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On New Generalizations of Hilbert's Inequality and Their Applications
Journal of Mathematical Analysis and Applications 245, 248᎐265 Ž2000. doi:10.1006rjmaa.2000.6766, available online at http:rrwww.idealibrary.com on
On New Generalizations of Hilbert’s Inequality and Their Applications Kuang Jichang Department of Mathematics, Hunan Normal Uni¨ ersity, Changsha, Hunan 410081 People’s Republic of China
and Lokenath Debnath Department of Mathematics, Uni¨ ersity of Central Florida, Orlando, Florida 32816 Communicated by William F. Ames Received January 4, 2000
This paper deals with new generalizations of Hilbert’s inequality and their applications. It is shown that results of this paper are significant extensions and improvements of many known results. 䊚 2000 Academic Press Key Words: Hilbert’s inequality; Euler᎐Maclaurin summation formula; beta function.
1. INTRODUCTION If a n , bn ) 0, p ) 1, 1p q 1q s 1, and 0 - Ý n a np - ⬁, 0 - Ý n bnq - ⬁, then the famous Hilbert’s inequality Žsee Hardy et al. w1x. is given by ⬁
⬁
Ý Ý ms1 ns1
a m bn mqn
-
sin Ž rp . 248
0022-247Xr00 $35.00 Copyright 䊚 2000 by Academic Press All rights of reproduction in any form reserved.
1rp
⬁
1rq
⬁
žÝ / žÝ / a np
ns1
bnq
ns1
.
Ž 1.1.
GENERALIZATIONS OF HILBERT’S INEQUALITY
249
This is a slightly sharper form of the inequality ⬁
⬁
a m bn
Ý Ý
mqnq1
ms0 ns0
-
1rp
⬁
1rq
⬁
žÝ / žÝ / a np
sin Ž rp .
bnq
ns0
,
Ž 1.2.
ns0
where the constant factor rsinŽrp . is the best possible. Inequalities Ž1.1. and Ž1.2. have been proved in many different ways and they have varied applications Žsee Hardy et al. w1x and Kuang w2x and references cited in w1, 2x.. Recently, Hsu w3, 4x first introduced the weight function in the form Ž p s q s 2.
Ž r , n. s
sin p
y Ž r , n. ,
Ž r , n. ) 0
Ž 1.3.
and replaced the right hand side of Ž1.1. by
½Ý n
1rp
y Ž q, n .
sin rp
a np
5 ½Ý
1rq
sin rp
n
y Ž p, n .
bnq
5
. Ž 1.4.
Gao w5, 7x gave some improvements of Ž1.3.. Later Gao and Yang w8, 9x found the best possible value of Ž r, n.,
Ž r , n. s
1yc n1y 1r r
,
where c is the Euler constant. Yang and Debnath w10x prove that
Ž p, n . s
1 1r q
2n
q ny1r p
.
The similar estimate of Ž r, n. in Ž1.3. corresponding to Ž1.2. can be found in Gao w11x and Yang w12x. Yang and Debnath w13x proved that ⬁
⬁
Ý Ý ms0 ns0
a m bn
Ž m q n q 1.
py2q qy2q -B , p q
ž
=
⬁
1
nq
2
/½ Ý ž ns0
nq
1 2
1rp
1y
/
a np
5
1rq
1y
½Ýž / 5 ns0
⬁
bnq
,
Ž 1.5.
250
JICHANG AND DEBNATH
where B Ž u, ¨ . is the beta function, a n , bn G 0, p ) 1, 1p q 1q s 1, 2 y min p, q4 - F 2, and 0 - Ý⬁ns 0 Ž n q 1r2.1y a np - ⬁, 0 - Ý⬁ns0 Ž n q 1r2.1y bnq - ⬁. Ingham w14x has proved that if a n G 0, 0 - Ý⬁ns0 a2n - ⬁, ) 0, then ⬁
⬁
Ý Ý ms0 ns0
am an mqnq
⬁
F M Ž .
Ý a2n ,
Ž 1.6.
ns0
where M Ž . s
½
rsin , M Ž 1r2 . s ,
0 - F 1r2 . ) 1r2
5
Ž 1.7.
In this work, it is interesting to consider the double series with two parameters, ⬁
⬁
Ý Ý ms0 ns0
a m bn
Ž m q n q 2 .
t
,
Ž 1.8.
and find the corresponding beta function B Ž u, ¨ ., which depends on t, , p, q. The major objective of this paper is to formulate a new inequality related to the general series ⬁
⬁
Ý Ý K Ž m q , n q . am bn ,
Ž 1.9.
ms0 ns0
with Ž1.8. as a special case. Our main results can be stated as follows: THEOREM 1.1. Let a n , bn G 0, p ) 1, 1p q 1q s 1, 0 - Ý⬁ns0 Ž n q 1y t p . a n - ⬁, 0 - Ý⬁ns0 Ž n q .1y t bnq - ⬁ 1r2 F F 1r2 min p, q4 , and let Ž K x, y . be a nonnegati¨ e and homogeneous function of degree yt Ž t ) 0.. If K Ž1, y . has its first four deri¨ ati¨ es continuous on Ž0, ⬁., and Žy1. n K Ž n. Ž1, y . G 0 for n s 0, 1, 2, 3, 4, K Ž m. Ž1, y . yy2 r r ª 0, y ª ⬁ for m s 0, 1, I Ž r , . s
⬁
y2 r r
H0 K Ž 1, u . u
du - ⬁,
r s p, q,
Ž 1.10.
GENERALIZATIONS OF HILBERT’S INEQUALITY
251
then ⬁
⬁
Ý Ý K Ž m q , n q . am bn ms0 ns0
-
1rp
⬁
½Ý ½Ý
I Ž q, . y Ž q, m, t , . Ž m q .
1y t
a mp
ms0
1rq
⬁
=
5
I Ž p, . y Ž p, m, t , . Ž m q .
1y t
5
bmq
ms0
, Ž 1.11.
where
Ž r , m, t , . s
ž
1y 2 rr
mq
½ž
/
K 1, y
mq
/
1 1 y 2 rr
1 24 Ž m q .
K ⬘ 1,
ž
1
y
2
mq
1q
ž /5
1 3r
/
) 0, Ž 1.12.
and r s p, q. Remark 1. When 0 - - 1r2, Theorem 1.1 fails. But we have the following result: THEOREM 1.2. Suppose that p, q, a n , bn and K Ž x, y . are defined as in Theorem 1.1. If ) 0, then N
N
Ý Ý K Ž m q , n q . am bn ms0 ns0 1rp
N
-
½Ý ½Ý
I Ž q, . y g Ž q, . Ž m q .
1y t
a mp
ms0
1rq
N
=
5
I Ž p, . y g Ž p, . Ž m q .
ms0
1y t
bmq
5
, Ž 1.13.
where g Ž r , . s
⬁
y2 r r
HŽ Nq2 .rŽ mq.K Ž 1, u . u
du,
r s p, q.
Ž 1.14.
252
JICHANG AND DEBNATH
Remark 2. When N ª ⬁, Ž1.13. reduces to the inequality ⬁
⬁
Ý Ý K Ž m q , n q . am bn ms0 ns0
- I Ž q, .
1rp
I Ž p, .
1rq
1rp
⬁
½Ý 5
Ž m q .
1yt
a mp
ms0
=
⬁
½Ý
5
1rq
Ž m q .
1y t
ms0
bmq
.
Ž 1.15.
Remark 3. If we replace the exponent Ž2 rpq . in Ž3.4. below by Ž2 y t .rpq, where 2 y min p, q4 - t F 2, we can similarly prove that N
N
Ý Ý K Ž m q , n q . am bn ms0 ns0 1rp
N
-
½Ý ½Ý
I1 Ž q, t . y g 1 Ž q, t . Ž m q .
1y t
a mp
ms0
1rq
N
=
5
I1 Ž p, t . y g 1 Ž p, t . Ž m q .
1y t
bmq
1yt
bmq
ms0
5
Ž 1.16.
and ⬁
⬁
Ý Ý K Ž m q , n q . am bn ms0 ns0
- I1 Ž q, t . =
⬁
½Ý
1rp
I1 Ž p, t .
1rq 1rp
Ž m q .
1y t
ms0
a mp
1rq
⬁
5 ½Ý
Ž m q .
ms0
5
. Ž 1.17.
where ⬁
K Ž 1, u .
I1 Ž r , t . s
H0
g 1Ž r , t . s
HŽ Nq2 .rŽ mq. u
⬁
uŽ2yt .r r
du - ⬁, K Ž 1, u . Ž2yt .r r
Ž 1.18. du.
Ž 1.19.
GENERALIZATIONS OF HILBERT’S INEQUALITY
253
In particular, when K Ž x, y . s Ž x q y .yt , s 1, 2, inequality Ž1.17. leads to Ž3.1. in Yang and Debnath w13x. In Section 4 we discuss some applications.
2. SOME LEMMAS For the proof of the above theorems, we need two lemmas: LEMMA 2.1. Let f ha¨ e its first four deri¨ ati¨ es continuous on Ž0, ⬁. and Žy1. n f Ž n. Ž x . G 0 Ž n s 0, 1, 2, 3, 4., and f Ž x ., f ⬘Ž x . ª 0 Ž x ª ⬁.; then ⬁
⬁
Ý f Ž k . - H f Ž x . dx q 0
ks0
1 2
f Ž 0. y
1 12
f ⬘ Ž 0. .
Ž 2.1.
Proof. By the Euler᎐Maclaurin summation formula Žsee Xu and Wang w15, p. 88x., we have N
N
Ý f Ž k . s H f Ž x . dx q
ks0
0
q
1
N Ž4.
H 24 0
f
1 2
f Ž N . q f Ž 0. q
Ž B4 y B4 Ž x y w x x .
1 12
dx . ,
f ⬘ Ž N . y f ⬘ Ž 0.
Ž 2.2.
where Bk , Bk Ž x . are the Bernoulli numbers and the Bernoulli polynomials, respectively. Since B4 s y1r30 and the sign of B4 y B4 Ž x y w x x. is the same as B4 , hence we obtain N
N
Ý f Ž k . - H f Ž x . dx q
ks0
0
1 2
f Ž N . q f Ž 0. q
1 12
f ⬘ Ž N . y f ⬘ Ž 0. .
Ž 2.3. Let N ª ⬁; Ž2.3. implies Ž2.1.. The lemma is proved. Remark 4. Lemma 2.1 shows that Ž2.1. holds under weaker conditions on f than those in w13x. LEMMA 2.2. Suppose that K Ž x, y . and are as in Theorem 1.1, r ) 1. Define a weight function by
Ž r , m, t , . s Ž m q .
2 rr
⬁
Ý K Ž m q , n q . Ž n q . y2 rr ; Ž 2.4. ns0
254
JICHANG AND DEBNATH
then
Ž r , m, t , . - Ž m q .
1y t
I Ž r , . y Ž r , m, t , . ,
Ž 2.5.
where I Ž r, . and Ž r, m, t, . are gi¨ en by Ž1.10. and Ž1.12., respecti¨ ely. Proof. Let f Ž x . s K Ž m q , x q .Ž x q .y2 r r ; then, by Lemma 2.1, we have ⬁
2 rr
Ž r , m, t , . s Ž m q .
Ý f Ž n. ns0
1y t
- Ž m q .
½
⬁
y2 r r
H0 K Ž 1, u . u
du y Ž r , m, t , . ,
5
Ž 2.6. where
H0r mq K Ž 1, u . u Ž
Ž r , m, t , . s
y q
ž
.
y2 r r
1y 2 rr
mq 1
12
2
ž
/
ž
1 2
du
q
1
/ ž
K 1,
6 r
2y 2 rr
mq
K ⬘ 1,
/
ž
mq
Integrating by parts twice gives
H0r mq K Ž 1, u . u Ž
.
s
y2 r r
1 1 y 2 rr
ž
du
K 1,
mq
/ž
mq
H0r mq K ⬘ Ž 1, u . u
y
Ž
.
1y 2 rr
/
1y 2 r r
du
mq
/
.
/ Ž 2.7.
GENERALIZATIONS OF HILBERT’S INEQUALITY
s
1
1y 2 rr
ž
1 y 2 rr m q y
/
ž
K 1,
1
2y 2 rr
K ⬘ 1,
/
Ž 1 y 2 rr . Ž 2 y 2 rr . m q
ž
rmq
y 1
1y 2 rr
ž
1 y 2 rr m q y
/
ž
K 1,
1
Ž 1 y 2 rr . Ž 2 y 2 rr
. ž
mq
mq
/
2y 2 rr
K ⬘ 1,
/
mq
/
K ⬙ Ž 1, u . u 2y 2 r r du
H0
)
/
mq
ž
255
ž
/
.
mq
/
mq
From this and equality Ž2.7., we obtain
Ž r , m, t , . )
1 1 y 2 rr
1
y
2
ž
1q
1 3r
/ž
1
y
Ž 1 y 2 rr . Ž 2 y 2 rr . =
ž
mq
ž
/
ž
K 1,
1 12 2
K ⬘ 1,
/
mq
y
2y 2 rr
1y 2 rr
/
mq
.
Ž 2.8.
Since 1r2 F - Ž1r2.min p, q4 , that is, 1 F 2 - r, r s p, q, we have
G
1 2
)
1 2
y
2r q 1 2 Ž 3r q 3r q 1 . 2
s
r Ž 3r q 1 . 6r2 q 6r q 2
that is, 1
-
2 r
q
6r 3r q 1
.
This implies that 1
y
2 r
-
ž
1 2
q
1 6r
y1
/
,
,
256
JICHANG AND DEBNATH
so that
ž
1
y
2
1
/ž
r
2
1
q
6r
s 1y
/ ž
2
1
/ ž 2
r
1q
1 3r
- 1.
/
This implies that 1 1 y 2 rr
1
y
2
ž
1
1q
3r
) 0.
/
Ž 2.9.
Next note that 0 - 1rr - 1 and 2rr - 1r F 2; these imply that 0-
ž
1
2
y
r
/ž
1
y
1
/
r
- 4,
that is, 1
Ž 1 y 2 rr . Ž 2 y 2 rr .
1
)
8 2
.
Thus, 1
Ž 1 y 2 rr . Ž 2 y 2 rr . By hypotheses, K ⬘Ž1, that
mq
y
1 12
2
)
1 8
2
y
1 12
2
s
1 242
. Ž 2.10.
. F 0. This and inequalities Ž2.8. ᎐ Ž2.10. imply
Ž r , m, t , . )
ž
mq
1y 2 rr
/
½ž
K 1,
mq
/ y
s Ž r , m, t , . ) 0.
1 1 y 2 rr
y
1 24 Ž m q .
1 2
ž
1q
K ⬘ 1,
ž
1 3r
/
mq
/5
Ž 2.11.
From Ž2.6., Ž2.7., and Ž2.11., we obtain inequality Ž2.5.. The lemma is proved.
GENERALIZATIONS OF HILBERT’S INEQUALITY
257
3. PROOFS Proof of Theorem 1.1. By Holder’s inequality, we obtain ¨ ⬁
⬁
Ý Ý K Ž m q , n q . am bn ms0 ns0
s
⬁
⬁
Ý Ý
1rp
am K Ž m q , n q .
ms0 ns0
= bn K Ž m q , n q . F
⬁
½Ý
1rq
Ž q, m, t , .
ms0
nq
nq
ž
1rp
a mp
ž
mq
mq ⬁
5 ½Ý
2 rp q
/
2 rp q
/ 1rq
Ž p, m, t , .
ms0
bmq
5
, Ž 3.1.
where
Ž r , m, t , . s Ž m q .
2 rr
⬁
Ý K Ž m q , n q . Ž n q . y2 rr , ns0
r s p, q.
Ž 3.2.
Thus, by Ž2.5., we have Ž1.11.. Theorem 1.1 is proved. Proof of Theorem 1.2. Define f by f Ž x . s K Ž m q , x q . Ž x q .
y2 rr
.
Note that f ⬘ Ž x . s K ⬘ Ž m q , x q . Ž x q . y Ž 2 rr . Ž x q .
y2 rry1
y2 rr
K Ž m q , x q . - 0
and f ⬙ Ž x . s K ⬙ Ž m q , x q . Ž x q . y Ž 4rr . Ž x q .
y2 rry1
y2 rr
K ⬘ Ž m q , x q .
q Ž 2 rr . Ž 2 rr q 1 . Ž x q .
y2 rry2
K Ž m q , x q . ) 0.
258
JICHANG AND DEBNATH
Hence, f Ž x . is strictly convex. Therefore, N
Ý f Ž m. - H
Nq
y
ms0
f Ž x . dx
s Ž m q .
H0 Nq2 r mq K Ž 1, u . u
1y ty2 rr
Ž
. Ž
.
y2 r r
du. Ž 3.3.
In view of Holder’s inequality, we get ¨ N
N
Ý Ý K Ž m q , n q . am bn ms0 ns0 N
s
N
Ý Ý am K Ž m q , n q . 1rp ms0 ns0
= bn K Ž m q , n q .
1rp
N
F
½Ý
1rq
Ž q, m, t , .
a mp
ms0
ž
ž
mq nq
nq mq
2 rp q
/
2 rp q
/ 1rq
N
5 ½Ý
Ž p, m, t , .
bmq
ms0
5
, Ž 3.4.
where
Ž r , m, t , . s Ž m q .
2 rr
N
Ý K Ž m q , n q . Ž n q . y2 rr . Ž 3.5. ns0
From this and inequality Ž3.3., we have
Ž r , m, t , . - Ž m q .
1y t
I Ž r , . y g Ž r , . 4 ,
Ž 3.6.
where I Ž r, . and g Ž r, . are defined by Ž1.10. and Ž1.14., respectively. Thus, Ž3.4. and Ž3.6. imply that Ž1.13. holds. The proof is complete.
4. SOME APPLICATIONS 4.1. We take K Ž x, y . as K Ž x, y . s Ž x q y .
yt
,
t ) 1 y 2 rr ) 0;
Ž 4.1.
GENERALIZATIONS OF HILBERT’S INEQUALITY
259
then I Ž r , . s
Ž r , m, t , . s
⬁
H0
ž
yt Ž 1 q t . uy2 r r du s B Ž 1 y 2 rr , t q 2 rr y 1 . , 1y 2 rr
mq
/
½ž
mq
t
mq2
/
1 1y2 rr q
)
ž
1y 2 rr
mq
/
1 2
1 1 y 2 rr
t
y
y
1 2
1
ž
2
1q
1 3r
t Ž m q .
t
24 Ž m q 2 .
ž
1q
1 3r
/
tq1
5
/
s h Ž r , m, t , . . Thus, by Theorem 1.1, we get the following theorem: THEOREM 4.1. Let a n , bn ) 0, p ) 1, 1r2 F - Ž1r2.min p, q4 , then ⬁
⬁
q 1q s 1, t ) 1 y 2 rr ) 0. If
a m bn
Ý Ý ms0 ns0
-
1 p
Ž m q n q 2 . ⬁
½Ý
t
B Ž 1 y 2 rq, t q 2 rq y 1 . y h Ž q, m, t , .
ms0 1rp
= Ž m q . =
⬁
½Ý
1yt
a mp
5
B Ž 1 y 2 rp, t q 2 rp y 1 .
ms0 1rq
yh Ž p, m, t , . Ž m q .
1yt
bmq
5
Ž 4.2.
unless the sequence a n4 or bn4 is null, where h Ž r , m, t , . s
ž
mq
1y 2 rr
/
1 2
t
1 1 y 2 rr
y
1 2
ž
1q
1 3r
/
) 0.
Ž 4.3.
260
JICHANG AND DEBNATH
In particular, when s 1r2, t s 1, we obtain ⬁
⬁
a m bn
Ý Ý ms0 ns0 Ž m q n q 1 . -
⬁
½Ý ½Ý
sin rp
ms0
⬁
=
y
ms0
sin rp
1
y
ž
1rp
2 Ž 2 m q 1.
pq
1 2 Ž 2 m q 1.
1rq
ž
1 3p
qq
4
y 1 3q
3 y
1rp
/ 5 / 5 a mp
4
1rq
bmq
3
.
Ž 4.4. On the other hand, when p s q s 2, Ž4.4. reduces to the form ⬁
⬁
a m bn
Ý Ý ms0 ns0 Ž m q n q 1 . -
⬁
y
žÝž žÝž ms0
⬁
=
1r2
5 12 Ž 2 m q 1 .
y
ms0
1r2
5 12 Ž 2 m q 1 .
/ / / / a2m
1r2
1r2
bm2
.
Ž 4.5.
This result is an impro¨ ement of Gau and Yang’s results w11, 12x. If 0 - - 1r2, then, by Theorem 1.2, we have Let a n , bn G 0, p ) 1,
THEOREM 4.2. ) 0; then N
N
1 p
q 1q s 1, ) 0, t ) 1 y 2 rr
a m bn
Ý Ý ms0 ns0
Ž m q n q 2 .
t
N
-
B Ž 1 y 2 rq, t q 2 rq y 1 .
Ý ms0
y
⬁
HŽ Nq2 .rŽ mq. Ž 1 q u .
=
½
y
yt
uy2 r q du Ž m q .
1yt
a mp 4
1rp
N
Ý
B Ž 1 y 2 rp, t q 2 rp y 1 .
ms0 ⬁
HŽ Nq2 .rŽ mq. Ž 1 q u .
1rq yt
y2 r p
u
du Ž m q .
1yt
bmq
5
, Ž 4.6.
GENERALIZATIONS OF HILBERT’S INEQUALITY ⬁
⬁
261
a m bn
Ý Ý
Ž m q n q 2 .
ms0 ns0
t
- B Ž 1 y 2 rq, t q 2 rq y 1 .
1rp
= B Ž 1 y 2 rp, t q 2 rp y 1 . =
⬁
½Ý
1rp
Ž m q .
1y t
a mp
ms0
⬁
5 ½Ý
1rq 1rq
Ž m q .
1yt
bmq
ms0
5
. Ž 4.7.
In particular, when s 1r2, inequality Ž4.7. reduces to the form ⬁
⬁
a m bn
Ý Ý ms0 ns0
Ž m q n q 1.
t
- B Ž 1rp, t y 1rp . 4 =
⬁
½Ý
1rp
B Ž 1rq, t y 1rq . 4 1rp
Ž m q .
1y t
a mp
ms0
⬁
5 ½Ý
1rq 1rq
Ž m q .
1yt
bmq
ms0
5
. Ž 4.8.
When t s 1, inequality Ž4.8. reduces to Ž1.2.. When s 1r2, t s 1, Ž4.6. reduces to the form N
N
a m bn
Ý Ý Ž m q n q 1.
- 2 tany1
ms0 ns0
ž
2N q 2 2m q 1 1r2
N
=
1r2
/ 1r2
N
žÝ / žÝ / a2m
ms0
bm2
.
Ž 4.9.
ms0
Since 2 tany1 ŽŽ2 N q 2.rŽ2 m q 1..1r2 - , Ž4.9. is an impro¨ ement of Cassel’s result w16x. 4.2. We next assume K Ž x, y . in the form K Ž x, y . s 1r Ž x t q y t . . Similar to the proof of Theorems 4.1 and 4.2, we have the following. THEOREM 4.3. Under the same conditions as those of Theorem 4.1, we ha¨ e ⬁
⬁
a m bn
Ý Ý ms0 ns0
-
t t Ž m q . q Ž n q .
⬁
½Ý
ms0
t sin Ž 1 y 2 rq . rt
1rp
y H Ž q, m, t , . Ž m q .
1y t
a mp
5
262
JICHANG AND DEBNATH
=
1rq
⬁
½Ý
y H Ž p, m, t , . Ž m q .
t sin Ž 1 y 2 rp . rt
ms0
1y t
bmq
5
Ž 4.10. unless the sequence a n4 or bn4 is null, where H Ž r , m, t , . s
1y 2 rr
ž
mq
/
1 2
1
ž
1 y 2 rr
y
1 2
ž
1
1q
//
3r
. Ž 4.11.
THEOREM 4.4. Under the same conditions as those of Theorem 4.2, we ha¨ e N
N
a m bn
Ý Ý
t t Ž m q . q Ž n q .
ms0 ns0
-
½
N
Ý
t sin Ž 1 y 2 rq . rt
ms0
y
HŽ Nq2 .rŽ mq. 1 q u
=
du Ž m q .
t
N
½
1rp
uy2 r q
⬁
Ý ms0
y
t sin Ž 1 y 2 rp . rt
1y t
5
a mp
uy2 r p
⬁
HŽ Nq2 .rŽ mq. 1 q u
t
du
1rq
= Ž m q .
1yt
bmq
5
Ž 4.12.
and ⬁
⬁
a m bn
Ý Ý
t t Ž m q . q Ž n q .
ms0 ns0
-
ž
t sin Ž 1 y 2 rq . rt
=
⬁
½Ý
1rp
/ ž 1rp
Ž m q .
1y t
ms0
a mp
1rq
t sin Ž 1 y 2 rp . rt ⬁
5 ½Ý
1rq
Ž m q .
1yt
bmq
ms0
4.3. We take K Ž x, y . in the form K Ž x, y . s
ln Ž yrx . yyx
;
then I Ž r , . s
⬁
H0
ln u uy1
uy2 r r du s
/
2 sin Ž 2rr .
.
5
. Ž 4.13.
GENERALIZATIONS OF HILBERT’S INEQUALITY
263
We next use the mean value theorem; we have ln ␣
s
␣y1
Ž ln 1 y ln ␣ .
s
1y␣
1
for 0 - ␣ - - 1.
)1
6
It follows that
ž
K 1,
ln
mq
/
s
ž
ž
/
mq ) 1. y1 mq1
/
From this and inequality Ž1.12., we obtain the following result:
Ž r , m, l, . )
ž
1y 2 rr
mq
/
1
1
2 1 y 2 rr
y
1
ž
2
1q
1 3r
/
s H Ž r , m, l, . ) 0. Thus, we get the following. Let a n , bn G 0, p ) 1,
THEOREM 4.5. ⬁
ln Ž n q . r Ž m q . 4
⬁
Ý Ý
nym
ms0 ns0
-
⬁
½Ý ½Ý =
ms0
a m bn
y H Ž q, m, 1, .
a mp
5 1rq
2
⬁
q 1q s 1, ) 0; then
1rp
2
sin 2rq
ms0
1 p
sin 2rp
y H Ž p, m, 1, .
bmq
5
.
Ž 4.14.
THEOREM 4.6. Under the same conditions as those of Theorem 4.5, we ha¨ e N
ln Ž n q . r Ž m q . 4
N
Ý Ý
nym
ms0 ns0
-
½
2
N
Ý
sin 2rq
ms0
=
½
N
Ý ms0
y
2 sin 2rp
a m bn ln u
⬁
1rp y2 r q
HŽ Nq2 .rŽ mq. u y 1 u y
⬁
ln u
du
a mp
5
du
bmq
1rq y2 r p
HŽ Nq2 .rŽ mq. u y 1 u
5
,
Ž 4.15.
264 ⬁
JICHANG AND DEBNATH
ln Ž n q . r Ž m q . 4
⬁
Ý Ý
nym
ms0 ns0
-
a m bn
2
Ž sin 2rq .
1rp
1rp
⬁
Ž sin 2rp .
1rq
1rq
⬁
½Ý 5 ½Ý 5 a mp
bmq
ms0
. Ž 4.16.
ms0
In particular, when s 1r2, inequality Ž4.14. reduces to ⬁
⬁
ln Ž 2 n q 1 . r Ž 2 m q 1 . 4
Ý Ý
nym
ms0 ns0
-
⬁
½Ý ½Ý
2
sin Ž rp .
ms0
=
ms0
y
2
⬁
1
sin Ž rp .
4
a m bn 1rp
G Ž p, m .
5 1rq
1
y
a mp
G Ž q, m .
4
bmq
5
,
Ž 4.17.
where rq GŽ r , m. s
1
y
4
3r 3 , 1rr Ž 2 m q 1.
r s p, q.
Ž 4.18.
Thus, inequality Ž4.17. is an impro¨ ement of w1, Theorem 342x. 4.4. Finally, we take K Ž x, y . as 1
K Ž x, y . s
max x t , y t 4
0 - 1 y 2 rr - t ;
,
then K Ž 1, u . s
1 max 1, u t 4
.
By Ž2.1., we have I Ž r , . s s
1 y2 r r
H0 u
du q
1 1 y Ž 2 rr .
⬁
yty2 r r
H1 u
y
du
1 1 y t y Ž 2 rr .
.
Ž 4.19.
Thus, integral Ž1.10. in Theorem 1.1 is replaced by Ž4.19.; we get the corresponding results. For example, since for t s 1, s 1r2, inequality Ž4.19. implies that I Ž p, 1r2 . s q q p s pq,
Ž 4.20.
GENERALIZATIONS OF HILBERT’S INEQUALITY
265
then, by Ž4.20. and Ž1.11., we obtain ⬁
⬁
a m bn
Ý Ý ms0 ns0
-
max m, n4 ⬁
Ý ms0
pq y G Ž p, m .
1rp a mp
4
⬁
½Ý
ms0
1rq
pq y G Ž q, m .
bmq
5
,
Ž 4.21. when GŽ r, m. is defined by Ž4.18.. Evidently, inequality Ž4.21. is an improvement of w1, Theorem 3.4x.
REFERENCES 1. G. H. Hardy, J. E. Littlewood, and G. Polya, ‘‘Inequalities,’’ Cambridge Univ. Press, Cambridge, UK, 1952. 2. K. Jichang, ‘‘Applied Inequalities,’’ 2nd ed., Hunan Education Press, Changsha, 1993. 3. L. C. Hsu and Y. J. Wang, A refinement of Hilbert’s double series theorem, J. Math. Res. Exposition 11, No. 1 Ž1991., 143᎐144. 4. L. C. Xu and Y. K. Gau, Note on Hardy᎐Riesz’s extension of Hilbert’s inequality, Chinese Quart. J. Math. 6, No. 1 Ž1991., 75᎐77. 5. M. Gao, A note on Hilbert double series theorem, Hunan Ann. Math. 12, No. 1-2 Ž1992., 142᎐147. 6. M. Gao, An improvement of Hardy᎐Riesz’s extension of the Hilbert inequality, J. Math. Res. Exposition 14, No. 2 Ž1994., 255᎐259. 7. M. Gao, A note on the Hardy᎐Hilbert inequality, J. Math. Anal. Appl. 204 Ž1996., 346᎐351. 8. B. C. Yang and M. Z. Gao, On a best value of Hardy᎐Hilbert’s inequality, Ad¨ . Math. 26, No. 2 Ž1997., 159᎐164. 9. M. Z. Gao and B. C. Yang, On the extended Hilbert’s inequality, Proc. Amer. Math. Soc. 126, No. 3 Ž1998., 751᎐759. 10. B. C. Yang and L. Debnath, On new strengthened Hardy᎐Hilbert’s inequality, Internat. J. Math. Math. Sci. 21, No. 1 Ž1998., 403᎐408. 11. M. Z. Gao, On Hilbert’s inequality and its applications, J. Math. Anal. Appl. 212 Ž1997., 316᎐323. 12. B. C. Yang, On a refinement of Hardy᎐Hilbert’s inequality and its applications, Northeast. Math J., to appear. 13. B. Yang and L. Debnath, On a new generalization of Hardy᎐Hilbert’s inequality, J. Math. Anal. Appl. 233 Ž1999., 484᎐497. 14. A. E. Ingham, A note on Hilbert’s inequality, J. London Math. Soc. 11 Ž1936., 237᎐240. 15. L. Xu and X. Wang, ‘‘Methods of Mathematical Analysis and Selected Examples,’’ revised ed., Higher Education Press, Beijing, 1984 Žin Chinese.. 16. J. W. S. Cassels, An elementary proof of some inequalities, J. London Math. Soc. 23 Ž1948., 285᎐290.
On New Generalizations of Hilbert’s Inequality and Their Applications Kuang Jichang Department of Mathematics, Hunan Normal Uni¨ ersity, Changsha, Hunan 410081 People’s Republic of China
and Lokenath Debnath Department of Mathematics, Uni¨ ersity of Central Florida, Orlando, Florida 32816 Communicated by William F. Ames Received January 4, 2000
This paper deals with new generalizations of Hilbert’s inequality and their applications. It is shown that results of this paper are significant extensions and improvements of many known results. 䊚 2000 Academic Press Key Words: Hilbert’s inequality; Euler᎐Maclaurin summation formula; beta function.
1. INTRODUCTION If a n , bn ) 0, p ) 1, 1p q 1q s 1, and 0 - Ý n a np - ⬁, 0 - Ý n bnq - ⬁, then the famous Hilbert’s inequality Žsee Hardy et al. w1x. is given by ⬁
⬁
Ý Ý ms1 ns1
a m bn mqn
-
sin Ž rp . 248
0022-247Xr00 $35.00 Copyright 䊚 2000 by Academic Press All rights of reproduction in any form reserved.
1rp
⬁
1rq
⬁
žÝ / žÝ / a np
ns1
bnq
ns1
.
Ž 1.1.
GENERALIZATIONS OF HILBERT’S INEQUALITY
249
This is a slightly sharper form of the inequality ⬁
⬁
a m bn
Ý Ý
mqnq1
ms0 ns0
-
1rp
⬁
1rq
⬁
žÝ / žÝ / a np
sin Ž rp .
bnq
ns0
,
Ž 1.2.
ns0
where the constant factor rsinŽrp . is the best possible. Inequalities Ž1.1. and Ž1.2. have been proved in many different ways and they have varied applications Žsee Hardy et al. w1x and Kuang w2x and references cited in w1, 2x.. Recently, Hsu w3, 4x first introduced the weight function in the form Ž p s q s 2.
Ž r , n. s
sin p
y Ž r , n. ,
Ž r , n. ) 0
Ž 1.3.
and replaced the right hand side of Ž1.1. by
½Ý n
1rp
y Ž q, n .
sin rp
a np
5 ½Ý
1rq
sin rp
n
y Ž p, n .
bnq
5
. Ž 1.4.
Gao w5, 7x gave some improvements of Ž1.3.. Later Gao and Yang w8, 9x found the best possible value of Ž r, n.,
Ž r , n. s
1yc n1y 1r r
,
where c is the Euler constant. Yang and Debnath w10x prove that
Ž p, n . s
1 1r q
2n
q ny1r p
.
The similar estimate of Ž r, n. in Ž1.3. corresponding to Ž1.2. can be found in Gao w11x and Yang w12x. Yang and Debnath w13x proved that ⬁
⬁
Ý Ý ms0 ns0
a m bn
Ž m q n q 1.
py2q qy2q -B , p q
ž
=
⬁
1
nq
2
/½ Ý ž ns0
nq
1 2
1rp
1y
/
a np
5
1rq
1y
½Ýž / 5 ns0
⬁
bnq
,
Ž 1.5.
250
JICHANG AND DEBNATH
where B Ž u, ¨ . is the beta function, a n , bn G 0, p ) 1, 1p q 1q s 1, 2 y min p, q4 - F 2, and 0 - Ý⬁ns 0 Ž n q 1r2.1y a np - ⬁, 0 - Ý⬁ns0 Ž n q 1r2.1y bnq - ⬁. Ingham w14x has proved that if a n G 0, 0 - Ý⬁ns0 a2n - ⬁, ) 0, then ⬁
⬁
Ý Ý ms0 ns0
am an mqnq
⬁
F M Ž .
Ý a2n ,
Ž 1.6.
ns0
where M Ž . s
½
rsin , M Ž 1r2 . s ,
0 - F 1r2 . ) 1r2
5
Ž 1.7.
In this work, it is interesting to consider the double series with two parameters, ⬁
⬁
Ý Ý ms0 ns0
a m bn
Ž m q n q 2 .
t
,
Ž 1.8.
and find the corresponding beta function B Ž u, ¨ ., which depends on t, , p, q. The major objective of this paper is to formulate a new inequality related to the general series ⬁
⬁
Ý Ý K Ž m q , n q . am bn ,
Ž 1.9.
ms0 ns0
with Ž1.8. as a special case. Our main results can be stated as follows: THEOREM 1.1. Let a n , bn G 0, p ) 1, 1p q 1q s 1, 0 - Ý⬁ns0 Ž n q 1y t p . a n - ⬁, 0 - Ý⬁ns0 Ž n q .1y t bnq - ⬁ 1r2 F F 1r2 min p, q4 , and let Ž K x, y . be a nonnegati¨ e and homogeneous function of degree yt Ž t ) 0.. If K Ž1, y . has its first four deri¨ ati¨ es continuous on Ž0, ⬁., and Žy1. n K Ž n. Ž1, y . G 0 for n s 0, 1, 2, 3, 4, K Ž m. Ž1, y . yy2 r r ª 0, y ª ⬁ for m s 0, 1, I Ž r , . s
⬁
y2 r r
H0 K Ž 1, u . u
du - ⬁,
r s p, q,
Ž 1.10.
GENERALIZATIONS OF HILBERT’S INEQUALITY
251
then ⬁
⬁
Ý Ý K Ž m q , n q . am bn ms0 ns0
-
1rp
⬁
½Ý ½Ý
I Ž q, . y Ž q, m, t , . Ž m q .
1y t
a mp
ms0
1rq
⬁
=
5
I Ž p, . y Ž p, m, t , . Ž m q .
1y t
5
bmq
ms0
, Ž 1.11.
where
Ž r , m, t , . s
ž
1y 2 rr
mq
½ž
/
K 1, y
mq
/
1 1 y 2 rr
1 24 Ž m q .
K ⬘ 1,
ž
1
y
2
mq
1q
ž /5
1 3r
/
) 0, Ž 1.12.
and r s p, q. Remark 1. When 0 - - 1r2, Theorem 1.1 fails. But we have the following result: THEOREM 1.2. Suppose that p, q, a n , bn and K Ž x, y . are defined as in Theorem 1.1. If ) 0, then N
N
Ý Ý K Ž m q , n q . am bn ms0 ns0 1rp
N
-
½Ý ½Ý
I Ž q, . y g Ž q, . Ž m q .
1y t
a mp
ms0
1rq
N
=
5
I Ž p, . y g Ž p, . Ž m q .
ms0
1y t
bmq
5
, Ž 1.13.
where g Ž r , . s
⬁
y2 r r
HŽ Nq2 .rŽ mq.K Ž 1, u . u
du,
r s p, q.
Ž 1.14.
252
JICHANG AND DEBNATH
Remark 2. When N ª ⬁, Ž1.13. reduces to the inequality ⬁
⬁
Ý Ý K Ž m q , n q . am bn ms0 ns0
- I Ž q, .
1rp
I Ž p, .
1rq
1rp
⬁
½Ý 5
Ž m q .
1yt
a mp
ms0
=
⬁
½Ý
5
1rq
Ž m q .
1y t
ms0
bmq
.
Ž 1.15.
Remark 3. If we replace the exponent Ž2 rpq . in Ž3.4. below by Ž2 y t .rpq, where 2 y min p, q4 - t F 2, we can similarly prove that N
N
Ý Ý K Ž m q , n q . am bn ms0 ns0 1rp
N
-
½Ý ½Ý
I1 Ž q, t . y g 1 Ž q, t . Ž m q .
1y t
a mp
ms0
1rq
N
=
5
I1 Ž p, t . y g 1 Ž p, t . Ž m q .
1y t
bmq
1yt
bmq
ms0
5
Ž 1.16.
and ⬁
⬁
Ý Ý K Ž m q , n q . am bn ms0 ns0
- I1 Ž q, t . =
⬁
½Ý
1rp
I1 Ž p, t .
1rq 1rp
Ž m q .
1y t
ms0
a mp
1rq
⬁
5 ½Ý
Ž m q .
ms0
5
. Ž 1.17.
where ⬁
K Ž 1, u .
I1 Ž r , t . s
H0
g 1Ž r , t . s
HŽ Nq2 .rŽ mq. u
⬁
uŽ2yt .r r
du - ⬁, K Ž 1, u . Ž2yt .r r
Ž 1.18. du.
Ž 1.19.
GENERALIZATIONS OF HILBERT’S INEQUALITY
253
In particular, when K Ž x, y . s Ž x q y .yt , s 1, 2, inequality Ž1.17. leads to Ž3.1. in Yang and Debnath w13x. In Section 4 we discuss some applications.
2. SOME LEMMAS For the proof of the above theorems, we need two lemmas: LEMMA 2.1. Let f ha¨ e its first four deri¨ ati¨ es continuous on Ž0, ⬁. and Žy1. n f Ž n. Ž x . G 0 Ž n s 0, 1, 2, 3, 4., and f Ž x ., f ⬘Ž x . ª 0 Ž x ª ⬁.; then ⬁
⬁
Ý f Ž k . - H f Ž x . dx q 0
ks0
1 2
f Ž 0. y
1 12
f ⬘ Ž 0. .
Ž 2.1.
Proof. By the Euler᎐Maclaurin summation formula Žsee Xu and Wang w15, p. 88x., we have N
N
Ý f Ž k . s H f Ž x . dx q
ks0
0
q
1
N Ž4.
H 24 0
f
1 2
f Ž N . q f Ž 0. q
Ž B4 y B4 Ž x y w x x .
1 12
dx . ,
f ⬘ Ž N . y f ⬘ Ž 0.
Ž 2.2.
where Bk , Bk Ž x . are the Bernoulli numbers and the Bernoulli polynomials, respectively. Since B4 s y1r30 and the sign of B4 y B4 Ž x y w x x. is the same as B4 , hence we obtain N
N
Ý f Ž k . - H f Ž x . dx q
ks0
0
1 2
f Ž N . q f Ž 0. q
1 12
f ⬘ Ž N . y f ⬘ Ž 0. .
Ž 2.3. Let N ª ⬁; Ž2.3. implies Ž2.1.. The lemma is proved. Remark 4. Lemma 2.1 shows that Ž2.1. holds under weaker conditions on f than those in w13x. LEMMA 2.2. Suppose that K Ž x, y . and are as in Theorem 1.1, r ) 1. Define a weight function by
Ž r , m, t , . s Ž m q .
2 rr
⬁
Ý K Ž m q , n q . Ž n q . y2 rr ; Ž 2.4. ns0
254
JICHANG AND DEBNATH
then
Ž r , m, t , . - Ž m q .
1y t
I Ž r , . y Ž r , m, t , . ,
Ž 2.5.
where I Ž r, . and Ž r, m, t, . are gi¨ en by Ž1.10. and Ž1.12., respecti¨ ely. Proof. Let f Ž x . s K Ž m q , x q .Ž x q .y2 r r ; then, by Lemma 2.1, we have ⬁
2 rr
Ž r , m, t , . s Ž m q .
Ý f Ž n. ns0
1y t
- Ž m q .
½
⬁
y2 r r
H0 K Ž 1, u . u
du y Ž r , m, t , . ,
5
Ž 2.6. where
H0r mq K Ž 1, u . u Ž
Ž r , m, t , . s
y q
ž
.
y2 r r
1y 2 rr
mq 1
12
2
ž
/
ž
1 2
du
q
1
/ ž
K 1,
6 r
2y 2 rr
mq
K ⬘ 1,
/
ž
mq
Integrating by parts twice gives
H0r mq K Ž 1, u . u Ž
.
s
y2 r r
1 1 y 2 rr
ž
du
K 1,
mq
/ž
mq
H0r mq K ⬘ Ž 1, u . u
y
Ž
.
1y 2 rr
/
1y 2 r r
du
mq
/
.
/ Ž 2.7.
GENERALIZATIONS OF HILBERT’S INEQUALITY
s
1
1y 2 rr
ž
1 y 2 rr m q y
/
ž
K 1,
1
2y 2 rr
K ⬘ 1,
/
Ž 1 y 2 rr . Ž 2 y 2 rr . m q
ž
rmq
y 1
1y 2 rr
ž
1 y 2 rr m q y
/
ž
K 1,
1
Ž 1 y 2 rr . Ž 2 y 2 rr
. ž
mq
mq
/
2y 2 rr
K ⬘ 1,
/
mq
/
K ⬙ Ž 1, u . u 2y 2 r r du
H0
)
/
mq
ž
255
ž
/
.
mq
/
mq
From this and equality Ž2.7., we obtain
Ž r , m, t , . )
1 1 y 2 rr
1
y
2
ž
1q
1 3r
/ž
1
y
Ž 1 y 2 rr . Ž 2 y 2 rr . =
ž
mq
ž
/
ž
K 1,
1 12 2
K ⬘ 1,
/
mq
y
2y 2 rr
1y 2 rr
/
mq
.
Ž 2.8.
Since 1r2 F - Ž1r2.min p, q4 , that is, 1 F 2 - r, r s p, q, we have
G
1 2
)
1 2
y
2r q 1 2 Ž 3r q 3r q 1 . 2
s
r Ž 3r q 1 . 6r2 q 6r q 2
that is, 1
-
2 r
q
6r 3r q 1
.
This implies that 1
y
2 r
-
ž
1 2
q
1 6r
y1
/
,
,
256
JICHANG AND DEBNATH
so that
ž
1
y
2
1
/ž
r
2
1
q
6r
s 1y
/ ž
2
1
/ ž 2
r
1q
1 3r
- 1.
/
This implies that 1 1 y 2 rr
1
y
2
ž
1
1q
3r
) 0.
/
Ž 2.9.
Next note that 0 - 1rr - 1 and 2rr - 1r F 2; these imply that 0-
ž
1
2
y
r
/ž
1
y
1
/
r
- 4,
that is, 1
Ž 1 y 2 rr . Ž 2 y 2 rr .
1
)
8 2
.
Thus, 1
Ž 1 y 2 rr . Ž 2 y 2 rr . By hypotheses, K ⬘Ž1, that
mq
y
1 12
2
)
1 8
2
y
1 12
2
s
1 242
. Ž 2.10.
. F 0. This and inequalities Ž2.8. ᎐ Ž2.10. imply
Ž r , m, t , . )
ž
mq
1y 2 rr
/
½ž
K 1,
mq
/ y
s Ž r , m, t , . ) 0.
1 1 y 2 rr
y
1 24 Ž m q .
1 2
ž
1q
K ⬘ 1,
ž
1 3r
/
mq
/5
Ž 2.11.
From Ž2.6., Ž2.7., and Ž2.11., we obtain inequality Ž2.5.. The lemma is proved.
GENERALIZATIONS OF HILBERT’S INEQUALITY
257
3. PROOFS Proof of Theorem 1.1. By Holder’s inequality, we obtain ¨ ⬁
⬁
Ý Ý K Ž m q , n q . am bn ms0 ns0
s
⬁
⬁
Ý Ý
1rp
am K Ž m q , n q .
ms0 ns0
= bn K Ž m q , n q . F
⬁
½Ý
1rq
Ž q, m, t , .
ms0
nq
nq
ž
1rp
a mp
ž
mq
mq ⬁
5 ½Ý
2 rp q
/
2 rp q
/ 1rq
Ž p, m, t , .
ms0
bmq
5
, Ž 3.1.
where
Ž r , m, t , . s Ž m q .
2 rr
⬁
Ý K Ž m q , n q . Ž n q . y2 rr , ns0
r s p, q.
Ž 3.2.
Thus, by Ž2.5., we have Ž1.11.. Theorem 1.1 is proved. Proof of Theorem 1.2. Define f by f Ž x . s K Ž m q , x q . Ž x q .
y2 rr
.
Note that f ⬘ Ž x . s K ⬘ Ž m q , x q . Ž x q . y Ž 2 rr . Ž x q .
y2 rry1
y2 rr
K Ž m q , x q . - 0
and f ⬙ Ž x . s K ⬙ Ž m q , x q . Ž x q . y Ž 4rr . Ž x q .
y2 rry1
y2 rr
K ⬘ Ž m q , x q .
q Ž 2 rr . Ž 2 rr q 1 . Ž x q .
y2 rry2
K Ž m q , x q . ) 0.
258
JICHANG AND DEBNATH
Hence, f Ž x . is strictly convex. Therefore, N
Ý f Ž m. - H
Nq
y
ms0
f Ž x . dx
s Ž m q .
H0 Nq2 r mq K Ž 1, u . u
1y ty2 rr
Ž
. Ž
.
y2 r r
du. Ž 3.3.
In view of Holder’s inequality, we get ¨ N
N
Ý Ý K Ž m q , n q . am bn ms0 ns0 N
s
N
Ý Ý am K Ž m q , n q . 1rp ms0 ns0
= bn K Ž m q , n q .
1rp
N
F
½Ý
1rq
Ž q, m, t , .
a mp
ms0
ž
ž
mq nq
nq mq
2 rp q
/
2 rp q
/ 1rq
N
5 ½Ý
Ž p, m, t , .
bmq
ms0
5
, Ž 3.4.
where
Ž r , m, t , . s Ž m q .
2 rr
N
Ý K Ž m q , n q . Ž n q . y2 rr . Ž 3.5. ns0
From this and inequality Ž3.3., we have
Ž r , m, t , . - Ž m q .
1y t
I Ž r , . y g Ž r , . 4 ,
Ž 3.6.
where I Ž r, . and g Ž r, . are defined by Ž1.10. and Ž1.14., respectively. Thus, Ž3.4. and Ž3.6. imply that Ž1.13. holds. The proof is complete.
4. SOME APPLICATIONS 4.1. We take K Ž x, y . as K Ž x, y . s Ž x q y .
yt
,
t ) 1 y 2 rr ) 0;
Ž 4.1.
GENERALIZATIONS OF HILBERT’S INEQUALITY
259
then I Ž r , . s
Ž r , m, t , . s
⬁
H0
ž
yt Ž 1 q t . uy2 r r du s B Ž 1 y 2 rr , t q 2 rr y 1 . , 1y 2 rr
mq
/
½ž
mq
t
mq2
/
1 1y2 rr q
)
ž
1y 2 rr
mq
/
1 2
1 1 y 2 rr
t
y
y
1 2
1
ž
2
1q
1 3r
t Ž m q .
t
24 Ž m q 2 .
ž
1q
1 3r
/
tq1
5
/
s h Ž r , m, t , . . Thus, by Theorem 1.1, we get the following theorem: THEOREM 4.1. Let a n , bn ) 0, p ) 1, 1r2 F - Ž1r2.min p, q4 , then ⬁
⬁
q 1q s 1, t ) 1 y 2 rr ) 0. If
a m bn
Ý Ý ms0 ns0
-
1 p
Ž m q n q 2 . ⬁
½Ý
t
B Ž 1 y 2 rq, t q 2 rq y 1 . y h Ž q, m, t , .
ms0 1rp
= Ž m q . =
⬁
½Ý
1yt
a mp
5
B Ž 1 y 2 rp, t q 2 rp y 1 .
ms0 1rq
yh Ž p, m, t , . Ž m q .
1yt
bmq
5
Ž 4.2.
unless the sequence a n4 or bn4 is null, where h Ž r , m, t , . s
ž
mq
1y 2 rr
/
1 2
t
1 1 y 2 rr
y
1 2
ž
1q
1 3r
/
) 0.
Ž 4.3.
260
JICHANG AND DEBNATH
In particular, when s 1r2, t s 1, we obtain ⬁
⬁
a m bn
Ý Ý ms0 ns0 Ž m q n q 1 . -
⬁
½Ý ½Ý
sin rp
ms0
⬁
=
y
ms0
sin rp
1
y
ž
1rp
2 Ž 2 m q 1.
pq
1 2 Ž 2 m q 1.
1rq
ž
1 3p
4
y 1 3q
3 y
1rp
/ 5 / 5 a mp
4
1rq
bmq
3
.
Ž 4.4. On the other hand, when p s q s 2, Ž4.4. reduces to the form ⬁
⬁
a m bn
Ý Ý ms0 ns0 Ž m q n q 1 . -
⬁
y
žÝž žÝž ms0
⬁
=
1r2
5 12 Ž 2 m q 1 .
y
ms0
1r2
5 12 Ž 2 m q 1 .
/ / / / a2m
1r2
1r2
bm2
.
Ž 4.5.
This result is an impro¨ ement of Gau and Yang’s results w11, 12x. If 0 - - 1r2, then, by Theorem 1.2, we have Let a n , bn G 0, p ) 1,
THEOREM 4.2. ) 0; then N
N
1 p
q 1q s 1, ) 0, t ) 1 y 2 rr
a m bn
Ý Ý ms0 ns0
Ž m q n q 2 .
t
N
-
B Ž 1 y 2 rq, t q 2 rq y 1 .
Ý ms0
y
⬁
HŽ Nq2 .rŽ mq. Ž 1 q u .
=
½
y
yt
uy2 r q du Ž m q .
1yt
a mp 4
1rp
N
Ý
B Ž 1 y 2 rp, t q 2 rp y 1 .
ms0 ⬁
HŽ Nq2 .rŽ mq. Ž 1 q u .
1rq yt
y2 r p
u
du Ž m q .
1yt
bmq
5
, Ž 4.6.
GENERALIZATIONS OF HILBERT’S INEQUALITY ⬁
⬁
261
a m bn
Ý Ý
Ž m q n q 2 .
ms0 ns0
t
- B Ž 1 y 2 rq, t q 2 rq y 1 .
1rp
= B Ž 1 y 2 rp, t q 2 rp y 1 . =
⬁
½Ý
1rp
Ž m q .
1y t
a mp
ms0
⬁
5 ½Ý
1rq 1rq
Ž m q .
1yt
bmq
ms0
5
. Ž 4.7.
In particular, when s 1r2, inequality Ž4.7. reduces to the form ⬁
⬁
a m bn
Ý Ý ms0 ns0
Ž m q n q 1.
t
- B Ž 1rp, t y 1rp . 4 =
⬁
½Ý
1rp
B Ž 1rq, t y 1rq . 4 1rp
Ž m q .
1y t
a mp
ms0
⬁
5 ½Ý
1rq 1rq
Ž m q .
1yt
bmq
ms0
5
. Ž 4.8.
When t s 1, inequality Ž4.8. reduces to Ž1.2.. When s 1r2, t s 1, Ž4.6. reduces to the form N
N
a m bn
Ý Ý Ž m q n q 1.
- 2 tany1
ms0 ns0
ž
2N q 2 2m q 1 1r2
N
=
1r2
/ 1r2
N
žÝ / žÝ / a2m
ms0
bm2
.
Ž 4.9.
ms0
Since 2 tany1 ŽŽ2 N q 2.rŽ2 m q 1..1r2 - , Ž4.9. is an impro¨ ement of Cassel’s result w16x. 4.2. We next assume K Ž x, y . in the form K Ž x, y . s 1r Ž x t q y t . . Similar to the proof of Theorems 4.1 and 4.2, we have the following. THEOREM 4.3. Under the same conditions as those of Theorem 4.1, we ha¨ e ⬁
⬁
a m bn
Ý Ý ms0 ns0
-
t t Ž m q . q Ž n q .
⬁
½Ý
ms0
t sin Ž 1 y 2 rq . rt
1rp
y H Ž q, m, t , . Ž m q .
1y t
a mp
5
262
JICHANG AND DEBNATH
=
1rq
⬁
½Ý
y H Ž p, m, t , . Ž m q .
t sin Ž 1 y 2 rp . rt
ms0
1y t
bmq
5
Ž 4.10. unless the sequence a n4 or bn4 is null, where H Ž r , m, t , . s
1y 2 rr
ž
mq
/
1 2
1
ž
1 y 2 rr
y
1 2
ž
1
1q
//
3r
. Ž 4.11.
THEOREM 4.4. Under the same conditions as those of Theorem 4.2, we ha¨ e N
N
a m bn
Ý Ý
t t Ž m q . q Ž n q .
ms0 ns0
-
½
N
Ý
t sin Ž 1 y 2 rq . rt
ms0
y
HŽ Nq2 .rŽ mq. 1 q u
=
du Ž m q .
t
N
½
1rp
uy2 r q
⬁
Ý ms0
y
t sin Ž 1 y 2 rp . rt
1y t
5
a mp
uy2 r p
⬁
HŽ Nq2 .rŽ mq. 1 q u
t
du
1rq
= Ž m q .
1yt
bmq
5
Ž 4.12.
and ⬁
⬁
a m bn
Ý Ý
t t Ž m q . q Ž n q .
ms0 ns0
-
ž
t sin Ž 1 y 2 rq . rt
=
⬁
½Ý
1rp
/ ž 1rp
Ž m q .
1y t
ms0
a mp
1rq
t sin Ž 1 y 2 rp . rt ⬁
5 ½Ý
1rq
Ž m q .
1yt
bmq
ms0
4.3. We take K Ž x, y . in the form K Ž x, y . s
ln Ž yrx . yyx
;
then I Ž r , . s
⬁
H0
ln u uy1
uy2 r r du s
/
2 sin Ž 2rr .
.
5
. Ž 4.13.
GENERALIZATIONS OF HILBERT’S INEQUALITY
263
We next use the mean value theorem; we have ln ␣
s
␣y1
Ž ln 1 y ln ␣ .
s
1y␣
1
for 0 - ␣ - - 1.
)1
6
It follows that
ž
K 1,
ln
mq
/
s
ž
ž
/
mq ) 1. y1 mq1
/
From this and inequality Ž1.12., we obtain the following result:
Ž r , m, l, . )
ž
1y 2 rr
mq
/
1
1
2 1 y 2 rr
y
1
ž
2
1q
1 3r
/
s H Ž r , m, l, . ) 0. Thus, we get the following. Let a n , bn G 0, p ) 1,
THEOREM 4.5. ⬁
ln Ž n q . r Ž m q . 4
⬁
Ý Ý
nym
ms0 ns0
-
⬁
½Ý ½Ý =
ms0
a m bn
y H Ž q, m, 1, .
a mp
5 1rq
2
⬁
q 1q s 1, ) 0; then
1rp
2
sin 2rq
ms0
1 p
sin 2rp
y H Ž p, m, 1, .
bmq
5
.
Ž 4.14.
THEOREM 4.6. Under the same conditions as those of Theorem 4.5, we ha¨ e N
ln Ž n q . r Ž m q . 4
N
Ý Ý
nym
ms0 ns0
-
½
2
N
Ý
sin 2rq
ms0
=
½
N
Ý ms0
y
2 sin 2rp
a m bn ln u
⬁
1rp y2 r q
HŽ Nq2 .rŽ mq. u y 1 u y
⬁
ln u
du
a mp
5
du
bmq
1rq y2 r p
HŽ Nq2 .rŽ mq. u y 1 u
5
,
Ž 4.15.
264 ⬁
JICHANG AND DEBNATH
ln Ž n q . r Ž m q . 4
⬁
Ý Ý
nym
ms0 ns0
-
a m bn
2
Ž sin 2rq .
1rp
1rp
⬁
Ž sin 2rp .
1rq
1rq
⬁
½Ý 5 ½Ý 5 a mp
bmq
ms0
. Ž 4.16.
ms0
In particular, when s 1r2, inequality Ž4.14. reduces to ⬁
⬁
ln Ž 2 n q 1 . r Ž 2 m q 1 . 4
Ý Ý
nym
ms0 ns0
-
⬁
½Ý ½Ý
2
sin Ž rp .
ms0
=
ms0
y
2
⬁
1
sin Ž rp .
4
a m bn 1rp
G Ž p, m .
5 1rq
1
y
a mp
G Ž q, m .
4
bmq
5
,
Ž 4.17.
where rq GŽ r , m. s
1
y
4
3r 3 , 1rr Ž 2 m q 1.
r s p, q.
Ž 4.18.
Thus, inequality Ž4.17. is an impro¨ ement of w1, Theorem 342x. 4.4. Finally, we take K Ž x, y . as 1
K Ž x, y . s
max x t , y t 4
0 - 1 y 2 rr - t ;
,
then K Ž 1, u . s
1 max 1, u t 4
.
By Ž2.1., we have I Ž r , . s s
1 y2 r r
H0 u
du q
1 1 y Ž 2 rr .
⬁
yty2 r r
H1 u
y
du
1 1 y t y Ž 2 rr .
.
Ž 4.19.
Thus, integral Ž1.10. in Theorem 1.1 is replaced by Ž4.19.; we get the corresponding results. For example, since for t s 1, s 1r2, inequality Ž4.19. implies that I Ž p, 1r2 . s q q p s pq,
Ž 4.20.
GENERALIZATIONS OF HILBERT’S INEQUALITY
265
then, by Ž4.20. and Ž1.11., we obtain ⬁
⬁
a m bn
Ý Ý ms0 ns0
-
max m, n4 ⬁
Ý ms0
pq y G Ž p, m .
1rp a mp
4
⬁
½Ý
ms0
1rq
pq y G Ž q, m .
bmq
5
,
Ž 4.21. when GŽ r, m. is defined by Ž4.18.. Evidently, inequality Ž4.21. is an improvement of w1, Theorem 3.4x.
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