ON A DECOMPOSITION OF HILBERT’S
INEQUALITY
Hilbert’s Inequality
BICHENG YANG
Department of Mathematics
Guangdong Education Institute
Guangzhou, Guangdong 510303
P.R. CHINA
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
Title Page
EMail: bcyang@pub.guangzhou.gd.cn
Contents
Received:
07 October, 2008
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Accepted:
20 March, 2009
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Communicated by:
B. Opic
Page 1 of 18
2000 AMS Sub. Class.:
26D15.
Key words:
Hilbert’s inequality; Weight coefficient; Equivalent form; Hilbert-type inequality.
Abstract:
By using the Euler-Maclaurin’s summation formula and the weight coefficient, a
pair of new inequalities is given, which is a decomposition of Hilbert’s inequality. The equivalent forms and the extended inequalities with a pair of conjugate
exponents are built.
Acknowledgement:
The author expresses his sincerest thanks to the referee for his thoughtful suggestions in improving this work.
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Contents
1
Introduction
3
2
Some Lemmas
6
3
Main Results and their Equivalent Forms
12
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
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1.
Introduction
In 1908, H. Weyl
the following
Hilbert inequality: If {an }, {bn } are real
Ppublished
P∞
∞
2
sequences, 0 < n=1 an < ∞ and 0 < n=1 b2n < ∞, then [1]
(1.1)
∞
∞ X
∞
∞
X
X
X
am bn
2
b2n
<π
an
m+n
n=1 m=1
n=1
n=1
! 12
,
Hilbert’s Inequality
Bicheng Yang
where the constant factor π is the best possible. In 1925, G. H. Hardy gave an
extension of (1.1) by introducing one pair of conjugate exponents (p, q) ( p1 + 1q = 1)
P
P∞ q
p
as [2]: If p > 1, an , bn ≥ 0, 0 < ∞
n=1 an < ∞ and 0 <
n=1 bn < ∞, then
(1.2)
∞ X
∞
X
am b n
π
<
m+n
sin( πp )
n=1 m=1
∞
X
! p1
apn
n=1
∞
X
(1.3)
∞
X
am b n
≤ K(p, q)
apn
λ
(m
+
n)
n=1 m=1
n=1
bqn
Contents
,
n=1
! p1
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! 1q
where the constant factor π/ sin( πp ) is the best possible. We refer to (1.2) as the
Hardy-Hilbert inequality. In 1934, Hardy et al. [3] gave some applications of (1.1)
and (1.2). By introducing a pair of non-conjugate exponents (p, q) in (1.1), Hardy et
al. [3] gave: If p, q > 1, p1 + 1q ≥ 1, 0 < λ = 2 − ( p1 + 1q ) ≤ 1, then
∞ X
∞
X
vol. 10, iss. 1, art. 25, 2009
∞
X
! 1q
bqn
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,
n=1
where the constant factor K(p, q) is the best value only for λ = 1. In 1951, Bonsall
[4] considered (1.3) in the case of a general kernel. In 1991, Mitrinović et al. [5]
summarized the above method and results.
Close
In 1997-1998, by using weight coefficients, Yang and Gao [6], [7] gave a strengthened version of (1.2) as:
(1.4)
∞
∞ X
X
am b n
m+n
n=1 m=1
(∞ "
X
<
n=1
# ) p1 ( ∞ "
# ) 1q
X
π
π
1−γ p
1−γ q
,
− 1/p an
− 1/q bn
sin( πp )
n
sin( πp )
n
n=1
where, 1 − γ = 0.42278433+ (γ is the Euler constant). In 2001, Yang [8] gave an
extension of (1.1) by introducing an independent parameter 0 < λ ≤ 4 as
(1.5)
∞ X
∞
X
am b n
<B
(m + n)λ
n=1 m=1
! 12
X
∞
∞
X
λ λ
,
,
n1−λ a2n
n1−λ b2n
2 2
n=1
n=1
where the constant factor B( λ2 , λ2 ) is the best possible (B(u, v) is the Beta function).
In 2004, Yang [9] published the dual form of (1.2) as follows
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
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Page 4 of 18
(1.6)
∞ X
∞
X
am b n
π
<
m+n
sin( πp )
n=1 m=1
∞
X
n=1
! p1
np−2 apn
∞
X
! 1q
nq−2 bqn
.
n=1
For p = q = 2, both (1.6) and (1.2) reduce to (1.1). It means that there are two different best extensions of (1.1). To generalize (1.2) and (1.6), in 2005, Yang [10] gave an
extension of (1.2) and (1.6) with two pairs of conjugate exponents (p, q), (r, s)(p, r >
P
p(1− αλ
−1 p
r )
1) and parameters α, λ > 0 (αλ ≤ min{r, s}) as: If 0 < ∞
an < ∞
n=1 n
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and 0 <
(1.7)
P∞
n=1
αλ
nq(1− s )−1 bqn < ∞, then
∞ X
∞
X
am b n
(mα + nα )λ
n=1 m=1
< kαλ (r)
(∞
X
n
p(1− αλ
)−1
r
apn
) p1 ( ∞
X
n=1
) 1q
n
q(1− αλ
)−1
s
bqn
,
Hilbert’s Inequality
n=1
Bicheng Yang
where the constant factor kαλ (r) = α1 B( λr , λs ) is the best possible. T. K. Pogány
[11] also considered a best extension of (1.2) with the non-homogeneous kernel as
1
(µ, λm , ρn > 0).
(λm +ρn )µ
We have a non-negative decomposition of kernel in (1.1):
1
max{m, n} min{m, n}
=
+
m+n
(m + n)2
(m + n)2
vol. 10, iss. 1, art. 25, 2009
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Contents
(m, n ∈ N)
(N is the set of positive integer numbers). In this paper, by using the Euler-Maclaurin
summation formula and the weight coefficient as in [8], we give a pair of new
Hilbert-type inequalities as
! 12
∞ X
∞
∞
∞
π
X
X
X
max{m, n}
am b n <
+1
a2n
b2n
(1.8)
;
2
(m
+
n)
2
n=1 m=1
n=1
n=1
! 12
∞
∞
∞ X
∞
π
X
X
X
min{m, n}
(1.9)
am b n <
−1
a2n
b2n
,
2
(m
+
n)
2
n=1
n=1
n=1 m=1
where the sum of two best constant factors is π. The equivalent forms and extended
inequalities with a pair of conjugate exponents are considered.
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2.
Some Lemmas
Lemma 2.1 (Euler-Maclaurin’s summation formula, cf. [8, 12, Lemma 1]). If
f (x) ∈ C 1 [1, ∞), then we have
Z ∞
Z ∞
∞
X
1
(2.1)
f (k) =
f (x)dx + f (1) +
P1 (x)f 0 (x)dx,
2
1
1
k=1
x−[x]− 21
(i)
3
where P1 (x) =
is the Bernoulli function of the first order; if g ∈ C [1, ∞),
i (i)
(−1) g (x) > 0, g (∞) = 0, (i = 0, 1, 2, 3), then
Z n
1
(2.2)
[g(n) − g(1)] <
P1 (x)g(x)dx < 0,
12
1
Z ∞
1
− g(n) <
P1 (x)g(x)dx < 0.
12
n
Lemma 2.2. If
1
2
≤ α < 1, setting the weight coefficient ω(α, m) as
ω(α, m) :=
(2.3)
∞
X
n=1
max{m, n}mα
(m + n)2 nα
(m ∈ N),
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k(α) = Aα (m) + ω(α, m);
where
1
k(α) :=
α
and Aα (m) = O(m
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
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then we have
(2.4)
Hilbert’s Inequality
α−1
Z
∞
0
), (m → ∞).
ω
1
,m
2
1
π
<k
= + 1,
2
2
max u1/α , 1 1 −2
u α du
(u1/α + 1)2
Close
Proof. For fixed 12 ≤ α < 1, m ∈ N, setting f (x) :=
(2.1), it follows that
ω(α, m) = m
(2.5)
α
∞
X
max{m,x}
,
(m+x)2 xα
x ∈ (0, ∞), then by
f (n)
n=1
=m
α
= mα
Z
∞
1
∞
Z
1
f (x)dx + f (1) +
2
Z
∞
0
P1 (x)f (x)dx
Hilbert’s Inequality
1
Bicheng Yang
f (x)dx − mα ρ(α, m), P1 (x)f 0 (x)dx.
vol. 10, iss. 1, art. 25, 2009
0
Z
ρ(α, m) :=
(2.6)
0
We find
1
1
f (x)dx − f (1) −
2
Z
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∞
0
P1 (x)f (x)dx.
Contents
1
1
−m
−1
1
− f (1) =
=
+
,
2
2
2(m + 1)
2(m + 1) 2(m + 1)2
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and
Z 1
m
m
1
f (x)dx =
dx ≥
dx =
;
2
α
2
m+1
0
0 (m + x) x
0 (m + x)
Z 1
Z 1
m
1
f (x)dx ≤
dx =
.
2
α
(1 − α)m
0
0 m x
Z
1
Z
For x ∈ (0, m), f (x) =
1
m
,
(m+x)2 xα
it follows f 0 (x) =
−2m
(m+x)3 xα
−
αm
;
(m+x)2 xα+1
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for
x ∈ (m, ∞), f (x) =
x1−α
,
(m+x)2
we find
−2x1−α
1−α
+
3
(m + x)
(m + x)2 xα
−2(x + m − m)
1−α
=
+
3
α
(m + x) x
(m + x)2 xα
−2
2m
1−α
=
+
+
.
(m + x)2 xα (m + x)3 xα (m + x)2 xα
f 0 (x) =
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
1
1
In the following, it is obvious that g1 (x) = (m+x)
3 xα , g2 (x) = (m+x)2 xα+1 and
1
g3 (x) = (m+x)
2 xα are suited to apply in (2.2). Then by (2.2), we obtain
Z m
−
(2.7)
P1 (x)f 0 (x)dx
1
Z m
Z m
αmP1 (x)dx
2mP1 (x)dx
=
+
3
α
(m + x) x
(m + x)2 xα+1
1
1
2m
1
1
αm
1
1
>
−
+
−
12 8m3+α (m + 1)3
12 4m3+α (m + 1)2
α+1
α
2−α
1
=
−
−
+
;
2+α
2
48m
12(m + 1) 12(m + 1)
6(m + 1)3
Z
m
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∞
(2.8) −
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P1 (x)f 0 (x)dx
Z ∞
Z ∞
Z ∞
2P1 (x)dx
2mP1 (x)dx
P1 (x)dx
=
−
− (1 − α)
2
α
3
α
2 α
m (m + x) x
m (m + x) x
m (m + x) x
Z ∞
−1
>
−
P1 (x)f 0 (x)dx
24m2+α
1
Z m
Z ∞
0
=−
P1 (x)f (x)dx −
P1 (x)f 0 (x)dx
1
m
α
2−α
1
α−1
>
−
−
+
.
2+α
2
48m
12(m + 1) 12(m + 1)
6(m + 1)3
Hence by (2.6), for α = 12 , it follows that
1
−1
1
1
(2.9) ρ
,m >
+
+
2
2
2(m + 1) 2(m + 1)
m+1
1
1
2 − 12
−1
1
2
+ 2 2+1/2 −
−
+
48m
12(m + 1) 12(m + 1)2 6(m + 1)3
9
−1
1
11
=
+
+
+
2
2+1/2
24(m + 1) 24(m + 1)
96m
6(m + 1)3
11
9
−1
1
≥
+
+
+
> 0.
2
2
24(m + 1)
96m
96m
6(m + 1)3
By (2.7) and (2.8), we obtain
Z ∞
Z
0
−
P1 (x)f (x)dx = −
1
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
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m
P1 (x)f 0 (x)dx −
1
1−α
1
+
<
2+α
48m
48m2+α
2−α
=
.
48m2+α
Z
∞
m
P1 (x)f 0 (x)dx
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Then by (2.6), it follows
0 < m1−α [mα ρ(α, m)]
−m
m
1
2−α
<
+
+
+
2(m + 1) 2(m + 1)2 1 − α 48m1+α
1
1
→
−
(m → ∞).
1−α 2
Setting u = (x/m)α , we find
Z ∞
Z ∞
max{m, x}
α
α
(2.11)
m
f (x)dx = m
dx
(m + x)2 xα
0
0
Z
1 ∞ max{u1/α , 1} 1 −2
u α du = k(α),
=
α 0
(u1/α + 1)2
Z ∞
Z 1
1
max{u2 , 1}
du
(2.12)
k
=2
du
=
4
2
2
2
(u2 + 1)2
0
0 (u + 1)
Z π
4
π
=4
cos2 θdθ = + 1.
2
0
Hence by (2.5), (2.9), (2.10) and (2.11), (2.4) is valid and the lemma is proved.
(2.10)
1
2
≤ α < 1, setting the weight coefficient $(α, m) as
$(α, m) :=
(2.13)
∞
X
min{m, n}mα
n=1
(m ∈ N),
then we have
(2.14)
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(m + n)2 nα
e
k(α) = Bα (m) + $(α, m);
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
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Similar to Lemma 2.2, we still have
Lemma 2.3. If
Hilbert’s Inequality
1
,m
2
1
π
e
<k
= − 1,
2
2
where
1
e
k(α) =
α
and Bα (m) = O(m
α−2
Z
∞
0
min{u1/α , 1} 1 −2
u α du
(u1/α + 1)2
), (m → ∞).
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
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3.
Main Results and their Equivalent Forms
P∞
p
−1 p
2
Theorem 3.1. If p > 1, p1 + 1q = 1, an , bn ≥ 0, 0 <
an < ∞ and
n=1 n
P∞
q
−1
q
0 < n=1 n 2 bn < ∞, then we have the following equivalent inequalities
∞ X
∞
X
max{m, n}
I :=
am b n
(m + n)2
n=1 m=1
! p1
∞
X
π
p
<
+1
n 2 −1 apn
2
n=1
(3.1)
Hilbert’s Inequality
∞
X
! 1q
n
q
−1
2
bqn
Bicheng Yang
;
vol. 10, iss. 1, art. 25, 2009
n=1
Title Page
(3.2)
J :=
∞
X
"
p
n 2 −1
n=1
where the constant factors
π
2
∞
X
max{m, n}
am
(m + n)2
m=1
+ 1 and
π
2
#p
<
π
2
∞
p X
q
+1
n 2 −1 apn ,
n=1
p
+ 1 are the best possible.
Proof. By Hölder’s inequality and (2.3) – (2.4), we find
" ∞
#p
X max{m, n}
(3.3)
am
(m + n)2
m=1
( ∞
)p
X max{m, n} m1/(2q) n1/(2p) =
am
(m + n)2
n1/(2p)
m1/(2q)
m=1
" ∞
#" ∞
#p−1
X max{m, n} mp/(2q)
X max{m, n} nq/(2p)
≤
apm
2
1/2
(m
+
n)
n
(m + n)2 m1/2
m=1
m=1
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∞
X
1
max{m, n} mp/(2q) p
1− p2
=ω
,n n
a
2
(m + n)2 n1/2 m
m=1
∞
π
p−1
X
max{m, n} mp/(2q) p
1− p2
≤
+1
n
a ;
2
(m + n)2 n1/2 m
m=1
p−1
J≤
=
=
π
2
π
2
π
2
∞ X
∞
p−1 X
max{m, n} mp/(2q) p
+1
a
(m + n)2 n1/2 m
n=1 m=1
"∞
#
∞
p−1 X
X max{m, n} mp/(2q)
+1
apm
2
1/2
(m + n)
n
m=1 n=1
∞
∞
p−1 X
π
p X
p
p
1
−1
p
+1
ω
, m m 2 am <
+1
m 2 −1 apm .
2
2
m=1
m=1
Therefore (3.2) is valid. By Hölder’s inequality, we find that
"
#
∞
∞
h −1 1 i
X
1
1 X max{m, n}
I=
n2−p
(3.4)
a
n 2 + p bn
m
2
(m
+
n)
n=1
m=1
! 1q
∞
X q
1
≤ Jp
n 2 −1 bqn
.
n=1
Then by (3.2), we have (3.1). On the other hand, suppose that (3.1) is valid. Setting
#p−1
" ∞
X max{m, n}
p
bn := n 2 −1
am
,
n ∈ N,
(m + n)2
m=1
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
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P
q
−1 q
2
then it follows ∞
bn = J. By (3.3), we confirm that J < ∞. If J = 0, then
n=1 n
(3.2) is naturally valid; if 0 < J < ∞, then by (3.1), we find
∞
X
n
q
−1
2
bqn = J = I <
n=1
∞
X
π
2
∞
X
p
+1
n 2 −1 apn
n=1
! p1
n
n=1
q
−1
2
bqn
! p1
1
p
=J <
π
2
∞
X
! 1q
n
q
−1
2
bqn
;
n=1
∞
X
p
+1
n 2 −1 apn
! p1
,
n=1
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
and inequality (3.2) is valid, which is equivalent to (3.1).
−1
−1
ε
− pε
2
ebn2 − q , for n ∈ N, if
eb = {ebn }∞ as e
,
For 0 < ε < 2q , setting e
a = {e
an } ∞
,
a
n
n=1
n=1
there exists a constant 0 < k ≤ π2 + 1, such that (3.1) is still valid when we replace
π
+ 1 by k, then we find
2
∞ X
∞
X
max{m, n}e
amebn
(m + n)2
n=1 m=1
! p1 ∞
! 1q
∞
∞
X
X q
X
p
1
−1 p
−1eq
2
2
<k
n e
an
n bn
;
=k
1+ε
n
n=1
n=1
n=1
Ie :=
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Ie =
∞
X
n=1
∞
X
"
∞
X
#
−1
ε
ε
max{m, n} −1
m 2 −p n 2 −q
2
(m + n)
m=1
1
ε
∞
∞
X
1 X max{m, n}m 2 + q
1
1 ε
=
=
ω
+ ,m .
m1+ε n=1 (m + n)2 n 12 + qε
m1+ε
2 q
m=1
m=1
Close
And then by (2.4) and the above results, it follows that


X
∞
∞
X
1 ε
1 
1
1
A 1 + ε (m)
1− >k
+
(3.5) k
1+ε
1+ε
2
q
1
ε
n
2
q
m
k
+
n=1
m=1
2
q


X
∞
∞
X
1 ε 
1
1
1
A 1 ε (m)
=k
−
+
1+ε 2 + q
1+ε
ε
1
m
2 q
m
k
+
m=1
2
q m=1


P
∞
X
1
∞
1
ε
m=1 m1+ε A 2 + q (m)]
1 
1 ε
;
P
=k
+
1
−
∞
1
1
ε
2 q m=1 m1+ε
k 2+q
m=1 m1+ε

1 ε 
P∞
1 2−q
1
O
m=1 m1+ε
m
1 ε 
.
P
k>k
+
1−
∞
1
2 q
k 1+ε
2
Setting α =
1
2
q
m=1 m1+ε
+ qε , by Fatou’s Lemma, it follows that
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
Title Page
Contents
JJ
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Page 15 of 18
lim k
ε→0+
1 ε
+
2 q
Z
1 ∞ max{u1/α , 1} 1 −2
= lim+
u α du
(u1/α + 1)2
α→ 12 α 0
Z ∞
max{u1/α , 1} 1 −2
π
1
α
u
du
=
k
=
+ 1.
≥2
lim+
(u1/α + 1)2
2
2
α→ 12
0
Then by (3.5), we have k ≥ π2 + 1 (ε → 0+ ). Hence k = π2 + 1 is the best value
of (3.1). We confirm that the constant factor in (3.2) is the best, otherwise we would
obtain a contradiction by (3.4) that the constant factor in (3.1) is not the best possible.
The theorem is proved.
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In the same manner, by Lemma 2.3, we have:
P∞
p
−1 p
2
an < ∞ and
Theorem 3.2. If p > 1, p1 + 1q = 1, an , bn ≥ 0, 0 <
n=1 n
P∞
q
−1
q
0 < n=1 n 2 bn < ∞, then we have the following equivalent inequalities
(3.6)
∞ X
∞
∞
π
X
X
p
min{m, n}
a
b
<
−
1
n 2 −1 apn
m n
2
(m + n)
2
n=1
n=1 m=1
∞
X
n=1
"
n
p
−1
2
∞
X
min{m, n}
am
(m + n)2
m=1
where the constant factors
π
2
#p
<
π
2
! p1
∞
X
! 1q
n
q
−1
2
bqn
n=1
∞
p X
q
−1
n 2 −1 apn ,
;
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
n=1
− 1 and ( π2 − 1)p are the best possible.
Title Page
Contents
Remark 1. For p = q = 2, (3.1) reduces to (1.8) and (3.6) reduces to (1.9).
JJ
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Page 16 of 18
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References
[1] H. WEYL, Singuläre Integralgleichungen mit besonderer Berücksichtigung des
Fourierschen Integraltheorems, Göttingen : Inaugural–Dissertation, 1908.
[2] G.H. HARDY, Note on a theorem of Hilbert concerning series of positive term,
Proceedings of the London Math. Society, 23 (1925),45–46.
[3] G.H. HARDY, J.E. LITTLEWOOD
University Press, Cambridge,1934.
AND
G. PÓLYA, Inequalities, Cambridge
[4] F.F. BONSALL, Inequalities with non-conjugate parameter, J. Math. Oxford
Ser., 2(2) (1951), 135–150.
[5] D. S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Inequalities Involving
Functions and their Integrals and Derivatives, Kluwer Academic Publishers,
Boston, 1991.
[6] BICHENG YANG AND MINGZHE GAO, On a best value of Hardy-Hilbert’s
inequality, Advances in Math. (China), 26(2) (1997), 159–164.
[7] MINGZHE GAO AND BICHENG YANG, On the extended Hilbert’s inequality, Proc. Amer. Math. Soc., 126(3) (1998), 751–759.
[8] BICHENG YANG, On a generalization of Hilbert double series theorem, Journal of Nanjing University Mathematical Biquarterly (China), 18(1) (2001),
145–152.
[9] BICHENG YANG, On new extensions of Hilbert’s inequality, Acta Math. Hungar., 104(4) (2004), 291–299.
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
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[10] BICHENG YANG, On best extensions of Hardy-Hilbert’s inequality with two
parameters, J. Ineq. Pure & Applied Math., 6(3) (2005), Art. 81. [ONLINE
http://jipam.vu.edu.au/article.php?sid=554].
[11] T.K. POGÁNY, Hilbert’s double series theorem extended to the case of nonhomogeneous kernels, J. Math. Anal. Appl., 342 (2008), 1485–1489.
[12] BICHENG YANG, On a strengthened version of the more accurate HardyHilbert’s inequality, Acta Mathematica Sinica (Chin. Ser.), 42(6) (1999), 1103–
1110.
Hilbert’s Inequality
Bicheng Yang
vol. 10, iss. 1, art. 25, 2009
Title Page
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