Journal of Inequalities in Pure and
Applied Mathematics
ON BEST EXTENSIONS OF HARDY-HILBERT’S INEQUALITY
WITH TWO PARAMETERS
volume 6, issue 3, article 81,
2005.
BICHENG YANG
Department of Mathematics
Guangdong Institute of Education
Guangzhou, Guangdong 510303
People’s Republic of China
Received 23 February, 2005;
accepted 17 June, 2005.
Communicated by: W.S. Cheung
EMail: bcyang@pub.guangzhou.gd.cn
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
055-05
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Abstract
This paper deals with some extensions of Hardy-Hilbert’s inequality with the
best constant factors by introducing two parameters λ and α and using the
Beta function. The equivalent form and some reversions are considered.
2000 Mathematics Subject Classification: 26D15.
Key words: Hardy-Hilbert’s inequality; Beta function; Hölder’s inequality.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Some Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4
Some Best Extensions of (1.3) . . . . . . . . . . . . . . . . . . . . . . . . . . 26
References
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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1.
Introduction
If an, bn ≥ 0 satisfy
0<
∞
X
a2n
< ∞ and
0<
n=1
∞
X
b2n < ∞,
n=1
then one has two equivalent inequalities as:
(∞
) 12
∞
∞
∞ X
X
X X
am b n
(1.1)
<π
a2n
b2n
m
+
n
n=1 m=1
n=1
n=1
and
∞
X
(1.2)
n=1
∞
X
am
m+n
m=1
!2
< π2
∞
X
Bicheng Yang
a2n ,
n=1
2
where the constant factors π and π are the best possible. Inequality (1.1) is
well known as Hilbert’s inequality (cf. Hardy et al. [1]). In 1925, Hardy [2]
gave some extensions of (1.1) and (1.2) by introducing the (p, q)−parameter as:
If p > 1, p1 + 1q = 1, an, bn ≥ 0 satisfy
0<
∞
X
n=1
apn
< ∞ and
0<
∞
X
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
bqn < ∞,
n=1
then one has the following two equivalent inequalities:
(∞
) p1 ( ∞ ) 1q
∞ X
∞
X
X
X
π
am b n
<
(1.3)
apn
bqn
π
m
+
n
sin p
n=1
n=1
n=1 m=1
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and
∞
X
(1.4)
n=1
∞
X
am
m+n
m=1
p

!p
<
sin
∞
π  X p
an ,
π
p
n=1
h
ip
π
π
where the constant factors sin(π/p)
and sin(π/p)
are the best possible. Inequality (1.3) is called Hardy-Hilbert’s inequality, and is important in analysis and its
applications (cf. Mitrinović et al. [3]).
In 1997-1998, by estimating the weight coefficient and introducing the Euler
constant γ, Yang and Gao [4, 5] gave a strengthened version of (1.3) as:
∞ X
∞
X
am bn
(1.5)
m+n
n=1 m=1


  p1 

  1q
∞
∞
X



X
 π − 1 −1 γ  apn
 π − 1 −1 γ  bqn
<
,
π
π

 

np
nq
n=1 sin p
n=1 sin p
where 1 − γ = 0.42278433+ is the best value. In 1998, Yang [6] first introduced an independent parameter λ and the Beta function to build an extension of Hilbert’s integral inequality. Recently, by introducing a parameter λ,
Yang [7] and Yang et al. [8] gave some extensions of (1.3) and (1.4) as: If
2 − min{p, q} < λ ≤ 2, an, bn ≥ 0 satisfy
0<
∞
X
n=1
n1−λ apn
< ∞ and
0<
∞
X
n=1
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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n1−λ bqn
< ∞,
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then one has the following two equivalent inequalities:
(∞
) p1 ( ∞
) 1q
∞ X
∞
X
X
X
am b n
< kλ (p)
(1.6)
n1−λ apn
n1−λ bqn
λ
(m
+
n)
n=1
n=1
n=1 m=1
and
#p
∞
X
a
m
p
(1.7)
n(p−1)(λ−1)
dy
<
[k
(p)]
n1−λ apn ,
λ
λ
(m
+
n)
n=1
m=1
n=1
where the constant factors kλ (p) = B p+λ−2
and [kλ (p)]p are the best
, q+λ−2
p
q
possible (B(u, v) is the β function). For λ = 1, inequalities (1.6) and (1.7)
reduce respectively to (1.3) and (1.4). By introducing a parameter α, Kuang [9]
gave an extension of (1.3), and Yang [10] gave an improvement of [9] as: If
0 < α ≤ min{p, q}, an, bn ≥ 0 satisfy
"
∞
X
0<
∞
X
∞
X
n(p−1)(1−α) apn < ∞ and
n=1
0<
∞
X
n(q−1)(1−α) bqn < ∞,
n=1
am b n
mα + nα
n=1 m=1
π
α sin πp
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∞ X
∞
X
<
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then one has two equivalent inequalities as:
(1.8)
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
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(∞
X
n=1
n(p−1)(1−α) apn
) p1 ( ∞
X
n=1
) 1q
n(q−1)(1−α) bqn
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and
(1.9)
∞
X
"
nα−1
n=1
∞
X
am
mα + nα
m=1
#p
p
∞
X
π


dy <
n(p−1)(1−α) apn ,
π
α sin p
n=1

h
ip
π
π
where the constant factors α sin(π/p)
and α sin(π/p)
are the best possible. For
α = 1, inequalities (1.8) and (1.9) reduce respectively to (1.3) and (1.4). Recently, Hong [11] gave an extension of (1.3) by introducing two parameters λ
1
and α as: If α ≥ 1, 1 − αr
< λ ≤ 1 (r = p, q), then
(1.10)
Bicheng Yang
∞ X
∞
X
am b n
α
(m + nα )λ
n=1 m=1
< Hλ,α (p)
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
(∞
X
n=1
nα(1−λ) apn
) p1 ( ∞
X
Title Page
) 1q
nα(1−λ) bpn
,
n=1
where
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p1 1q
1
1
1
1
,λ +
−1
,λ +
−1
.
Hλ,α (p) = B 1 −
B 1−
αq
αq
αp
αp
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For λ = α = 1, (1.10) reduces to (1.3). However, it is obvious that (1.10) is not
an extension of (1.6) or (1.8).
In 2003, Yang et al. [12] provided an extensive account of the above results.
More recently, Yang [13] gave some extensions of (1.1) and (1.2) as: If 0 <
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λ ≤ min{p, q}, satisfy
0<
∞
X
np−1−λ apn
< ∞ and
0<
n=1
∞
X
nq−1−λ apn < ∞,
n=1
then one has the following two equivalent inequalities:
(1.11)
(∞
) p1 ( ∞
) 1q
X
X
am b n
< Kλ (p)
np−1−λ apn
nq−1−λ bpn
λ
(m
+
n)
n=1
n=1
n=1 m=1
∞ X
∞
X
and
(1.12)
∞
X
"
n(p−1)λ−1
n=1
∞
X
am
(m + n)λ
m=1
#p
dy < [Kλ (p)]p
∞
X
Bicheng Yang
np−1−λ apn ,
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n=1
where the constants Kλ (p) = B λp , λq and [Kλ (p)]p are the best possible. For
λ = 1, (1.11) and (1.12) reduce to the following two equivalent inequalities:
(1.13)
∞ X
∞
X
am b n
π
<
π
m
+
n
sin
n=1 m=1
p
(∞
X
np−2 apn
) p1 ( ∞
X
n=1
) 1q
nq−2 bqn
n=1
and
(1.14)
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∞
X
n=1
np−2
∞
X
am
m+n
m=1
!p
p

<
sin
∞
π  X p−2 p
n an .
Page 7 of 35
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π
p
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For p = q = 2, inequalities (1.13) and (1.14) reduce respectively to (1.1) and
(1.2). We find that inequalities (1.3) and (1.13) are different, although both of
them are the best extensions of (1.1) with the (p, q)−parameter.
The main objective of this paper is to obtain some extensions of (1.3) with
the best constant factors, by introducing two parameters
using the
P∞λ and aαm band
P
n
Beta function, related to the double series as ∞
n=1
m=1 (mα +nα )λ (λ, α >
0), so that inequality (1.10) can be improved. The equivalent form and some
reversions are considered.
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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2.
Some Lemmas
First, we need the form of the Beta function as (cf. Wang et al. [14]):
Z ∞
1
(2.1)
B(u, v) :=
tu−1 dt = B(v, u)
(u, v > 0).
u+v
(1
+
t)
0
Lemma 2.1. If p > 0 (p 6= 1), p1 + 1q = 1, λ, α > 0, φr = φr (λ, α) > 0 (r =
p, q), satisfy φp + φq = λα, define the weight function ωr (x) as
∞
Z
(2.2)
ωr (x) :=
0
1−φr
1
x
dy
(xα + y α )λ y
λα−φr
(x > 0; r = p, q).
Bicheng Yang
Then for x > 0, each ωr (x) is constant, that is
φp φq
1
(2.3)
ωr (x) = B
,
(x > 0; r = p, q).
α
α α
y α
x
Proof. Setting u =
ωr (x) = x
λα−φr
1
=
α
Z
0
∞
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in the integral (2.2), one has dy = αx u α −1 du and
Z
∞
1
α
(x + xα u)λ
0
φr
1
u α −1 du
λ
(1 + u)
1
xu1/α
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
1−φr
x 1 −1
u α du
α
(r = p, q).
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By (2.1), since φp + φq = λα, one has (2.3). The lemma is proved.
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Lemma 2.2. If p > 1, p1 + 1q = 1, λ, α > 0, φr > 0 (r = p, q), satisfy
φp + φq = λα, and 0 < ε < qφp , then one has
(2.4)
ε
x−1+φq − p −1+φp − qε
I1 :=
y
dxdy
(xα + y α )λ
1
1
1
φp
ε φq
ε
>
B
−
,
+
− O(1).
εα
α
qα α
qα
Z
∞
Z
∞
If 0 < p < 1 and 0 < ε < −qφq , with the above assumption, then one has
∞ Z
X
(2.5)
ε
m−1+φq − p −1+φp − qε
I2 :=
y
dy
(mα + y α )λ
m=1 0
∞
1
φp
ε φq
ε X 1
= B
−
,
+
.
α
α
qα α
qα m=1 m1+ε
Proof. Setting u =
y α
x
∞
in the integral I1 , one has
Z ∞
Z ∞
1
−1+φq − pε
−1+φp − qε
y
I1 =
x
dy dx
(xα + y α )λ
1
1
Z
Z
φp
ε
1 ∞ −1−ε ∞
1
− qα
−1
α
=
x
u
dudx
λ
1
α 1
(1 + u)
xα
Z ∞ φαp − qαε −1
Z
Z 1 φp ε
u
1 ∞ −1−ε xα u α − qα −1
1
dudx
du −
x
=
(1 + u)λ
εα 0 (1 + u)λ
α 1
0
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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(2.6)
1
>
εα
Z
1
=
εα
Z
∞
0
0
∞
φp
ε
Z
Z 1
u α − qα −1
1 ∞ −1 xα φαp − qαε −1
du −
u
dudx
x
(1 + u)λ
α 1
0
φp
ε
−2
u α − qα −1
ε
du − φp −
.
(1 + u)λ
q
y α
) in the
By (2.1), it follows that (2.4) is valid. For 0 < p < 1, setting u = ( m
integral of I2 , in the same manner, one has (2.5). The lemma is thus proved.
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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3.
Main Results
Theorem 3.1. If p > 1, p1 + 1q = 1, λ, α > 0, 0 < φr ≤ 1 (r = p, q),
φp + φq = λα and an , bn ≥ 0 satisfy
0<
∞
X
np(1−φq )−1 apn
< ∞ and
n=1
0<
∞
X
nq(1−φp )−1 bqn < ∞,
n=1
then one has
(3.1)
∞ X
∞
X
n=1 m=1
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
am b n
+ nα )λ
(mα
Bicheng Yang
) p1 ( ∞
) 1q
(X
∞
X
φp φq
1
,
< B
,
np(1−φq )−1 apn
nq(1−φp )−1 bqn
α
α α
n=1
n=1
where the constant factor α1 B
φp φq
,
α α
is the best possible.
Proof. By Hölder’s inequality with weight (see [15]), one has
∞ X
∞
X
am b n
H(am , bn ) :=
α
(m + nα )λ
n=1 m=1
∞ X
∞
X
(1−φq )/q (1−φp )/p 1
m
n
=
am
bn
α
α
λ
(1−φ
)/p
p
(m + n ) n
m(1−φq )/q
n=1 m=1
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(
≤
) p1
#
mλα−φp
1
mp(1−φq )−1 apm
·
α + nα )λ n1−φp
(m
n=1
(∞ " ∞
#
) 1q
X X nλα−φq
1
×
· 1−φq nq(1−φp )−1 bqn
.
α
α
λ
(m + n ) m
n=1 m=1
"∞
∞
X
X
m=1
(3.2)
Since λ, α > 0, and 1 − φr ≥ 0 (r = p, q), in view of (2.2), we rewrite (3.2) as
(
H(am , bn ) <
∞
X
ωp (m)mp(1−φq )−1 apm
m=1
) p1 ( ∞
X
ωq (n)nq(1−φp )−1 bqn
(3.3)
,
n=1
Bicheng Yang
and then by (2.3), one has (3.1). For 0 < ε < qφp , setting
ε
ε
n−1+φq − p , b0n = n−1+φp − q , n ∈ N, then we find
(∞
X
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
) 1q
np(1−φq )−1 a0p
n
) p1 ( ∞
X
n=1
n=1
) 1q
nq(1−φp )−1 b0q
n
a0n and
b0n
as:
a0n
=
∞
X
1
=1+
n1+ε
n=2
Z
∞
1
<1+
dt
t1+ε
1
1
= (1 + ε).
ε
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If the constant factor
1
B
α
φp φq
,
α α
in (3.1) is not the best possible, then there
exists a positive constant k (with k < α1 B φαp , φαq ), such that (3.1) is still valid
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if one replaces α1 B
1
B
α
φp φq
,
α α
by k. In particular, by (2.4) and (3.3),
φp
ε φq
ε
−
,
+
− εO(1)
α
qα α
qα
< εI1
< εH(a0m , b0n )
(∞
) 1q
) p1 ( ∞
X
X
< εk
nq(1−φp )−1 b0q
np(1−φq )−1 a0p
n
n
n=1
n=1
= k(1 + ε),
φp φq
1
and then α B α , α ≤ k (ε → 0+ ). This contradicts the fact that k <
φp φq
φp φq
1
1
B
,
.
Hence
the
constant
factor
B
,
in (3.1) is the best possiα
α α
α
α α
ble. The theorem is proved.
Theorem 3.2. If p > 1, p1 + 1q = 1, λ, α > 0, 0 < φr ≤ 1 (r = p, q),
φp + φq = λα and an ≥ 0 satisfy
0<
∞
X
np(1−φq )−1 apn < ∞,
n=1
(3.4)
n=1
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then one has
∞
X
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
"
n
pφp −1
∞
X
am
α
(m + nα )λ
m=1
#p
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1
<
B
α
where the constant factor
is equivalent to (3.1).
h
1
B
α
φp φq
,
α α
ip
φp φq
,
α α
p X
∞
np(1−φq )−1 apn ,
n=1
is the best possible. Inequality (3.4)
Proof. Set
"
∞
X
bn := npφp −1
m=1
(mα
am
+ nα )λ
#p−1
,
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
and use (3.1) to obtain
(3.5)
0<
=
=
∞
X
n=1
∞
X
nq(1−φp )−1 bqn
"
n
pφp −1
n=1
∞ X
∞
X
n=1 m=1
≤
1
B
α
Bicheng Yang
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∞
X
am
α
(m + nα )λ
m=1
am b n
+ nα )λ
(X
∞
φq
#p
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(mα
φp
,
α α
n=1
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np(1−φq )−1 apn
) p1 ( ∞
X
n=1
) 1q
nq(1−φp )−1 bqn
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and
0<
(∞
X
) p1
nq(1−φp )−1 bqn
n=1
# ) p1
a
m
p
=
npφp −1
α + nα )λ
(m
n=1
m=1
) p1
(X
∞
φp φq
1
< ∞.
≤ B
,
np(1−φq )−1 apn
α
α α
n=1
(∞
X
(3.6)
"
∞
X
It follows that (3.5) takes the form of strict inequality by using (3.1); so does
(3.6). Hence, one has (3.4).
On the other hand, if (3.4) is valid, by Hölder’s inequality, one has
(3.7)
∞ X
∞
X
am b n
+ nα )λ
n=1 m=1
"
#
1
∞
∞
−1+φp
X
X
q
n
am h 1−φp − 1q i
bn
=
n
(mα + nα )λ
n=1 m=1
(∞
" ∞
#p ) p1 ( ∞
) 1q
X
X
X
am
≤
npφp −1
nq(1−φp )−1 bqn
.
α
α
λ
(m + n )
n=1
m=1
n=1
(mα
By (3.4), one has (3.1). It follows that inequalities (3.4) and (3.1) are equivalent.
If the constant factor in (3.4) is not the best possible, one can obtain a contradiction that the constant factor in (3.1) is not the best possible by using (3.7).
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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Hence the constant factor in (3.4) is still the best possible. Thus the theorem is
proved.
1
p
Theorem 3.3. If 0 < p < 1,
+
1
q
= 1,
A = {(λ, α); λ, α > 0, 0 < φr ≤ 1 (r = p, q), φp + φq = λα} =
6 Φ,
and an , bn ≥ 0 satisfy
0<
∞
X
np(1−φq )−1 apn
< ∞ and
0<
n=1
∞
X
nq(1−φp )−1 bqn < ∞,
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
n=1
then for (λ, α) ∈ A, one has
Bicheng Yang
∞ X
∞
X
am b n
+ nα )λ
n=1 m=1
) p1 ( ∞
) 1q
(X
∞
X
φp φq
1
,
nq(1−φp )−1 bqn
,
> B
[1 − θp (n)]np(1−φq )−1 apn
α
α α
n=1
n=1
(3.8)
Title Page
(mα
where 0 < θp (n) = O
1
nφp
< 1; the constant α1 B
φp φq
,
α α
Contents
JJ
J
is the best possible.
Proof. By the reverse of Hölder’s inequality (see [15]), following the method of
proof in Theorem 3.1, since 0 < p < 1 and q < 0, one has
(3.9) H(am , bn ) >
(∞
X
n=1
$p (n)np(1−φq )−1 apn
) p1 ( ∞
X
n=1
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) 1q
ωq (n)nq(1−φp )−1 bqn
II
I
,
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where ωq (n) is defined as in (2.2) and
(3.10)
$p (n) :=
∞
X
k=1
nλα−φp
(nα + k α )λ
1−φp
1
k
(n ∈ N).
Define θp (n) as
(3.11)
nλα−φp
θp (n) :=
ωp (n)
1
Z
0
Since
Z
ωp (n) >
0
1
1
α
(n + y α )λ
1−φp
1
dy
y
(n ∈ N).
1−φp
nλα−φp
1
dy,
α
α
λ
(n + y )
y
then we find 0 < θp (n) < 1, and
1−φp
Z ∞
nλα−φp
1
dy = ωp (n) [1 − θp (n)] .
(3.12)
$p (n) >
(nα + y α )λ y
1
By (3.12), (2.3) and (3.9), one has (3.8). Since
1−φp
Z
nλα−φp 1 1
1
1
1
(3.13)
0 < θp (n) <
dy =
· φp ,
λα
ωp (n) 0 n
y
ωp (n)φp n
and ωp (n) is a constant, we have θp (n) = O n1φp (n → ∞).
ε
For 0 < ε < min{q(φp − 1), −qφq }, setting a0n and b0n as: a0n = n−1+φq − p ,
ε
b0n = n−1+φp − q , n ∈ N, since φp > 0, then
∞
X
1
O
= O(1) (ε → 0+ ),
φ
p+1+ε
n
n=1
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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and
(3.14)
(∞
X
[1 − θp (n)]np(1−φq )−1 a0p
n
) p1 ( ∞
X
n=1
) 1q
nq(1−φp )−1 b0q
n
n=1
 p1

P∞
1
∞


O
X
n=1
1
nφp+1+ε
P
=
1
−
∞
1

n1+ε 
n=1 n1+ε
n=1
∞
X
1
1
=
(1 − o(1)) p .
1+ε
n
n=1
φp φq
1
If the constant α B α , α in (3.8) is not the best possible, then there exists a
positive number K (with K > α1 B φαp , φαq ), such that (3.8) is still valid if one
φp φq
1
replaces α B α , α by K. In particular, by (3.14) and (2.5), one has
∞
X
1
1
{1 − o(1)} p
K
1+ε
n
n=1
(∞
) 1q
) p1 ( ∞
X
X
=K
[1 − θp (n)]np(1−φq )−1 a0p
nq(1−φp )−1 b0q
n
n
<
n=1
H(a0m , b0n )
1
< I2 = B
α
n=1
φp
ε φq
ε
−
,
+
α
qα α
qα
X
∞
n=1
1
n
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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,
1+ε
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and then K ≤ α1 B
φp φq
,
α α
(ε → 0+ ). By this contradiction we can conclude
that the constant α1 B φαp , φαq in (3.8) is the best possible. Thus the theorem is
proved.
Theorem 3.4. If 0 < p < 1,
1
p
+
1
q
= 1,
A = {(λ, α); λ, α > 0, 0 < φr ≤ 1 (r = p, q), φp + φq = λα} =
6 Φ,
and an , bn ≥ 0 satisfy
0<
∞
X
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
np(1−φq )−1 apn < ∞,
n=1
Bicheng Yang
for (λ, α) ∈ A, one has
" ∞
∞
X
X
pφp −1
(3.15)
n
#p
am
(mα + nα )λ
n=1
m=1
p X
∞
1
φp φq
>
B
,
[1 − θp (n)]np(1−φq )−1 apn ,
α
α α
n=1
h
ip
φp φq
1
1
where 0 < θp (n) = O nφp < 1, and the constant factor α B α , α
is
the best possible. Inequality (3.15) is equivalent to (3.8).
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Proof. Still setting
"
bn := n
pφp −1
∞
X
am
α
(m + nα )λ
m=1
Page 20 of 35
#p−1
,
J. Ineq. Pure and Appl. Math. 6(3) Art. 81, 2005
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by (3.8), one has
(3.16)
0<
∞
X
nq(1−φp )−1 bqn
n=1
#p
∞ X
∞
X
a
am bn
m
pφp −1
=
=
n
α
α
λ
α
(m + n )
(m + nα )λ
n=1 m=1
n=1
m=1
) p1
(X
∞
φp φq
1
≥ B
,
[1 − θp (n)]np(1−φq )−1 apn
α
α α
n=1
(∞
) 1q
X
×
nq(1−φp )−1 bqn
"
∞
X
∞
X
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
n=1
and
0<
(∞
X
Title Page
) p1
Contents
nq(1−φp )−1 bqn
JJ
J
n=1
=
(∞
X
"
npφp −1
n=1
(3.17)
P∞
≥
1
B
α
∞
X
m=1
φp φq
,
α α
(mα
(X
∞
am
+ nα )λ
#p ) p1
Go Back
) p1
[1 − θp (n)]np(1−φq )−1 apn
II
I
.
n=1
If n=1 nq(1−φp )−1 bqn < ∞, by
(3.8), (3.16) takes the form of strict inP∞usingq(1−φ
p )−1 q
equality; so does (3.17). If n=1 n
bn = ∞, (3.17) takes naturally
strict inequality. Hence we have (3.15).
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On the other hand, if (3.15) is valid, by the reverse of Hölder’s inequality,
(3.18)
∞ X
∞
X
am b n
+ nα )λ
n=1 m=1
" ∞
#
∞
X
X n 1q −1+φp am
1
=
[n1−φp − q bn ]
α
α
λ
(m + n )
n=1 m=1
) 1q
(∞
" ∞
#p ) p1 ( ∞
X
X
X
a
m
.
≥
nq(1−φp )−1 bqn
npφp −1
α + nα )λ
(m
n=1
n=1
m=1
(mα
Hence by (3.15), one has (3.8). If the constant factor in (3.15) is not the best
possible, we can conclude that the constant factor in (3.8) is not the best possible
by using (3.18). The theorem is proved.
Note: In view of (3.1), if φr = φr (λ, α) (r = p, q) satisfy
B(φp (1, 1), φq (1, 1)) =
sin
π
,
π
p
and rφr (1, 1) = 1 (r = p, q), one can get a best extension of (1.3); if
π
B(φp (1, 1), φq (1, 1)) = sin(π/p)
(or π2 ), and rφr (1, 1) 6= 1 (r = p, q), one can get
a best extension of (1.1)
but not a best extension of (1.3). For example, setting
1
φr = r (α − 2) + 1 λ (r = p, q), then rφr (1, 1) = r − 1 6= 1, by Theorems
3.1 – 3.4, one can get a best extension of (1.13) and (1.1) as follows:
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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Corollary 3.5. If p > 1, p1 + 1q = 1, λ > 0, α > 2−min{p, q},
1 (r = p, q), an, bn ≥ 0, satisfy
0<
∞
X
1
r
(α − 2) + 1 λ ≤
np[1−λ(α−1)]+(α−2)λ−1 apn < ∞
n=1
and
0<
∞
X
nq[1−λ(α−1)]+(α−2)λ−1 bqn < ∞,
n=1
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
one has equivalent inequalities as:
(3.19)
∞
∞ X
X
n=1 m=1
am b n
< Kλ,α (p)
+ nα )λ
(∞
) p1
X
×
np[1−λ(α−1)]+(α−2)λ−1 apn
Bicheng Yang
(mα
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n=1
×
(∞
X
) 1q
nq[1−λ(α−1)]+(α−2)λ−1 bqn
n=1
II
I
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and
(3.20)
JJ
J
∞
X
n=1
"
n
(p+α−2)λ−1
∞
X
am
α
(m + nα )λ
m=1
Close
#p
< [Kλ,α (p)]p
Quit
∞
X
n=1
Page 23 of 35
np[1−λ(α−1)]+(α−2)λ−1 apn ,
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where
1
p+α−2 q+α−2
Kλ,α (p) = B λ
,λ
α
αp
αq
and [Kλ,α (p)]p are the best possible. In particular,
(i) for α = 1, one has 0 < λ ≤ min{p, q} and (1.11);
(ii) for λ = 1, one has 2 − min{p, q} < α ≤ 2 and
(3.21)
∞ X
∞
X
am b n
mα + nα
n=1 m=1
(∞
) p1 ( ∞
) 1q
X
X
< K1,α (p)
n(p−1)(2−α)−1 apn
n(q−1)(2−α)−1 bqn
n=1
Contents
∞
X
n=1
"
n
p+α−3
∞
X
am
α
m + nα
m=1
#p
p
< [K1,α (p)]
∞
X
n(p−1)(2−α)−1 apn .
n=1
If 0 < p < 1, for (λ, α) = (1, 2) ∈ A(6= Φ), by (3.8), (3.15) and (3.13),
one can obtain two equivalent reversions as
(3.23)
Bicheng Yang
Title Page
n=1
and
(3.22)
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
∞ X
∞
X
am b n
π
>
2
2
m +n
2
n=1 m=1
(∞ X
n=1
1−
2
πn
apn
n
) p1 ( ∞
)1
X bq q
n
n=1
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n
J. Ineq. Pure and Appl. Math. 6(3) Art. 81, 2005
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and
(3.24)
∞
X
n=1
"
n
p−1
∞
X
am
2
m + n2
m=1
#p
>
∞ π p X
2
n=1
2
1−
πn
apn
,
n
where the constant factors in the above inequalities are the best possible.
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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4.
Some Best Extensions of (1.3)
Setting φr =
λα
r
(r = p, q), by Theorems 3.1 – 3.4, one has
Corollary 4.1. If p > 1,
satisfy
0<
∞
X
1
p
+
n(p−1)(1−λα) apn
1
q
= 1, λ, α > 0, λα ≤ min{p, q}, an , bn ≥ 0,
<∞
and 0 <
n=1
∞
X
n(q−1)(1−λα) bqn < ∞,
n=1
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
then one has the following equivalent inequalities:
(4.1)
∞ X
∞
X
Bicheng Yang
am b n
(mα + nα )λ
n=1 m=1
Kλ (p)
<
α
(∞
X
n(p−1)(1−λα) apn
) p1 ( ∞
X
n=1
) 1q
n(q−1)(1−λα) bqn
n=1
(4.2)
"
n
λα−1
n=1
where Kλ (p) = B
∞
X
am
α
(m + nα )λ
m=1
λ λ
,
p q
#p
Kλ (p)
<
α
p X
∞
Contents
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and
∞
X
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n(p−1)(1−λα) apn ,
n=1
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. In particular,
Page 26 of 35
(i) for α = 1, one has 0 < λ ≤ min{p, q} and the following two equivalent
J. Ineq. Pure and Appl. Math. 6(3) Art. 81, 2005
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inequalities:
(4.3)
∞ X
∞
X
am b n
(m + n)λ
n=1 m=1
< Kλ (p)
(∞
X
n(p−1)(1−λ) apn
n=1
) p1 ( ∞
X
) 1q
n(q−1)(1−λ) bqn
n=1
and
(4.4)
∞
X
"
nλ−1
n=1
∞
X
am
(m + n)λ
m=1
#p
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
∞
X
p
< [Kλ (p)]
n(p−1)(1−λ) apn ;
n=1
Bicheng Yang
(ii) for λ = 1, 0 < α ≤ min{p, q} one has two equivalent inequalities as:
(4.5)
Title Page
∞ X
∞
X
am b n
mα + nα
n=1 m=1
<
π
α sin πp
Contents
(∞
X
n(p−1)(1−α) apn
) p1 ( ∞
X
) 1q
n(q−1)(1−α) bqn
n=1
n=1
and
(4.6)
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∞
X
n=1
"
n
α−1
∞
X
am
α
m + nα
m=1
#p

<
p
∞
X
π

n(p−1)(1−α) apn ,
π
α sin p
n=1
where the constant factors in the above inequalities are the best possible.
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Note: Since for 0 < p < 1, φq = λα
< 0, then A = Φ. It follows that both
q
+ 1r (r = p, q), by
(4.1) and (4.2) do not possess reversions. Setting φr = λα−1
2
Theorems 3.1 – 3.4, one has
n
o
Corollary 4.2. If p > 1, p1 + 1q = 1, λ, α > 0, 1 − 2 min p1 , 1q < λα ≤
n
o
1 + 2 min p1 , 1q , an , bn ≥ 0, satisfy
0<
∞
X
n
p
(1−λα)
2
apn
<∞
and
n=1
0<
∞
X
q
n 2 (1−λα) bqn < ∞,
n=1
then one has the following equivalent inequalities:
(4.7)
∞
∞ X
X
am b n
α
(m + nα )λ
n=1 m=1
pλα − p + 2 qλα − q + 2
1
< B
,
α
2pα
2qα
(∞
) p1 ( ∞
) 1q
X p
X q
×
n 2 (1−λα) apn
n 2 (1−λα) bqn
n=1
Bicheng Yang
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n=1
Go Back
and
(4.8)
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
∞
X
n=1
#p
am
dy
(mα + nα )λ
m=1
p X
∞
p
1
pλα − p + 2 qλα − q + 2
<
B
,
n 2 (1−λα) apn .
α
2pα
2qα
n=1
"
n
p
(λα−1)
2
∞
X
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In particular,
n
o
n
o
(i) for α = 1, one has 1 − 2 min p1 , 1q < λ ≤ 1 + 2 min p1 , 1q , and the
following two equivalent inequalities:
∞ X
∞
X
pλ − p + 2 qλ − q + 2
am b n
(4.9)
<B
,
λ
(m
+
n)
2p
2q
n=1 m=1
) 1q
) p1 ( ∞
(∞
X q
X p
n 2 (1−λ) bqn
×
n 2 (1−λ) apn
n=1
n=1
and
∞
X
"
∞
X
Bicheng Yang
#p
am
dy
(m + n)λ
m=1
n=1
p X
∞
p
pλ − p + 2 qλ − q + 2
< B
,
n 2 (1−λα) apn ;
2p
2q
n=1
n
o
n
o
(ii) for λ = 1, one has 1 − 2 min p1 , 1q < α ≤ 1 + 2 min p1 , 1q and two
equivalent inequalities as:
∞ X
∞
X
1
pα − p + 2 qα − q + 2
am b n
(4.11)
< B
,
α + nα
m
α
2pα
2qα
n=1 m=1
(∞
) p1 ( ∞
) 1q
X p
X q
×
n 2 (1−α) apn
n 2 (1−α) bqn
(4.10)
p
n 2 (λ−1)
n=1
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
n=1
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and
(4.12)
∞
X
#p
am
dy
mα + nα
m=1
p X
∞
p
1
pα − p + 2 qα − q + 2
<
B
,
n 2 (1−α) apn ,
α
2pα
2qα
n=1
"
n
p
(α−1)
2
n=1
∞
X
where the constant factors in the above inequalities are the best possible.
Note: Since for 0 < p < 1, we find
λα − 1 1
A = (λ, α); λ, α > 0, 0 <
+ ≤ 1 (r = p, q) = Φ,
2
r
it follows
that both (4.7) and (4.8) do not possess reversions. Setting φr =
1
1 − r (λα − 2) + 1 (r = p, q), by Theorems 3.1 – 3.4, one has
Corollary 4.3. If p > 1, p1 + 1q = 1, λ, α > 0, 2 − min{p, q} < λα ≤ 2,
an , bn ≥ 0, satisfy
∞
∞
X
X
0<
n1−λα apn < ∞ and 0 <
n1−λα bqn < ∞,
n=1
n=1
then one has the following equivalent inequalities:
∞ X
∞
X
1
p + λα − 2 q + λα − 2
am bn
(4.13)
< B
,
α + nα )λ
(m
α
pα
qα
n=1 m=1
(∞
) p1 ( ∞
) 1q
X
X
×
n1−λα apn
n1−λα bqn
n=1
n=1
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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and
(4.14)
#p
a
m
n(p−1)(λα−1)
α + nα )λ
(m
n=1
m=1
p X
∞
1
p + λα − 2 q + λα − 2
<
B
,
n1−λα apn ,
α
pα
qα
n=1
∞
X
"
∞
X
where the constant factors in the above inequalities are the best possible.
If 0 < p < 1, one has (α/2, α) ∈ A and two equivalent reversions as:
) 1q
(∞ ) p1 ( ∞
∞ X
∞
X1
X
X
am b n
1
1 p
bqn
(4.15)
> kα
1−
an
α + nα )2/α
n
(m
k
n
n
α
n=1 m=1
n=1
n=1
and
Bicheng Yang
Title Page
∞
X
"
∞
X
#p
∞ X
am
1
1 p
p
>
k
1
−
an ,
α
α + nα )2/α
(m
k
n
n
α
n=1
m=1
n=1
where kα = α1 B α1 , α1 , and the constant factors are the best possible.
(4.16)
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
np−1
Proof. For 0 < p < 1,
1
(λα − 2) + 1 ≤ 1
φr = 1 −
r
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(r = p, q),
we obtain λα = 2 and φr = 1. By (3.13), we find
0 < θp (n) <
Contents
1
1
1
· φp =
.
ωp (n)φp n
kα n
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Hence by (3.8) and (3.15), one has (4.15) and (4.16). The corollary is proved.
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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Remark 1.
(i) For α = 1, by (4.13) and (4.14), one has (1.6) and (1.7); for λ = 1,
by (4.13) and (4.14), one has (1.8) and (1.9). It follows that (4.13) is an
extension of (1.6) and (1.8), which is an improvement of (1.10), and (4.14)
is an extension of (1.7) and (1.9). (4.11) and (4.12) are extensions of (3.23)
and (3.24).
(ii) Inequalities (1.6), (4.3) and (4.9) are different extensions of (1.3) with a
parameter λ; inequalities (1.8), (4.5) and (4.11) are different extensions
of (1.3) with a parameter α and inequalities (4.1), (4.7) and (4.13) are
different extensions of (1.3) with two parameters λ and α.
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
(iii) Inequalities (1.7), (4.4) and (4.10) are different extensions of (1.4) with a
parameter λ; inequalities (1.9), (4.6) and (4.12) are different extensions
of (1.4) with a parameter α and inequalities (4.2), (4.8) and (4.14) are
different extensions of (1.4) with two parameters λ and α. Since the above
inequalities and some reversions are all with the best constant factors, one
gives some new results.
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References
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge University Press, Cambridge, 1952.
[2] G.H. HARDY, Note on a theorem of Hilbert concerning series of positive
terms, Proc. Math. Soc., 23(2) (1925), Records of Proc. XLV-XLVI.
[3] D.S. MINTRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Inequalities Involving Functions and Their Integrals and Derivertives, Kluwer Academic
Publishers, Boston, 1991.
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
[4] BICHENG YANG AND MINGZHE GAO, On a best value of hardyHilbert’s inequality, Advances in Math., 26(2) (1997), 159–164.
Bicheng Yang
[5] MINGZHE GAO AND BICHENG YANG, On the extended Hilbert’s inequality, Proc. Amer. Math. Soc., 126(3) (1998), 751–759.
Title Page
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[6] BICHENG YANG, On Hilbert’s integral inequality, J. Math. Anal. Appl.,
220 (1998), 778–785.
[7] BICHENG YANG AND L. DEBNATH, On the extended Hardy-Hilbert’s
inequality, J. Math. Anal. Appl., 272 (2002), 187–199.
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[8] BICHENG YANG, On generalizations of Hardy-Hilbert’s inequality and
their equivalent forms, J. of Math., 24(1) (2004), 24–30.
Close
[9] JICHANG KUANG, On a new extension of Hilbert’s integral inequality,
J. Math. Anal. Appl., 235 (1999), 608–614.
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[10] BICHENG YANG, On an extension of Hardy-Hilbert’s inequality, Chinese Annals of Math., 23A(2) (2002), 247–254.
[11] YONG HONG, A extension and improvement of Hardy-Hilbert’s double
series inequality, Mathematics in Practice and Theory, 32(5) (2002), 850–
854.
[12] BICHENG YANG AND Th. M. RASSIAS, On the way of weight coefficient and research for the Hilbert-type inequalities, Math. Ineq. Appl., 6(4)
(2003), 625–658.
[13] BICHENG YANG, On new extensions of Hilbert’s inequality, Acta Math.
Hungar., 104(3) (2004), 293–301.
[14] ZHUXI WANG AND DUNRIN GUO, An Introduction to Special Functions, Science Press, Beijing, 1979.
[15] JICHANG KUANG, Applied Inequalities, Hunan Education Press,
Changsha, 2004.
On Best Extensions of
Hardy-Hilbert’s Inequality with
Two Parameters
Bicheng Yang
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