119902
119889119909
1119902
lt 119896119886119902120595
(23)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof By the Lebesgue term-by-term integration theoremwe find that there are two expressions of 119868 in (21) In viewof (15) 120603
120572(119909) lt 119896 and 0 lt 119891
119901120601lt infin we have (22) By
Holderrsquos inequality we find that
119868 =
infin
sum
119899=1
[
[
[
(119899 minus 1205742)120572minus1119901
int
infin
1205741
119891 (119909) 119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(119899 minus 1205742)1119901minus120572
119886119899]
le 119869
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119869119886119902120595
(24)
Hence (21) is valid by (22) On the other hand assuming that(21) is valid and setting
119886119899= (119899 minus 120574
2)119901120572minus1[
[
[
int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
119901minus1
(119899 isin N) (25)
then we have
119886119902
119902120595=
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899= 119869119901
= 119868 (26)
Journal of Function Spaces and Applications 5
By (15) it follows that 119869 lt infin If 119869 = 0 then (22) is triviallyvalid If 119869 gt 0 then 0 lt 119886
119902120595= 119869119901minus1
lt infin By (21) we have
119886119902
119902120595= 119869119901
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595lt 119896
10038171003817100381710038171198911003817100381710038171003817119901120601
(27)
Hence (22) is valid which is equivalent to (21)In view of (16) and120603
120572(119909) lt 119896 we obtain (23) By Holderrsquos
inequality again we have
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
le 119871int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909
1119901
= 11987110038171003817100381710038171198911003817100381710038171003817119901120601
(28)
Then (21) is valid by using (23) Supposing that (21) is validand setting
119891 (119909) = (119909 minus 1205741)119902120572minus1[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(29)
then it follows that
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909) 119889119909 = 119871119902
= 119868 (30)
In view of (16) and (9) we find 119871 lt infin If 119871 = 0 then (23) istrivially valid if 119871 gt 0 that is 0 lt 119891
119901120601lt infin then by (21)
we have
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= 119871119902
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119871 =10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601lt 119896119886
119902120595
(31)
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 0 lt 120576 lt 119902120572 setting 119886 = 119886
119899infin
119899=1
and 119891(119909) as follows (119901 119902 gt 1)
119886119899= (119899 minus 120574
2)120572minus120576119902minus1
119891 (119909) =
(119909 minus 1205741)120572+120576119901minus1
119909 isin (1205741 1 + 120574
1)
0 119909 isin [1 + 1205741infin)
(32)
if there exists a positive number 1198961015840 le 119896 such that (21) is stillvalid as we replace 119896 by 1198961015840 then substitution of 119886 and
119891(119909)we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
(33)
For 119901 119902 gt 1 letting = 120572 minus 120576119902 we find by Lemma 1 that
119868 = int
1205741+1
1205741
(119909 minus 1205741)120576minus1
(119909 minus 1205741)120572minus120576119902
times
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
120603(119909) 119889119909
gt 119896 ( 120573 120582) int
1205741+1
1205741
(119909 minus 1205741)120576minus1
[1 minus 119874((119909 minus 1205741)
)] 119889119909
= 119896(120572 minus
120576
119902
120573 120582) (
1
120576
minus 119874 (1))
(34)
By simple calculation we have
119886119902
119902120595=
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899=
infin
sum
119899=1
(119899 minus 1205742)minus1minus120576
lt (1 minus 1205742)minus1minus120576
+ int
infin
1
(119910 minus 1205742)minus1minus120576
119889119910 =
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1
(35)
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
)
1119901
(36)
In view of (33) (34) and 119902 gt 1 it follows that
119896(120572 minus
120576
119902
120573 120582) [1 minus 120576119874 (1)]
lt 120576119868 lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
lt 1198961015840
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]
1119902
(37)
Thenwe have 119896(120572 120573 120582) = 119896 le 1198961015840 for 120576 rarr 0+ Hence 1198961015840 = 119896 is
the best value of (21) By the equivalency the constant factor kin (22) ((23)) is the best possible Otherwise we would reacha contradiction by (24) ((28)) that the constant factor in (21)is not the best possible
Remark 4 (i) Define a half-discrete Hilbert operator 119879
119871119901120601(1205741infin) rarr 119897
1199011205951minus119901 as follows For 119891 isin 119871
119901120601(1205741infin) we
define 119879119891 isin 1198971199011205951minus119901 satisfying
119879119891 (119899) = int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909 (119899 isin N) (38)
6 Journal of Function Spaces and Applications
Then by (22) it follows that 1198791198911199011205951minus119901 le 119896119891
119901120601 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (22) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)(ii) Define a half-discrete Hilbert operator 119879 119897
119902120595rarr
1198711199021206011minus119902(1205741infin) in the following way For 119886 isin 119897
119902120595 we define
119879119886 isin 119871
1199021206011minus119902 satisfying
119879119886 (119909) =
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119909 isin (1205741infin))
(39)
Then by (23) it follows that 1198791198861199021206011minus119902 le 119896119886
119902120595 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (23) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)
Theorem 5 Suppose that 0 lt 119901 lt 1 1119901 + 1119902 = 1 120572 120573 120582 gt0 120572 lt 120582120573 120572 le min1 2 minus 120573 120574
1isin (minusinfininfin) 0 le 120574
2le 12
and 119891(119909) 119886119899ge 0 such that 119891 isin 119871
119901120601(1205741infin) 119886 = 119886
119899infin
119899=1isin
119897119902120595
119891119901120601gt 0 and 119886
119902120595gt 0 Then one has the following
equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
(40)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891(119909)
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
119889119909]
]
]
119901
1119901
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
(41)
= int
infin
1205741
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]120582
]
119902
119889119909
1119902
gt 119896119886119902120595
(42)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof In view of (9) the reverse of (15) and 0 lt 119891119901120601lt infin
we have (41) Using the reverse Holder inequality we obtainthe reverse form of (24) as follows
119868 ge 119869119886119902120595 (43)
Then by (41) (40) is valid
On the other hand if (40) is valid setting 119886119899as in (25)
then (26) still holds with 0 lt 119901 lt 1 By (40) it follows that119869 gt 0 If 119869 = infin then (41) is trivially valid if 119869 lt infin then0 lt 119886
119902120595= 119869119901minus1
lt infin and we have
119886119902
119902120595= 119869119901
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595gt 119896
10038171003817100381710038171198911003817100381710038171003817119901 120601
(44)
Then (41) is valid which is equivalent to (40)By the reverse of (16) and120603(119909) gt 119896(1minus120579(119909)) and by 119902 lt 0
we have
gt 119896(119902minus1)119902
1198711ge 119896(119902minus1)119902
119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119896119886119902120595
(45)
Hence (42) is valid By the reverse Holder inequality againwe obtain
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
(1 minus 120579 (119909))1119901
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(1 minus 120579 (119909))1119901
(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
ge 10038171003817100381710038171198911003817100381710038171003817119901 120601
(46)
Then (40) is valid by (42) On the other hand if (40) is validsetting
119891 (119909) =
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(47)
then 119891119901119901120601
= int
infin
1205741
[1 minus 120579(119909)](119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909 = 119902
= 119868
By the reverse of (16) we have gt 0 If = infin then (42) istrivially valid if 0 lt lt infin then by (40) we obtain
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601
= 119902
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is = 10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601
gt 119896119886119902120595
(48)
Hence (42) is valid which is equivalent to (40) It follows that(40) (41) and (42) are equivalent
Journal of Function Spaces and Applications 7
If there exists a positive number 1198961015840 ge 119896 such that (40) isstill valid as we replace 119896 by 1198961015840 then in particular we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
gt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
119886119902120595
(49)
where 119886 = 119886119899infin
119899=1and
119891 are taken as in (32) (0 lt 120576 lt 119901(120573120582 minus120572)) Since by (35) we find in
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
= int
1+1205741
1205741
[1 minus 119874 ((119909 minus 1205741)120572
)] (119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
minus 119874(1))
1119901
119868 le
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
int
infin
1205741
(119909 minus 1205741)120572+120576119901minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
=
1
120573
infin
sum
119899=1
(119899 minus 1205742)minus120576minus1
int
infin
0
119906120572120573+120576120573119901minus1
(1 + 119906)120582
119889119906
le
1
120573
[
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
(50)
then by (35) (49) and the above results we obtain (noticethat 119902 lt 0)
1
120573
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
ge 120576119868 gt 1205761198961015840
(
1
120576
minus 119874(1))
1119901
(
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1)
1119902
= 1198961015840
(1 minus 120576119874(1))1119901
(
120576 + 1 minus 1205742
(1 minus 1205742)120576+1)
1119902
(51)
For 120576 rarr 0+ we have 119896 = (1120573)119861(120572120573 120582 minus 120572120573) ge 119896
1015840Hence 119896 = 119896
1015840 is the best value of (40) By the equivalencythe constant factor k in (41) ((42)) is the best possibleOtherwise wewould reach a contradiction by (43) ((46)) thatthe constant factor in (40) is not the best possible
In the same way for 119901 lt 0 we also have the followingresult
Theorem 6 If the assumption of 119901 gt 1 in Theorem 3 isreplaced by 119901 lt 0 then the reverses of (21) (22) and (23)are valid and equivalent Moreover the same constant factor isthe best possible
Remark 7 (i) If 1205741= 1205742= 0 120582 = 120573 = 1 and 120572 = 1119901
then 119896 = 119861(1119901 1119902) and (23) reduces to (3) In particularfor 119901 = 119902 = 2 (21) reduces to (4)
(ii) For 1205741= 0 120574
2= 12 and 120572 = 1119901 in (21) and (23) it
followsinfin
sum
119899=1
119886119899int
infin
0
119891 (119909)
(1 + (119909119899 minus 1199092)120573
)
120582
119889119909
lt 119896int
infin
0
119909119901minus2
119891119901
(119909)119889119909
1119901
infin
sum
119899=1
119886119902
119899
1119902
(52)
int
infin
0
119909119902minus2
(
infin
sum
119899=1
119886119899
(1 + (119909119899 minus 1199092)120573
)
120582
)
119902
119889119909 lt 119896119902
infin
sum
119899=1
119886119902
119899 (53)
where the constant factors both 119896 = (1120573)119861(1119901120573 120582 minus 1119901120573)and 119896119902 are the best possibles In particular for 120582 = 120573 = 1(53) comes to (5) for 120582 = 120573 = 1 and 119901 = 119902 = 2 in (52) weobtain (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 ConstructionSpecial Foundation Item of Guangdong Institution of HigherLearning College and University (no 2012KJCX0079)
References
[1] H Weyl Singulare integral gleichungen mit besonderer be-rucksichtigung des fourierschen integral theorems [Inaugeral-Dissertation] Gottingen 1908
[2] I Schur ldquoBernerkungen sur Theorie der beschrankten Bilin-earformenmit unendlich vielen veranderlichenrdquo Journal fur dieReine und Angewandte Mathematik vol 140 pp 1ndash28 1911
[3] D S Mitrinovic J E Pecaric and A M Fink InequalitiesInvolving Functions and Their Integrals and Derivatives KluwerAcaremic Boston Mass USA 1991
[4] Y Bicheng ldquoOn Hilbertrsquos Integral Inequalityrdquo Journal of Math-ematical Analysis and Applications vol 220 no 2 pp 778ndash7851998
[5] B Yang and L Debnath ldquoOn the extended Hardy-Hilbertrsquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 272 no 1 pp 187ndash199 2002
[6] J Jin and L Debnath ldquoOn a Hilbert-type linear series operatorand its applicationsrdquo Journal of Mathematical Analysis andApplications vol 371 no 2 pp 691ndash704 2010
[7] B Yang and T M Rassias ldquoOn a new extension of Hilbertrsquosinequalityrdquo Mathematical Inequalities and Applications vol 8no 4 pp 575ndash582 2005
[8] M Krnic and J Pecaric ldquoHilbertrsquos inequalities and theirreversesrdquo Publicationes Mathematicae Debrecen vol 67 no 3-4 pp 315ndash331 2005
[9] L E Azar ldquoOn some extensions of hardy-hilbertrsquos inequalityand applicationsrdquo Journal of Inequalities and Applications vol2008 Article ID 546829 2008
8 Journal of Function Spaces and Applications
[10] Q Huang and B Yang ldquoOn a multiple hilbert-type integraloperator and applicationsrdquo Journal of Inequalities and Applica-tions vol 2009 Article ID 192197 2009
[11] Q Huang ldquoOn a Multiple hilbertrsquos inequality with parametersrdquoJournal of Inequalities and Applications vol 2010 Article ID309319 2010
[12] B Yang The Norm of Operator and Hilbert-Type InequalitiesScience Press Beijing China 2009 (Chinese)
[13] B Yang Hilbert-Type Integral Inequalities Bentham ScienceDubai UAE 2009
[14] B Yang Discrete Hilbert-Type Inequalities Bentham ScienceDubai UAE 2011
[15] J Kuang Applied Inequalities Shangdong Science TechnicPress Jinan China 2010 (Chinese)
[16] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1934
[17] B Yang ldquoA mixed Hilbert-type inequality with a best constantfactorrdquo International Journal of Pure and Applied Mathematicsvol 20 no 3 pp 319ndash328 2005
[18] B Yang ldquoOn a half-discrete reverse hilbert-yype inequality witha non-homogeneous kernelrdquo Journal of Inner Mongolia NormalUniversity (Natural Science Edition) vol 40 no 5 pp 433ndash4362011 (Chinese)
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Stochastic AnalysisInternational Journal of
Journal of Function Spaces and Applications 3
1198711=
int
infin
1205741
(119909 minus 1205741)119902120572minus1
120603119902minus1
120572 (119909)
times[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902
119889119909
1119901
le 119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
(16)
where 120603120572(119909) and 120596
120572(119899) are indicated by (7) and (8)
(ii) For 119901 lt 1 (119901 = 0) one has the reverses of (15) and (16)
Proof (i) By (7)ndash(9) andHolderrsquos inequality (cf [15]) we findthat
[
[
[
int
infin
1205741
119891(119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
119901
=
int
infin
1205741
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times [
(119909 minus 1205741)(1minus120572)119902
(119899 minus 1205742)(1minus120572)119901
119891 (119909)]
times[
(119899 minus 1205742)(1minus120572)119901
(119909 minus 1205741)(1minus120572)119902
]119889119909
119901
le int
infin
1205741
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119909 minus 1205741)(1minus120572)(119901minus1)
(119899 minus 1205742)1minus120572
119891119901
(119909) 119889119909
times
[
[
[
[
[
int
infin
1205741
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119899 minus 1205742)(1minus120572)(119902minus1)
(119909 minus 1205741)1minus120572
119889119909
]
]
]
]
]
119901minus1
= int
infin
1205741
119891119901
(119909) (119909 minus 1205741)(1minus120572)(119901minus1)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
(119899 minus 1205742)1minus120572
times [(119899 minus 1205742)119902(1minus120572)minus1
120596120572(119899)]
119901minus1
= (119899 minus 1205742)1minus119901120572
119896119901minus1
times int
infin
1205741
119891119901
(119909) (119909 minus 1205741)(1minus120572)(119901minus1)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
1
(119899 minus 1205742)1minus120572
119889119909
(17)
119869119901
le 119896119901minus1
infin
sum
119899=1
int
infin
1205741
119891119901
(119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119909 minus 1205741)(1minus120572)(119901minus1)
(119899 minus 1205742)1minus120572
119889119909
= 119896119901minus1
int
infin
1205741
infin
sum
119899=1
(119899 minus 1205742)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times (119909 minus 1205741)120572+119901(1minus120572)minus1
119891119901
(119909) 119889119909
= 119896119901minus1
int
infin
1205741
120603120572(119909) (119909 minus 120574
1)119901(1minus120572)minus1
119891119901
(119909) 119889119909
(18)
Hence (15) is valid Using Holderrsquos inequality again we have
[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902
=
infin
sum
119899=1
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times[
(119909 minus 1205741)(1minus120572)119902
(119899 minus 1205742)(1minus120572)119901
][
(119899 minus 1205742)(1minus120572)119901
(119909 minus 1205741)(1minus120572)119902
119886119899]
119902
le [120603120572(119909) (119909 minus 120574
1)119901(1minus120572)minus1
]
119902minus1
times
infin
sum
119899=1
119886119902
119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119899 minus 1205742)(1minus120572)119902119901
(119909 minus 1205741)1minus120572
= 120603119902minus1
120572(119909) (119909 minus 120574
1)1minus119902120572
times
infin
sum
119899=1
(119899 minus 1205742)(1minus120572)(119902minus1)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
1
(119909 minus 1205741)1minus120572
119886119902
119899
4 Journal of Function Spaces and Applications
119871119902
1le int
infin
1205741
infin
sum
119899=1
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119899 minus 1205742)(1minus120572)(119902minus1)
(119909 minus 1205741)1minus120572
119886119902
119899119889119909
=
infin
sum
119899=1
[
[
[
(119899 minus 1205742)
120572
int
infin
1205741
(119909 minus 1205741)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
times (119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
=
infin
sum
119899=1
120596120572(119899) (119899 minus 120574
2)119902(1minus120572)minus1
119886119902
119899
= 119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
(19)
Hence (16) is valid(ii) For 0 lt 119901 lt 1 (119902 lt 0) or 119901 lt 0 (0 lt 119902 lt 1) using the
reverse Holder inequality and in the same way we have thereverses of (15) and (16)
3 Main Results
Some functions and spaces are introduced as follow
120601 (119909) = (119909 minus 1205741)119901(1minus120572)minus1
120601 (119909) = (1 minus 120579 (119909)) 120601 (119909)
120595 (119899) = (119899 minus 1205742)119902(1minus120572)minus1
(119909 isin (1205741infin) 119899 isin N 120579 (119909) is indicated as Lemma 1)
119871119901120601(1205741infin) =119891
10038171003817100381710038171198911003817100381710038171003817119901120601
=int
infin
1205741
120601 (119909)1003816100381610038161003816119891 (119909)
1003816100381610038161003816
119901
119889119909
1119901
ltinfin
119897119902120595
= 119886=119886119899infin
119899=1 119886119902120595=
infin
sum
119899=1
120595 (119899)1003816100381610038161003816119886119899
1003816100381610038161003816
119902
1119902
ltinfin
(20)
Note If 119901 gt 1 then 119871119901120601(1205741infin) and 119897
119902120595are normal spaces
if 0 lt 119901 lt 1 or 119901 lt 0 then both 119871119901120601(1205741infin) and 119897
119902120595are
not normal spaces but we still use the formal symbols in thefollowing
Theorem 3 Suppose that 119901 gt 1 1119901 + 1119902 = 1 120572 120573 120582 gt
0 120572 lt 120582120573 120572 le min1 2 minus 120573 1205741
isin (minusinfininfin) 0 le
1205742le 12 and 119891(119909) 119886
119899ge 0 such that 119891 isin 119871
119901120601(1205741infin)
119886 = 119886119899infin
119899=1isin 119897119902120595
119891119901120601
gt 0 and 119886119902120595
gt 0 then we havethe following equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (x)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
(21)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891 (119909) 119889119909
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
]
]
]
119901
1119901
lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
(22)
119871 =
int
infin
1205741
(119909 minus 1205741)119902120572minus1[
[
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
]
]
]
119902
119889119909
1119902
lt 119896119886119902120595
(23)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof By the Lebesgue term-by-term integration theoremwe find that there are two expressions of 119868 in (21) In viewof (15) 120603
120572(119909) lt 119896 and 0 lt 119891
119901120601lt infin we have (22) By
Holderrsquos inequality we find that
119868 =
infin
sum
119899=1
[
[
[
(119899 minus 1205742)120572minus1119901
int
infin
1205741
119891 (119909) 119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(119899 minus 1205742)1119901minus120572
119886119899]
le 119869
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119869119886119902120595
(24)
Hence (21) is valid by (22) On the other hand assuming that(21) is valid and setting
119886119899= (119899 minus 120574
2)119901120572minus1[
[
[
int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
119901minus1
(119899 isin N) (25)
then we have
119886119902
119902120595=
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899= 119869119901
= 119868 (26)
Journal of Function Spaces and Applications 5
By (15) it follows that 119869 lt infin If 119869 = 0 then (22) is triviallyvalid If 119869 gt 0 then 0 lt 119886
119902120595= 119869119901minus1
lt infin By (21) we have
119886119902
119902120595= 119869119901
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595lt 119896
10038171003817100381710038171198911003817100381710038171003817119901120601
(27)
Hence (22) is valid which is equivalent to (21)In view of (16) and120603
120572(119909) lt 119896 we obtain (23) By Holderrsquos
inequality again we have
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
le 119871int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909
1119901
= 11987110038171003817100381710038171198911003817100381710038171003817119901120601
(28)
Then (21) is valid by using (23) Supposing that (21) is validand setting
119891 (119909) = (119909 minus 1205741)119902120572minus1[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(29)
then it follows that
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909) 119889119909 = 119871119902
= 119868 (30)
In view of (16) and (9) we find 119871 lt infin If 119871 = 0 then (23) istrivially valid if 119871 gt 0 that is 0 lt 119891
119901120601lt infin then by (21)
we have
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= 119871119902
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119871 =10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601lt 119896119886
119902120595
(31)
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 0 lt 120576 lt 119902120572 setting 119886 = 119886
119899infin
119899=1
and 119891(119909) as follows (119901 119902 gt 1)
119886119899= (119899 minus 120574
2)120572minus120576119902minus1
119891 (119909) =
(119909 minus 1205741)120572+120576119901minus1
119909 isin (1205741 1 + 120574
1)
0 119909 isin [1 + 1205741infin)
(32)
if there exists a positive number 1198961015840 le 119896 such that (21) is stillvalid as we replace 119896 by 1198961015840 then substitution of 119886 and
119891(119909)we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
(33)
For 119901 119902 gt 1 letting = 120572 minus 120576119902 we find by Lemma 1 that
119868 = int
1205741+1
1205741
(119909 minus 1205741)120576minus1
(119909 minus 1205741)120572minus120576119902
times
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
120603(119909) 119889119909
gt 119896 ( 120573 120582) int
1205741+1
1205741
(119909 minus 1205741)120576minus1
[1 minus 119874((119909 minus 1205741)
)] 119889119909
= 119896(120572 minus
120576
119902
120573 120582) (
1
120576
minus 119874 (1))
(34)
By simple calculation we have
119886119902
119902120595=
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899=
infin
sum
119899=1
(119899 minus 1205742)minus1minus120576
lt (1 minus 1205742)minus1minus120576
+ int
infin
1
(119910 minus 1205742)minus1minus120576
119889119910 =
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1
(35)
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
)
1119901
(36)
In view of (33) (34) and 119902 gt 1 it follows that
119896(120572 minus
120576
119902
120573 120582) [1 minus 120576119874 (1)]
lt 120576119868 lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
lt 1198961015840
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]
1119902
(37)
Thenwe have 119896(120572 120573 120582) = 119896 le 1198961015840 for 120576 rarr 0+ Hence 1198961015840 = 119896 is
the best value of (21) By the equivalency the constant factor kin (22) ((23)) is the best possible Otherwise we would reacha contradiction by (24) ((28)) that the constant factor in (21)is not the best possible
Remark 4 (i) Define a half-discrete Hilbert operator 119879
119871119901120601(1205741infin) rarr 119897
1199011205951minus119901 as follows For 119891 isin 119871
119901120601(1205741infin) we
define 119879119891 isin 1198971199011205951minus119901 satisfying
119879119891 (119899) = int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909 (119899 isin N) (38)
6 Journal of Function Spaces and Applications
Then by (22) it follows that 1198791198911199011205951minus119901 le 119896119891
119901120601 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (22) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)(ii) Define a half-discrete Hilbert operator 119879 119897
119902120595rarr
1198711199021206011minus119902(1205741infin) in the following way For 119886 isin 119897
119902120595 we define
119879119886 isin 119871
1199021206011minus119902 satisfying
119879119886 (119909) =
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119909 isin (1205741infin))
(39)
Then by (23) it follows that 1198791198861199021206011minus119902 le 119896119886
119902120595 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (23) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)
Theorem 5 Suppose that 0 lt 119901 lt 1 1119901 + 1119902 = 1 120572 120573 120582 gt0 120572 lt 120582120573 120572 le min1 2 minus 120573 120574
1isin (minusinfininfin) 0 le 120574
2le 12
and 119891(119909) 119886119899ge 0 such that 119891 isin 119871
119901120601(1205741infin) 119886 = 119886
119899infin
119899=1isin
119897119902120595
119891119901120601gt 0 and 119886
119902120595gt 0 Then one has the following
equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
(40)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891(119909)
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
119889119909]
]
]
119901
1119901
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
(41)
= int
infin
1205741
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]120582
]
119902
119889119909
1119902
gt 119896119886119902120595
(42)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof In view of (9) the reverse of (15) and 0 lt 119891119901120601lt infin
we have (41) Using the reverse Holder inequality we obtainthe reverse form of (24) as follows
119868 ge 119869119886119902120595 (43)
Then by (41) (40) is valid
On the other hand if (40) is valid setting 119886119899as in (25)
then (26) still holds with 0 lt 119901 lt 1 By (40) it follows that119869 gt 0 If 119869 = infin then (41) is trivially valid if 119869 lt infin then0 lt 119886
119902120595= 119869119901minus1
lt infin and we have
119886119902
119902120595= 119869119901
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595gt 119896
10038171003817100381710038171198911003817100381710038171003817119901 120601
(44)
Then (41) is valid which is equivalent to (40)By the reverse of (16) and120603(119909) gt 119896(1minus120579(119909)) and by 119902 lt 0
we have
gt 119896(119902minus1)119902
1198711ge 119896(119902minus1)119902
119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119896119886119902120595
(45)
Hence (42) is valid By the reverse Holder inequality againwe obtain
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
(1 minus 120579 (119909))1119901
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(1 minus 120579 (119909))1119901
(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
ge 10038171003817100381710038171198911003817100381710038171003817119901 120601
(46)
Then (40) is valid by (42) On the other hand if (40) is validsetting
119891 (119909) =
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(47)
then 119891119901119901120601
= int
infin
1205741
[1 minus 120579(119909)](119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909 = 119902
= 119868
By the reverse of (16) we have gt 0 If = infin then (42) istrivially valid if 0 lt lt infin then by (40) we obtain
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601
= 119902
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is = 10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601
gt 119896119886119902120595
(48)
Hence (42) is valid which is equivalent to (40) It follows that(40) (41) and (42) are equivalent
Journal of Function Spaces and Applications 7
If there exists a positive number 1198961015840 ge 119896 such that (40) isstill valid as we replace 119896 by 1198961015840 then in particular we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
gt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
119886119902120595
(49)
where 119886 = 119886119899infin
119899=1and
119891 are taken as in (32) (0 lt 120576 lt 119901(120573120582 minus120572)) Since by (35) we find in
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
= int
1+1205741
1205741
[1 minus 119874 ((119909 minus 1205741)120572
)] (119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
minus 119874(1))
1119901
119868 le
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
int
infin
1205741
(119909 minus 1205741)120572+120576119901minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
=
1
120573
infin
sum
119899=1
(119899 minus 1205742)minus120576minus1
int
infin
0
119906120572120573+120576120573119901minus1
(1 + 119906)120582
119889119906
le
1
120573
[
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
(50)
then by (35) (49) and the above results we obtain (noticethat 119902 lt 0)
1
120573
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
ge 120576119868 gt 1205761198961015840
(
1
120576
minus 119874(1))
1119901
(
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1)
1119902
= 1198961015840
(1 minus 120576119874(1))1119901
(
120576 + 1 minus 1205742
(1 minus 1205742)120576+1)
1119902
(51)
For 120576 rarr 0+ we have 119896 = (1120573)119861(120572120573 120582 minus 120572120573) ge 119896
1015840Hence 119896 = 119896
1015840 is the best value of (40) By the equivalencythe constant factor k in (41) ((42)) is the best possibleOtherwise wewould reach a contradiction by (43) ((46)) thatthe constant factor in (40) is not the best possible
In the same way for 119901 lt 0 we also have the followingresult
Theorem 6 If the assumption of 119901 gt 1 in Theorem 3 isreplaced by 119901 lt 0 then the reverses of (21) (22) and (23)are valid and equivalent Moreover the same constant factor isthe best possible
Remark 7 (i) If 1205741= 1205742= 0 120582 = 120573 = 1 and 120572 = 1119901
then 119896 = 119861(1119901 1119902) and (23) reduces to (3) In particularfor 119901 = 119902 = 2 (21) reduces to (4)
(ii) For 1205741= 0 120574
2= 12 and 120572 = 1119901 in (21) and (23) it
followsinfin
sum
119899=1
119886119899int
infin
0
119891 (119909)
(1 + (119909119899 minus 1199092)120573
)
120582
119889119909
lt 119896int
infin
0
119909119901minus2
119891119901
(119909)119889119909
1119901
infin
sum
119899=1
119886119902
119899
1119902
(52)
int
infin
0
119909119902minus2
(
infin
sum
119899=1
119886119899
(1 + (119909119899 minus 1199092)120573
)
120582
)
119902
119889119909 lt 119896119902
infin
sum
119899=1
119886119902
119899 (53)
where the constant factors both 119896 = (1120573)119861(1119901120573 120582 minus 1119901120573)and 119896119902 are the best possibles In particular for 120582 = 120573 = 1(53) comes to (5) for 120582 = 120573 = 1 and 119901 = 119902 = 2 in (52) weobtain (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 ConstructionSpecial Foundation Item of Guangdong Institution of HigherLearning College and University (no 2012KJCX0079)
References
[1] H Weyl Singulare integral gleichungen mit besonderer be-rucksichtigung des fourierschen integral theorems [Inaugeral-Dissertation] Gottingen 1908
[2] I Schur ldquoBernerkungen sur Theorie der beschrankten Bilin-earformenmit unendlich vielen veranderlichenrdquo Journal fur dieReine und Angewandte Mathematik vol 140 pp 1ndash28 1911
[3] D S Mitrinovic J E Pecaric and A M Fink InequalitiesInvolving Functions and Their Integrals and Derivatives KluwerAcaremic Boston Mass USA 1991
[4] Y Bicheng ldquoOn Hilbertrsquos Integral Inequalityrdquo Journal of Math-ematical Analysis and Applications vol 220 no 2 pp 778ndash7851998
[5] B Yang and L Debnath ldquoOn the extended Hardy-Hilbertrsquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 272 no 1 pp 187ndash199 2002
[6] J Jin and L Debnath ldquoOn a Hilbert-type linear series operatorand its applicationsrdquo Journal of Mathematical Analysis andApplications vol 371 no 2 pp 691ndash704 2010
[7] B Yang and T M Rassias ldquoOn a new extension of Hilbertrsquosinequalityrdquo Mathematical Inequalities and Applications vol 8no 4 pp 575ndash582 2005
[8] M Krnic and J Pecaric ldquoHilbertrsquos inequalities and theirreversesrdquo Publicationes Mathematicae Debrecen vol 67 no 3-4 pp 315ndash331 2005
[9] L E Azar ldquoOn some extensions of hardy-hilbertrsquos inequalityand applicationsrdquo Journal of Inequalities and Applications vol2008 Article ID 546829 2008
8 Journal of Function Spaces and Applications
[10] Q Huang and B Yang ldquoOn a multiple hilbert-type integraloperator and applicationsrdquo Journal of Inequalities and Applica-tions vol 2009 Article ID 192197 2009
[11] Q Huang ldquoOn a Multiple hilbertrsquos inequality with parametersrdquoJournal of Inequalities and Applications vol 2010 Article ID309319 2010
[12] B Yang The Norm of Operator and Hilbert-Type InequalitiesScience Press Beijing China 2009 (Chinese)
[13] B Yang Hilbert-Type Integral Inequalities Bentham ScienceDubai UAE 2009
[14] B Yang Discrete Hilbert-Type Inequalities Bentham ScienceDubai UAE 2011
[15] J Kuang Applied Inequalities Shangdong Science TechnicPress Jinan China 2010 (Chinese)
[16] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1934
[17] B Yang ldquoA mixed Hilbert-type inequality with a best constantfactorrdquo International Journal of Pure and Applied Mathematicsvol 20 no 3 pp 319ndash328 2005
[18] B Yang ldquoOn a half-discrete reverse hilbert-yype inequality witha non-homogeneous kernelrdquo Journal of Inner Mongolia NormalUniversity (Natural Science Edition) vol 40 no 5 pp 433ndash4362011 (Chinese)
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Journal of Function Spaces and Applications
119871119902
1le int
infin
1205741
infin
sum
119899=1
1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
times
(119899 minus 1205742)(1minus120572)(119902minus1)
(119909 minus 1205741)1minus120572
119886119902
119899119889119909
=
infin
sum
119899=1
[
[
[
(119899 minus 1205742)
120572
int
infin
1205741
(119909 minus 1205741)120572minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
times (119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
=
infin
sum
119899=1
120596120572(119899) (119899 minus 120574
2)119902(1minus120572)minus1
119886119902
119899
= 119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
(19)
Hence (16) is valid(ii) For 0 lt 119901 lt 1 (119902 lt 0) or 119901 lt 0 (0 lt 119902 lt 1) using the
reverse Holder inequality and in the same way we have thereverses of (15) and (16)
3 Main Results
Some functions and spaces are introduced as follow
120601 (119909) = (119909 minus 1205741)119901(1minus120572)minus1
120601 (119909) = (1 minus 120579 (119909)) 120601 (119909)
120595 (119899) = (119899 minus 1205742)119902(1minus120572)minus1
(119909 isin (1205741infin) 119899 isin N 120579 (119909) is indicated as Lemma 1)
119871119901120601(1205741infin) =119891
10038171003817100381710038171198911003817100381710038171003817119901120601
=int
infin
1205741
120601 (119909)1003816100381610038161003816119891 (119909)
1003816100381610038161003816
119901
119889119909
1119901
ltinfin
119897119902120595
= 119886=119886119899infin
119899=1 119886119902120595=
infin
sum
119899=1
120595 (119899)1003816100381610038161003816119886119899
1003816100381610038161003816
119902
1119902
ltinfin
(20)
Note If 119901 gt 1 then 119871119901120601(1205741infin) and 119897
119902120595are normal spaces
if 0 lt 119901 lt 1 or 119901 lt 0 then both 119871119901120601(1205741infin) and 119897
119902120595are
not normal spaces but we still use the formal symbols in thefollowing
Theorem 3 Suppose that 119901 gt 1 1119901 + 1119902 = 1 120572 120573 120582 gt
0 120572 lt 120582120573 120572 le min1 2 minus 120573 1205741
isin (minusinfininfin) 0 le
1205742le 12 and 119891(119909) 119886
119899ge 0 such that 119891 isin 119871
119901120601(1205741infin)
119886 = 119886119899infin
119899=1isin 119897119902120595
119891119901120601
gt 0 and 119886119902120595
gt 0 then we havethe following equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (x)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
(21)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891 (119909) 119889119909
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
]
]
]
119901
1119901
lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
(22)
119871 =
int
infin
1205741
(119909 minus 1205741)119902120572minus1[
[
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
]
]
]
119902
119889119909
1119902
lt 119896119886119902120595
(23)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof By the Lebesgue term-by-term integration theoremwe find that there are two expressions of 119868 in (21) In viewof (15) 120603
120572(119909) lt 119896 and 0 lt 119891
119901120601lt infin we have (22) By
Holderrsquos inequality we find that
119868 =
infin
sum
119899=1
[
[
[
(119899 minus 1205742)120572minus1119901
int
infin
1205741
119891 (119909) 119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(119899 minus 1205742)1119901minus120572
119886119899]
le 119869
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119869119886119902120595
(24)
Hence (21) is valid by (22) On the other hand assuming that(21) is valid and setting
119886119899= (119899 minus 120574
2)119901120572minus1[
[
[
int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909]
]
]
119901minus1
(119899 isin N) (25)
then we have
119886119902
119902120595=
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899= 119869119901
= 119868 (26)
Journal of Function Spaces and Applications 5
By (15) it follows that 119869 lt infin If 119869 = 0 then (22) is triviallyvalid If 119869 gt 0 then 0 lt 119886
119902120595= 119869119901minus1
lt infin By (21) we have
119886119902
119902120595= 119869119901
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595lt 119896
10038171003817100381710038171198911003817100381710038171003817119901120601
(27)
Hence (22) is valid which is equivalent to (21)In view of (16) and120603
120572(119909) lt 119896 we obtain (23) By Holderrsquos
inequality again we have
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
le 119871int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909
1119901
= 11987110038171003817100381710038171198911003817100381710038171003817119901120601
(28)
Then (21) is valid by using (23) Supposing that (21) is validand setting
119891 (119909) = (119909 minus 1205741)119902120572minus1[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(29)
then it follows that
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909) 119889119909 = 119871119902
= 119868 (30)
In view of (16) and (9) we find 119871 lt infin If 119871 = 0 then (23) istrivially valid if 119871 gt 0 that is 0 lt 119891
119901120601lt infin then by (21)
we have
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= 119871119902
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119871 =10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601lt 119896119886
119902120595
(31)
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 0 lt 120576 lt 119902120572 setting 119886 = 119886
119899infin
119899=1
and 119891(119909) as follows (119901 119902 gt 1)
119886119899= (119899 minus 120574
2)120572minus120576119902minus1
119891 (119909) =
(119909 minus 1205741)120572+120576119901minus1
119909 isin (1205741 1 + 120574
1)
0 119909 isin [1 + 1205741infin)
(32)
if there exists a positive number 1198961015840 le 119896 such that (21) is stillvalid as we replace 119896 by 1198961015840 then substitution of 119886 and
119891(119909)we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
(33)
For 119901 119902 gt 1 letting = 120572 minus 120576119902 we find by Lemma 1 that
119868 = int
1205741+1
1205741
(119909 minus 1205741)120576minus1
(119909 minus 1205741)120572minus120576119902
times
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
120603(119909) 119889119909
gt 119896 ( 120573 120582) int
1205741+1
1205741
(119909 minus 1205741)120576minus1
[1 minus 119874((119909 minus 1205741)
)] 119889119909
= 119896(120572 minus
120576
119902
120573 120582) (
1
120576
minus 119874 (1))
(34)
By simple calculation we have
119886119902
119902120595=
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899=
infin
sum
119899=1
(119899 minus 1205742)minus1minus120576
lt (1 minus 1205742)minus1minus120576
+ int
infin
1
(119910 minus 1205742)minus1minus120576
119889119910 =
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1
(35)
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
)
1119901
(36)
In view of (33) (34) and 119902 gt 1 it follows that
119896(120572 minus
120576
119902
120573 120582) [1 minus 120576119874 (1)]
lt 120576119868 lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
lt 1198961015840
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]
1119902
(37)
Thenwe have 119896(120572 120573 120582) = 119896 le 1198961015840 for 120576 rarr 0+ Hence 1198961015840 = 119896 is
the best value of (21) By the equivalency the constant factor kin (22) ((23)) is the best possible Otherwise we would reacha contradiction by (24) ((28)) that the constant factor in (21)is not the best possible
Remark 4 (i) Define a half-discrete Hilbert operator 119879
119871119901120601(1205741infin) rarr 119897
1199011205951minus119901 as follows For 119891 isin 119871
119901120601(1205741infin) we
define 119879119891 isin 1198971199011205951minus119901 satisfying
119879119891 (119899) = int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909 (119899 isin N) (38)
6 Journal of Function Spaces and Applications
Then by (22) it follows that 1198791198911199011205951minus119901 le 119896119891
119901120601 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (22) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)(ii) Define a half-discrete Hilbert operator 119879 119897
119902120595rarr
1198711199021206011minus119902(1205741infin) in the following way For 119886 isin 119897
119902120595 we define
119879119886 isin 119871
1199021206011minus119902 satisfying
119879119886 (119909) =
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119909 isin (1205741infin))
(39)
Then by (23) it follows that 1198791198861199021206011minus119902 le 119896119886
119902120595 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (23) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)
Theorem 5 Suppose that 0 lt 119901 lt 1 1119901 + 1119902 = 1 120572 120573 120582 gt0 120572 lt 120582120573 120572 le min1 2 minus 120573 120574
1isin (minusinfininfin) 0 le 120574
2le 12
and 119891(119909) 119886119899ge 0 such that 119891 isin 119871
119901120601(1205741infin) 119886 = 119886
119899infin
119899=1isin
119897119902120595
119891119901120601gt 0 and 119886
119902120595gt 0 Then one has the following
equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
(40)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891(119909)
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
119889119909]
]
]
119901
1119901
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
(41)
= int
infin
1205741
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]120582
]
119902
119889119909
1119902
gt 119896119886119902120595
(42)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof In view of (9) the reverse of (15) and 0 lt 119891119901120601lt infin
we have (41) Using the reverse Holder inequality we obtainthe reverse form of (24) as follows
119868 ge 119869119886119902120595 (43)
Then by (41) (40) is valid
On the other hand if (40) is valid setting 119886119899as in (25)
then (26) still holds with 0 lt 119901 lt 1 By (40) it follows that119869 gt 0 If 119869 = infin then (41) is trivially valid if 119869 lt infin then0 lt 119886
119902120595= 119869119901minus1
lt infin and we have
119886119902
119902120595= 119869119901
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595gt 119896
10038171003817100381710038171198911003817100381710038171003817119901 120601
(44)
Then (41) is valid which is equivalent to (40)By the reverse of (16) and120603(119909) gt 119896(1minus120579(119909)) and by 119902 lt 0
we have
gt 119896(119902minus1)119902
1198711ge 119896(119902minus1)119902
119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119896119886119902120595
(45)
Hence (42) is valid By the reverse Holder inequality againwe obtain
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
(1 minus 120579 (119909))1119901
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(1 minus 120579 (119909))1119901
(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
ge 10038171003817100381710038171198911003817100381710038171003817119901 120601
(46)
Then (40) is valid by (42) On the other hand if (40) is validsetting
119891 (119909) =
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(47)
then 119891119901119901120601
= int
infin
1205741
[1 minus 120579(119909)](119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909 = 119902
= 119868
By the reverse of (16) we have gt 0 If = infin then (42) istrivially valid if 0 lt lt infin then by (40) we obtain
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601
= 119902
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is = 10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601
gt 119896119886119902120595
(48)
Hence (42) is valid which is equivalent to (40) It follows that(40) (41) and (42) are equivalent
Journal of Function Spaces and Applications 7
If there exists a positive number 1198961015840 ge 119896 such that (40) isstill valid as we replace 119896 by 1198961015840 then in particular we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
gt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
119886119902120595
(49)
where 119886 = 119886119899infin
119899=1and
119891 are taken as in (32) (0 lt 120576 lt 119901(120573120582 minus120572)) Since by (35) we find in
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
= int
1+1205741
1205741
[1 minus 119874 ((119909 minus 1205741)120572
)] (119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
minus 119874(1))
1119901
119868 le
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
int
infin
1205741
(119909 minus 1205741)120572+120576119901minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
=
1
120573
infin
sum
119899=1
(119899 minus 1205742)minus120576minus1
int
infin
0
119906120572120573+120576120573119901minus1
(1 + 119906)120582
119889119906
le
1
120573
[
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
(50)
then by (35) (49) and the above results we obtain (noticethat 119902 lt 0)
1
120573
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
ge 120576119868 gt 1205761198961015840
(
1
120576
minus 119874(1))
1119901
(
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1)
1119902
= 1198961015840
(1 minus 120576119874(1))1119901
(
120576 + 1 minus 1205742
(1 minus 1205742)120576+1)
1119902
(51)
For 120576 rarr 0+ we have 119896 = (1120573)119861(120572120573 120582 minus 120572120573) ge 119896
1015840Hence 119896 = 119896
1015840 is the best value of (40) By the equivalencythe constant factor k in (41) ((42)) is the best possibleOtherwise wewould reach a contradiction by (43) ((46)) thatthe constant factor in (40) is not the best possible
In the same way for 119901 lt 0 we also have the followingresult
Theorem 6 If the assumption of 119901 gt 1 in Theorem 3 isreplaced by 119901 lt 0 then the reverses of (21) (22) and (23)are valid and equivalent Moreover the same constant factor isthe best possible
Remark 7 (i) If 1205741= 1205742= 0 120582 = 120573 = 1 and 120572 = 1119901
then 119896 = 119861(1119901 1119902) and (23) reduces to (3) In particularfor 119901 = 119902 = 2 (21) reduces to (4)
(ii) For 1205741= 0 120574
2= 12 and 120572 = 1119901 in (21) and (23) it
followsinfin
sum
119899=1
119886119899int
infin
0
119891 (119909)
(1 + (119909119899 minus 1199092)120573
)
120582
119889119909
lt 119896int
infin
0
119909119901minus2
119891119901
(119909)119889119909
1119901
infin
sum
119899=1
119886119902
119899
1119902
(52)
int
infin
0
119909119902minus2
(
infin
sum
119899=1
119886119899
(1 + (119909119899 minus 1199092)120573
)
120582
)
119902
119889119909 lt 119896119902
infin
sum
119899=1
119886119902
119899 (53)
where the constant factors both 119896 = (1120573)119861(1119901120573 120582 minus 1119901120573)and 119896119902 are the best possibles In particular for 120582 = 120573 = 1(53) comes to (5) for 120582 = 120573 = 1 and 119901 = 119902 = 2 in (52) weobtain (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 ConstructionSpecial Foundation Item of Guangdong Institution of HigherLearning College and University (no 2012KJCX0079)
References
[1] H Weyl Singulare integral gleichungen mit besonderer be-rucksichtigung des fourierschen integral theorems [Inaugeral-Dissertation] Gottingen 1908
[2] I Schur ldquoBernerkungen sur Theorie der beschrankten Bilin-earformenmit unendlich vielen veranderlichenrdquo Journal fur dieReine und Angewandte Mathematik vol 140 pp 1ndash28 1911
[3] D S Mitrinovic J E Pecaric and A M Fink InequalitiesInvolving Functions and Their Integrals and Derivatives KluwerAcaremic Boston Mass USA 1991
[4] Y Bicheng ldquoOn Hilbertrsquos Integral Inequalityrdquo Journal of Math-ematical Analysis and Applications vol 220 no 2 pp 778ndash7851998
[5] B Yang and L Debnath ldquoOn the extended Hardy-Hilbertrsquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 272 no 1 pp 187ndash199 2002
[6] J Jin and L Debnath ldquoOn a Hilbert-type linear series operatorand its applicationsrdquo Journal of Mathematical Analysis andApplications vol 371 no 2 pp 691ndash704 2010
[7] B Yang and T M Rassias ldquoOn a new extension of Hilbertrsquosinequalityrdquo Mathematical Inequalities and Applications vol 8no 4 pp 575ndash582 2005
[8] M Krnic and J Pecaric ldquoHilbertrsquos inequalities and theirreversesrdquo Publicationes Mathematicae Debrecen vol 67 no 3-4 pp 315ndash331 2005
[9] L E Azar ldquoOn some extensions of hardy-hilbertrsquos inequalityand applicationsrdquo Journal of Inequalities and Applications vol2008 Article ID 546829 2008
8 Journal of Function Spaces and Applications
[10] Q Huang and B Yang ldquoOn a multiple hilbert-type integraloperator and applicationsrdquo Journal of Inequalities and Applica-tions vol 2009 Article ID 192197 2009
[11] Q Huang ldquoOn a Multiple hilbertrsquos inequality with parametersrdquoJournal of Inequalities and Applications vol 2010 Article ID309319 2010
[12] B Yang The Norm of Operator and Hilbert-Type InequalitiesScience Press Beijing China 2009 (Chinese)
[13] B Yang Hilbert-Type Integral Inequalities Bentham ScienceDubai UAE 2009
[14] B Yang Discrete Hilbert-Type Inequalities Bentham ScienceDubai UAE 2011
[15] J Kuang Applied Inequalities Shangdong Science TechnicPress Jinan China 2010 (Chinese)
[16] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1934
[17] B Yang ldquoA mixed Hilbert-type inequality with a best constantfactorrdquo International Journal of Pure and Applied Mathematicsvol 20 no 3 pp 319ndash328 2005
[18] B Yang ldquoOn a half-discrete reverse hilbert-yype inequality witha non-homogeneous kernelrdquo Journal of Inner Mongolia NormalUniversity (Natural Science Edition) vol 40 no 5 pp 433ndash4362011 (Chinese)
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Journal of Function Spaces and Applications 5
By (15) it follows that 119869 lt infin If 119869 = 0 then (22) is triviallyvalid If 119869 gt 0 then 0 lt 119886
119902120595= 119869119901minus1
lt infin By (21) we have
119886119902
119902120595= 119869119901
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595lt 119896
10038171003817100381710038171198911003817100381710038171003817119901120601
(27)
Hence (22) is valid which is equivalent to (21)In view of (16) and120603
120572(119909) lt 119896 we obtain (23) By Holderrsquos
inequality again we have
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
le 119871int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909
1119901
= 11987110038171003817100381710038171198911003817100381710038171003817119901120601
(28)
Then (21) is valid by using (23) Supposing that (21) is validand setting
119891 (119909) = (119909 minus 1205741)119902120572minus1[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(29)
then it follows that
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= int
infin
1205741
(119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909) 119889119909 = 119871119902
= 119868 (30)
In view of (16) and (9) we find 119871 lt infin If 119871 = 0 then (23) istrivially valid if 119871 gt 0 that is 0 lt 119891
119901120601lt infin then by (21)
we have
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601= 119871119902
= 119868 lt 11989610038171003817100381710038171198911003817100381710038171003817119901120601
119886119902120595
that is 119871 =10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601lt 119896119886
119902120595
(31)
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 0 lt 120576 lt 119902120572 setting 119886 = 119886
119899infin
119899=1
and 119891(119909) as follows (119901 119902 gt 1)
119886119899= (119899 minus 120574
2)120572minus120576119902minus1
119891 (119909) =
(119909 minus 1205741)120572+120576119901minus1
119909 isin (1205741 1 + 120574
1)
0 119909 isin [1 + 1205741infin)
(32)
if there exists a positive number 1198961015840 le 119896 such that (21) is stillvalid as we replace 119896 by 1198961015840 then substitution of 119886 and
119891(119909)we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
(33)
For 119901 119902 gt 1 letting = 120572 minus 120576119902 we find by Lemma 1 that
119868 = int
1205741+1
1205741
(119909 minus 1205741)120576minus1
(119909 minus 1205741)120572minus120576119902
times
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
120603(119909) 119889119909
gt 119896 ( 120573 120582) int
1205741+1
1205741
(119909 minus 1205741)120576minus1
[1 minus 119874((119909 minus 1205741)
)] 119889119909
= 119896(120572 minus
120576
119902
120573 120582) (
1
120576
minus 119874 (1))
(34)
By simple calculation we have
119886119902
119902120595=
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899=
infin
sum
119899=1
(119899 minus 1205742)minus1minus120576
lt (1 minus 1205742)minus1minus120576
+ int
infin
1
(119910 minus 1205742)minus1minus120576
119889119910 =
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1
(35)
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
= int
1205741+1
1205741
(119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
)
1119901
(36)
In view of (33) (34) and 119902 gt 1 it follows that
119896(120572 minus
120576
119902
120573 120582) [1 minus 120576119874 (1)]
lt 120576119868 lt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901120601
119886119902120595
lt 1198961015840
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]
1119902
(37)
Thenwe have 119896(120572 120573 120582) = 119896 le 1198961015840 for 120576 rarr 0+ Hence 1198961015840 = 119896 is
the best value of (21) By the equivalency the constant factor kin (22) ((23)) is the best possible Otherwise we would reacha contradiction by (24) ((28)) that the constant factor in (21)is not the best possible
Remark 4 (i) Define a half-discrete Hilbert operator 119879
119871119901120601(1205741infin) rarr 119897
1199011205951minus119901 as follows For 119891 isin 119871
119901120601(1205741infin) we
define 119879119891 isin 1198971199011205951minus119901 satisfying
119879119891 (119899) = int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909 (119899 isin N) (38)
6 Journal of Function Spaces and Applications
Then by (22) it follows that 1198791198911199011205951minus119901 le 119896119891
119901120601 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (22) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)(ii) Define a half-discrete Hilbert operator 119879 119897
119902120595rarr
1198711199021206011minus119902(1205741infin) in the following way For 119886 isin 119897
119902120595 we define
119879119886 isin 119871
1199021206011minus119902 satisfying
119879119886 (119909) =
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119909 isin (1205741infin))
(39)
Then by (23) it follows that 1198791198861199021206011minus119902 le 119896119886
119902120595 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (23) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)
Theorem 5 Suppose that 0 lt 119901 lt 1 1119901 + 1119902 = 1 120572 120573 120582 gt0 120572 lt 120582120573 120572 le min1 2 minus 120573 120574
1isin (minusinfininfin) 0 le 120574
2le 12
and 119891(119909) 119886119899ge 0 such that 119891 isin 119871
119901120601(1205741infin) 119886 = 119886
119899infin
119899=1isin
119897119902120595
119891119901120601gt 0 and 119886
119902120595gt 0 Then one has the following
equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
(40)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891(119909)
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
119889119909]
]
]
119901
1119901
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
(41)
= int
infin
1205741
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]120582
]
119902
119889119909
1119902
gt 119896119886119902120595
(42)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof In view of (9) the reverse of (15) and 0 lt 119891119901120601lt infin
we have (41) Using the reverse Holder inequality we obtainthe reverse form of (24) as follows
119868 ge 119869119886119902120595 (43)
Then by (41) (40) is valid
On the other hand if (40) is valid setting 119886119899as in (25)
then (26) still holds with 0 lt 119901 lt 1 By (40) it follows that119869 gt 0 If 119869 = infin then (41) is trivially valid if 119869 lt infin then0 lt 119886
119902120595= 119869119901minus1
lt infin and we have
119886119902
119902120595= 119869119901
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595gt 119896
10038171003817100381710038171198911003817100381710038171003817119901 120601
(44)
Then (41) is valid which is equivalent to (40)By the reverse of (16) and120603(119909) gt 119896(1minus120579(119909)) and by 119902 lt 0
we have
gt 119896(119902minus1)119902
1198711ge 119896(119902minus1)119902
119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119896119886119902120595
(45)
Hence (42) is valid By the reverse Holder inequality againwe obtain
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
(1 minus 120579 (119909))1119901
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(1 minus 120579 (119909))1119901
(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
ge 10038171003817100381710038171198911003817100381710038171003817119901 120601
(46)
Then (40) is valid by (42) On the other hand if (40) is validsetting
119891 (119909) =
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(47)
then 119891119901119901120601
= int
infin
1205741
[1 minus 120579(119909)](119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909 = 119902
= 119868
By the reverse of (16) we have gt 0 If = infin then (42) istrivially valid if 0 lt lt infin then by (40) we obtain
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601
= 119902
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is = 10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601
gt 119896119886119902120595
(48)
Hence (42) is valid which is equivalent to (40) It follows that(40) (41) and (42) are equivalent
Journal of Function Spaces and Applications 7
If there exists a positive number 1198961015840 ge 119896 such that (40) isstill valid as we replace 119896 by 1198961015840 then in particular we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
gt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
119886119902120595
(49)
where 119886 = 119886119899infin
119899=1and
119891 are taken as in (32) (0 lt 120576 lt 119901(120573120582 minus120572)) Since by (35) we find in
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
= int
1+1205741
1205741
[1 minus 119874 ((119909 minus 1205741)120572
)] (119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
minus 119874(1))
1119901
119868 le
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
int
infin
1205741
(119909 minus 1205741)120572+120576119901minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
=
1
120573
infin
sum
119899=1
(119899 minus 1205742)minus120576minus1
int
infin
0
119906120572120573+120576120573119901minus1
(1 + 119906)120582
119889119906
le
1
120573
[
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
(50)
then by (35) (49) and the above results we obtain (noticethat 119902 lt 0)
1
120573
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
ge 120576119868 gt 1205761198961015840
(
1
120576
minus 119874(1))
1119901
(
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1)
1119902
= 1198961015840
(1 minus 120576119874(1))1119901
(
120576 + 1 minus 1205742
(1 minus 1205742)120576+1)
1119902
(51)
For 120576 rarr 0+ we have 119896 = (1120573)119861(120572120573 120582 minus 120572120573) ge 119896
1015840Hence 119896 = 119896
1015840 is the best value of (40) By the equivalencythe constant factor k in (41) ((42)) is the best possibleOtherwise wewould reach a contradiction by (43) ((46)) thatthe constant factor in (40) is not the best possible
In the same way for 119901 lt 0 we also have the followingresult
Theorem 6 If the assumption of 119901 gt 1 in Theorem 3 isreplaced by 119901 lt 0 then the reverses of (21) (22) and (23)are valid and equivalent Moreover the same constant factor isthe best possible
Remark 7 (i) If 1205741= 1205742= 0 120582 = 120573 = 1 and 120572 = 1119901
then 119896 = 119861(1119901 1119902) and (23) reduces to (3) In particularfor 119901 = 119902 = 2 (21) reduces to (4)
(ii) For 1205741= 0 120574
2= 12 and 120572 = 1119901 in (21) and (23) it
followsinfin
sum
119899=1
119886119899int
infin
0
119891 (119909)
(1 + (119909119899 minus 1199092)120573
)
120582
119889119909
lt 119896int
infin
0
119909119901minus2
119891119901
(119909)119889119909
1119901
infin
sum
119899=1
119886119902
119899
1119902
(52)
int
infin
0
119909119902minus2
(
infin
sum
119899=1
119886119899
(1 + (119909119899 minus 1199092)120573
)
120582
)
119902
119889119909 lt 119896119902
infin
sum
119899=1
119886119902
119899 (53)
where the constant factors both 119896 = (1120573)119861(1119901120573 120582 minus 1119901120573)and 119896119902 are the best possibles In particular for 120582 = 120573 = 1(53) comes to (5) for 120582 = 120573 = 1 and 119901 = 119902 = 2 in (52) weobtain (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 ConstructionSpecial Foundation Item of Guangdong Institution of HigherLearning College and University (no 2012KJCX0079)
References
[1] H Weyl Singulare integral gleichungen mit besonderer be-rucksichtigung des fourierschen integral theorems [Inaugeral-Dissertation] Gottingen 1908
[2] I Schur ldquoBernerkungen sur Theorie der beschrankten Bilin-earformenmit unendlich vielen veranderlichenrdquo Journal fur dieReine und Angewandte Mathematik vol 140 pp 1ndash28 1911
[3] D S Mitrinovic J E Pecaric and A M Fink InequalitiesInvolving Functions and Their Integrals and Derivatives KluwerAcaremic Boston Mass USA 1991
[4] Y Bicheng ldquoOn Hilbertrsquos Integral Inequalityrdquo Journal of Math-ematical Analysis and Applications vol 220 no 2 pp 778ndash7851998
[5] B Yang and L Debnath ldquoOn the extended Hardy-Hilbertrsquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 272 no 1 pp 187ndash199 2002
[6] J Jin and L Debnath ldquoOn a Hilbert-type linear series operatorand its applicationsrdquo Journal of Mathematical Analysis andApplications vol 371 no 2 pp 691ndash704 2010
[7] B Yang and T M Rassias ldquoOn a new extension of Hilbertrsquosinequalityrdquo Mathematical Inequalities and Applications vol 8no 4 pp 575ndash582 2005
[8] M Krnic and J Pecaric ldquoHilbertrsquos inequalities and theirreversesrdquo Publicationes Mathematicae Debrecen vol 67 no 3-4 pp 315ndash331 2005
[9] L E Azar ldquoOn some extensions of hardy-hilbertrsquos inequalityand applicationsrdquo Journal of Inequalities and Applications vol2008 Article ID 546829 2008
8 Journal of Function Spaces and Applications
[10] Q Huang and B Yang ldquoOn a multiple hilbert-type integraloperator and applicationsrdquo Journal of Inequalities and Applica-tions vol 2009 Article ID 192197 2009
[11] Q Huang ldquoOn a Multiple hilbertrsquos inequality with parametersrdquoJournal of Inequalities and Applications vol 2010 Article ID309319 2010
[12] B Yang The Norm of Operator and Hilbert-Type InequalitiesScience Press Beijing China 2009 (Chinese)
[13] B Yang Hilbert-Type Integral Inequalities Bentham ScienceDubai UAE 2009
[14] B Yang Discrete Hilbert-Type Inequalities Bentham ScienceDubai UAE 2011
[15] J Kuang Applied Inequalities Shangdong Science TechnicPress Jinan China 2010 (Chinese)
[16] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1934
[17] B Yang ldquoA mixed Hilbert-type inequality with a best constantfactorrdquo International Journal of Pure and Applied Mathematicsvol 20 no 3 pp 319ndash328 2005
[18] B Yang ldquoOn a half-discrete reverse hilbert-yype inequality witha non-homogeneous kernelrdquo Journal of Inner Mongolia NormalUniversity (Natural Science Edition) vol 40 no 5 pp 433ndash4362011 (Chinese)
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Journal of Function Spaces and Applications
Then by (22) it follows that 1198791198911199011205951minus119901 le 119896119891
119901120601 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (22) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)(ii) Define a half-discrete Hilbert operator 119879 119897
119902120595rarr
1198711199021206011minus119902(1205741infin) in the following way For 119886 isin 119897
119902120595 we define
119879119886 isin 119871
1199021206011minus119902 satisfying
119879119886 (119909) =
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
(119909 isin (1205741infin))
(39)
Then by (23) it follows that 1198791198861199021206011minus119902 le 119896119886
119902120595 that is 119879
is the bounded operator with 119879 le 119896 Since the constantfactor 119896 in (23) is the best possible then we have 119879 = 119896 =
(1120573)119861(120572120573 120582 minus 120572120573)
Theorem 5 Suppose that 0 lt 119901 lt 1 1119901 + 1119902 = 1 120572 120573 120582 gt0 120572 lt 120582120573 120572 le min1 2 minus 120573 120574
1isin (minusinfininfin) 0 le 120574
2le 12
and 119891(119909) 119886119899ge 0 such that 119891 isin 119871
119901120601(1205741infin) 119886 = 119886
119899infin
119899=1isin
119897119902120595
119891119901120601gt 0 and 119886
119902120595gt 0 Then one has the following
equivalent inequalities
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
= int
infin
1205741
119891 (119909)
infin
sum
119899=1
119886119899119889119909
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
(40)
119869 =
infin
sum
119899=1
(119899 minus 1205742)119901120572minus1[
[
[
int
infin
1205741
119891(119909)
[1+(119909minus1205741)120573
(119899minus1205742)120573
]
120582
119889119909]
]
]
119901
1119901
gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
(41)
= int
infin
1205741
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
infin
sum
119899=1
119886119899
[1+(119909minus1205741)120573
(119899minus1205742)120573
]120582
]
119902
119889119909
1119902
gt 119896119886119902120595
(42)
where the constant factor 119896 = (1120573)119861(120572120573 120582 minus 120572120573) is the bestpossible
Proof In view of (9) the reverse of (15) and 0 lt 119891119901120601lt infin
we have (41) Using the reverse Holder inequality we obtainthe reverse form of (24) as follows
119868 ge 119869119886119902120595 (43)
Then by (41) (40) is valid
On the other hand if (40) is valid setting 119886119899as in (25)
then (26) still holds with 0 lt 119901 lt 1 By (40) it follows that119869 gt 0 If 119869 = infin then (41) is trivially valid if 119869 lt infin then0 lt 119886
119902120595= 119869119901minus1
lt infin and we have
119886119902
119902120595= 119869119901
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is 119869 = 119886119902minus1
119902120595gt 119896
10038171003817100381710038171198911003817100381710038171003817119901 120601
(44)
Then (41) is valid which is equivalent to (40)By the reverse of (16) and120603(119909) gt 119896(1minus120579(119909)) and by 119902 lt 0
we have
gt 119896(119902minus1)119902
1198711ge 119896(119902minus1)119902
119896
infin
sum
119899=1
(119899 minus 1205742)119902(1minus120572)minus1
119886119902
119899
1119902
= 119896119886119902120595
(45)
Hence (42) is valid By the reverse Holder inequality againwe obtain
119868 = int
infin
1205741
[
[
[
(119909 minus 1205741)120572minus1119902
(1 minus 120579 (119909))1119901
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
times [(1 minus 120579 (119909))1119901
(119909 minus 1205741)1119902minus120572
119891 (119909)] 119889119909
ge 10038171003817100381710038171198911003817100381710038171003817119901 120601
(46)
Then (40) is valid by (42) On the other hand if (40) is validsetting
119891 (119909) =
(119909 minus 1205741)119902120572minus1
[1 minus 120579 (119909)]119902minus1
[
[
[
infin
sum
119899=1
119886119899
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
]
]
]
119902minus1
(119909 isin (1205741infin))
(47)
then 119891119901119901120601
= int
infin
1205741
[1 minus 120579(119909)](119909 minus 1205741)119901(1minus120572)minus1
119891119901
(119909)119889119909 = 119902
= 119868
By the reverse of (16) we have gt 0 If = infin then (42) istrivially valid if 0 lt lt infin then by (40) we obtain
10038171003817100381710038171198911003817100381710038171003817
119901
119901120601
= 119902
= 119868 gt 11989610038171003817100381710038171198911003817100381710038171003817119901 120601
119886119902120595
that is = 10038171003817100381710038171198911003817100381710038171003817
119901minus1
119901120601
gt 119896119886119902120595
(48)
Hence (42) is valid which is equivalent to (40) It follows that(40) (41) and (42) are equivalent
Journal of Function Spaces and Applications 7
If there exists a positive number 1198961015840 ge 119896 such that (40) isstill valid as we replace 119896 by 1198961015840 then in particular we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
gt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
119886119902120595
(49)
where 119886 = 119886119899infin
119899=1and
119891 are taken as in (32) (0 lt 120576 lt 119901(120573120582 minus120572)) Since by (35) we find in
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
= int
1+1205741
1205741
[1 minus 119874 ((119909 minus 1205741)120572
)] (119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
minus 119874(1))
1119901
119868 le
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
int
infin
1205741
(119909 minus 1205741)120572+120576119901minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
=
1
120573
infin
sum
119899=1
(119899 minus 1205742)minus120576minus1
int
infin
0
119906120572120573+120576120573119901minus1
(1 + 119906)120582
119889119906
le
1
120573
[
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
(50)
then by (35) (49) and the above results we obtain (noticethat 119902 lt 0)
1
120573
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
ge 120576119868 gt 1205761198961015840
(
1
120576
minus 119874(1))
1119901
(
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1)
1119902
= 1198961015840
(1 minus 120576119874(1))1119901
(
120576 + 1 minus 1205742
(1 minus 1205742)120576+1)
1119902
(51)
For 120576 rarr 0+ we have 119896 = (1120573)119861(120572120573 120582 minus 120572120573) ge 119896
1015840Hence 119896 = 119896
1015840 is the best value of (40) By the equivalencythe constant factor k in (41) ((42)) is the best possibleOtherwise wewould reach a contradiction by (43) ((46)) thatthe constant factor in (40) is not the best possible
In the same way for 119901 lt 0 we also have the followingresult
Theorem 6 If the assumption of 119901 gt 1 in Theorem 3 isreplaced by 119901 lt 0 then the reverses of (21) (22) and (23)are valid and equivalent Moreover the same constant factor isthe best possible
Remark 7 (i) If 1205741= 1205742= 0 120582 = 120573 = 1 and 120572 = 1119901
then 119896 = 119861(1119901 1119902) and (23) reduces to (3) In particularfor 119901 = 119902 = 2 (21) reduces to (4)
(ii) For 1205741= 0 120574
2= 12 and 120572 = 1119901 in (21) and (23) it
followsinfin
sum
119899=1
119886119899int
infin
0
119891 (119909)
(1 + (119909119899 minus 1199092)120573
)
120582
119889119909
lt 119896int
infin
0
119909119901minus2
119891119901
(119909)119889119909
1119901
infin
sum
119899=1
119886119902
119899
1119902
(52)
int
infin
0
119909119902minus2
(
infin
sum
119899=1
119886119899
(1 + (119909119899 minus 1199092)120573
)
120582
)
119902
119889119909 lt 119896119902
infin
sum
119899=1
119886119902
119899 (53)
where the constant factors both 119896 = (1120573)119861(1119901120573 120582 minus 1119901120573)and 119896119902 are the best possibles In particular for 120582 = 120573 = 1(53) comes to (5) for 120582 = 120573 = 1 and 119901 = 119902 = 2 in (52) weobtain (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 ConstructionSpecial Foundation Item of Guangdong Institution of HigherLearning College and University (no 2012KJCX0079)
References
[1] H Weyl Singulare integral gleichungen mit besonderer be-rucksichtigung des fourierschen integral theorems [Inaugeral-Dissertation] Gottingen 1908
[2] I Schur ldquoBernerkungen sur Theorie der beschrankten Bilin-earformenmit unendlich vielen veranderlichenrdquo Journal fur dieReine und Angewandte Mathematik vol 140 pp 1ndash28 1911
[3] D S Mitrinovic J E Pecaric and A M Fink InequalitiesInvolving Functions and Their Integrals and Derivatives KluwerAcaremic Boston Mass USA 1991
[4] Y Bicheng ldquoOn Hilbertrsquos Integral Inequalityrdquo Journal of Math-ematical Analysis and Applications vol 220 no 2 pp 778ndash7851998
[5] B Yang and L Debnath ldquoOn the extended Hardy-Hilbertrsquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 272 no 1 pp 187ndash199 2002
[6] J Jin and L Debnath ldquoOn a Hilbert-type linear series operatorand its applicationsrdquo Journal of Mathematical Analysis andApplications vol 371 no 2 pp 691ndash704 2010
[7] B Yang and T M Rassias ldquoOn a new extension of Hilbertrsquosinequalityrdquo Mathematical Inequalities and Applications vol 8no 4 pp 575ndash582 2005
[8] M Krnic and J Pecaric ldquoHilbertrsquos inequalities and theirreversesrdquo Publicationes Mathematicae Debrecen vol 67 no 3-4 pp 315ndash331 2005
[9] L E Azar ldquoOn some extensions of hardy-hilbertrsquos inequalityand applicationsrdquo Journal of Inequalities and Applications vol2008 Article ID 546829 2008
8 Journal of Function Spaces and Applications
[10] Q Huang and B Yang ldquoOn a multiple hilbert-type integraloperator and applicationsrdquo Journal of Inequalities and Applica-tions vol 2009 Article ID 192197 2009
[11] Q Huang ldquoOn a Multiple hilbertrsquos inequality with parametersrdquoJournal of Inequalities and Applications vol 2010 Article ID309319 2010
[12] B Yang The Norm of Operator and Hilbert-Type InequalitiesScience Press Beijing China 2009 (Chinese)
[13] B Yang Hilbert-Type Integral Inequalities Bentham ScienceDubai UAE 2009
[14] B Yang Discrete Hilbert-Type Inequalities Bentham ScienceDubai UAE 2011
[15] J Kuang Applied Inequalities Shangdong Science TechnicPress Jinan China 2010 (Chinese)
[16] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1934
[17] B Yang ldquoA mixed Hilbert-type inequality with a best constantfactorrdquo International Journal of Pure and Applied Mathematicsvol 20 no 3 pp 319ndash328 2005
[18] B Yang ldquoOn a half-discrete reverse hilbert-yype inequality witha non-homogeneous kernelrdquo Journal of Inner Mongolia NormalUniversity (Natural Science Edition) vol 40 no 5 pp 433ndash4362011 (Chinese)
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Journal of Function Spaces and Applications 7
If there exists a positive number 1198961015840 ge 119896 such that (40) isstill valid as we replace 119896 by 1198961015840 then in particular we have
119868 =
infin
sum
119899=1
119886119899int
infin
1205741
119891 (119909)
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
gt 119896101584010038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
119886119902120595
(49)
where 119886 = 119886119899infin
119899=1and
119891 are taken as in (32) (0 lt 120576 lt 119901(120573120582 minus120572)) Since by (35) we find in
10038171003817100381710038171003817
119891
10038171003817100381710038171003817119901 120601
= int
1+1205741
1205741
[1 minus 119874 ((119909 minus 1205741)120572
)] (119909 minus 1205741)120576minus1
119889119909
1119901
= (
1
120576
minus 119874(1))
1119901
119868 le
infin
sum
119899=1
(119899 minus 1205742)120572minus120576119902minus1
int
infin
1205741
(119909 minus 1205741)120572+120576119901minus1
[1 + (119909 minus 1205741)120573
(119899 minus 1205742)120573
]
120582
119889119909
=
1
120573
infin
sum
119899=1
(119899 minus 1205742)minus120576minus1
int
infin
0
119906120572120573+120576120573119901minus1
(1 + 119906)120582
119889119906
le
1
120573
[
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
(50)
then by (35) (49) and the above results we obtain (noticethat 119902 lt 0)
1
120573
[
120576 + 1 minus 1205742
(1 minus 1205742)120576+1]119861(
120572
120573
+
120576
120573119901
120582 minus
120572
120573
minus
120576
120573119901
)
ge 120576119868 gt 1205761198961015840
(
1
120576
minus 119874(1))
1119901
(
120576 + 1 minus 1205742
120576(1 minus 1205742)120576+1)
1119902
= 1198961015840
(1 minus 120576119874(1))1119901
(
120576 + 1 minus 1205742
(1 minus 1205742)120576+1)
1119902
(51)
For 120576 rarr 0+ we have 119896 = (1120573)119861(120572120573 120582 minus 120572120573) ge 119896
1015840Hence 119896 = 119896
1015840 is the best value of (40) By the equivalencythe constant factor k in (41) ((42)) is the best possibleOtherwise wewould reach a contradiction by (43) ((46)) thatthe constant factor in (40) is not the best possible
In the same way for 119901 lt 0 we also have the followingresult
Theorem 6 If the assumption of 119901 gt 1 in Theorem 3 isreplaced by 119901 lt 0 then the reverses of (21) (22) and (23)are valid and equivalent Moreover the same constant factor isthe best possible
Remark 7 (i) If 1205741= 1205742= 0 120582 = 120573 = 1 and 120572 = 1119901
then 119896 = 119861(1119901 1119902) and (23) reduces to (3) In particularfor 119901 = 119902 = 2 (21) reduces to (4)
(ii) For 1205741= 0 120574
2= 12 and 120572 = 1119901 in (21) and (23) it
followsinfin
sum
119899=1
119886119899int
infin
0
119891 (119909)
(1 + (119909119899 minus 1199092)120573
)
120582
119889119909
lt 119896int
infin
0
119909119901minus2
119891119901
(119909)119889119909
1119901
infin
sum
119899=1
119886119902
119899
1119902
(52)
int
infin
0
119909119902minus2
(
infin
sum
119899=1
119886119899
(1 + (119909119899 minus 1199092)120573
)
120582
)
119902
119889119909 lt 119896119902
infin
sum
119899=1
119886119902
119899 (53)
where the constant factors both 119896 = (1120573)119861(1119901120573 120582 minus 1119901120573)and 119896119902 are the best possibles In particular for 120582 = 120573 = 1(53) comes to (5) for 120582 = 120573 = 1 and 119901 = 119902 = 2 in (52) weobtain (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 ConstructionSpecial Foundation Item of Guangdong Institution of HigherLearning College and University (no 2012KJCX0079)
References
[1] H Weyl Singulare integral gleichungen mit besonderer be-rucksichtigung des fourierschen integral theorems [Inaugeral-Dissertation] Gottingen 1908
[2] I Schur ldquoBernerkungen sur Theorie der beschrankten Bilin-earformenmit unendlich vielen veranderlichenrdquo Journal fur dieReine und Angewandte Mathematik vol 140 pp 1ndash28 1911
[3] D S Mitrinovic J E Pecaric and A M Fink InequalitiesInvolving Functions and Their Integrals and Derivatives KluwerAcaremic Boston Mass USA 1991
[4] Y Bicheng ldquoOn Hilbertrsquos Integral Inequalityrdquo Journal of Math-ematical Analysis and Applications vol 220 no 2 pp 778ndash7851998
[5] B Yang and L Debnath ldquoOn the extended Hardy-Hilbertrsquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 272 no 1 pp 187ndash199 2002
[6] J Jin and L Debnath ldquoOn a Hilbert-type linear series operatorand its applicationsrdquo Journal of Mathematical Analysis andApplications vol 371 no 2 pp 691ndash704 2010
[7] B Yang and T M Rassias ldquoOn a new extension of Hilbertrsquosinequalityrdquo Mathematical Inequalities and Applications vol 8no 4 pp 575ndash582 2005
[8] M Krnic and J Pecaric ldquoHilbertrsquos inequalities and theirreversesrdquo Publicationes Mathematicae Debrecen vol 67 no 3-4 pp 315ndash331 2005
[9] L E Azar ldquoOn some extensions of hardy-hilbertrsquos inequalityand applicationsrdquo Journal of Inequalities and Applications vol2008 Article ID 546829 2008
8 Journal of Function Spaces and Applications
[10] Q Huang and B Yang ldquoOn a multiple hilbert-type integraloperator and applicationsrdquo Journal of Inequalities and Applica-tions vol 2009 Article ID 192197 2009
[11] Q Huang ldquoOn a Multiple hilbertrsquos inequality with parametersrdquoJournal of Inequalities and Applications vol 2010 Article ID309319 2010
[12] B Yang The Norm of Operator and Hilbert-Type InequalitiesScience Press Beijing China 2009 (Chinese)
[13] B Yang Hilbert-Type Integral Inequalities Bentham ScienceDubai UAE 2009
[14] B Yang Discrete Hilbert-Type Inequalities Bentham ScienceDubai UAE 2011
[15] J Kuang Applied Inequalities Shangdong Science TechnicPress Jinan China 2010 (Chinese)
[16] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1934
[17] B Yang ldquoA mixed Hilbert-type inequality with a best constantfactorrdquo International Journal of Pure and Applied Mathematicsvol 20 no 3 pp 319ndash328 2005
[18] B Yang ldquoOn a half-discrete reverse hilbert-yype inequality witha non-homogeneous kernelrdquo Journal of Inner Mongolia NormalUniversity (Natural Science Edition) vol 40 no 5 pp 433ndash4362011 (Chinese)
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Journal of Function Spaces and Applications
[10] Q Huang and B Yang ldquoOn a multiple hilbert-type integraloperator and applicationsrdquo Journal of Inequalities and Applica-tions vol 2009 Article ID 192197 2009
[11] Q Huang ldquoOn a Multiple hilbertrsquos inequality with parametersrdquoJournal of Inequalities and Applications vol 2010 Article ID309319 2010
[12] B Yang The Norm of Operator and Hilbert-Type InequalitiesScience Press Beijing China 2009 (Chinese)
[13] B Yang Hilbert-Type Integral Inequalities Bentham ScienceDubai UAE 2009
[14] B Yang Discrete Hilbert-Type Inequalities Bentham ScienceDubai UAE 2011
[15] J Kuang Applied Inequalities Shangdong Science TechnicPress Jinan China 2010 (Chinese)
[16] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1934
[17] B Yang ldquoA mixed Hilbert-type inequality with a best constantfactorrdquo International Journal of Pure and Applied Mathematicsvol 20 no 3 pp 319ndash328 2005
[18] B Yang ldquoOn a half-discrete reverse hilbert-yype inequality witha non-homogeneous kernelrdquo Journal of Inner Mongolia NormalUniversity (Natural Science Edition) vol 40 no 5 pp 433ndash4362011 (Chinese)
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
