Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 3, Article 81, 2005
ON BEST EXTENSIONS OF HARDY-HILBERT’S INEQUALITY WITH TWO
PARAMETERS
BICHENG YANG
D EPARTMENT OF M ATHEMATICS
G UANGDONG I NSTITUTE OF E DUCATION
G UANGZHOU , G UANGDONG 510303
P EOPLE ’ S R EPUBLIC OF C HINA
bcyang@pub.guangzhou.gd.cn
Received 23 February, 2005; accepted 17 June, 2005
Communicated by W.S. Cheung
A BSTRACT. 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.
Key words and phrases: Hardy-Hilbert’s inequality; Beta function; Hölder’s inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
If an, bn ≥ 0 satisfy
0<
∞
X
a2n
<∞
and 0 <
n=1
∞
X
b2n < ∞,
n=1
then one has two equivalent inequalities as:
(1.1)
∞
∞ X
X
am b n
<π
m
+
n
n=1 m=1
(∞
X
a2n
n=1
∞
X
) 12
b2n
n=1
and
(1.2)
∞
X
n=1
∞
X
am
m+n
m=1
!2
<π
2
∞
X
a2n ,
n=1
where the constant factors π and π 2 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
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
055-05
2
B ICHENG YANG
(1.2) by introducing the (p, q)−parameter as: If p > 1,
0<
∞
X
apn < ∞
and 0 <
n=1
1
p
1
q
+
∞
X
= 1, an, bn ≥ 0 satisfy
bqn < ∞,
n=1
then one has the following two equivalent inequalities:
(∞
) p1 ( ∞ ) 1q
∞ X
∞
X
X
X
am b n
π
(1.3)
<
apn
bqn
π
m
+
n
sin
n=1 m=1
n=1
n=1
p
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:


  p1 

  1q
∞
∞
∞
∞




X X am b n
X
X
 π − 1 −1 γ  apn
 π − 1 −1 γ  bqn
(1.5)
<
,
π
π




m
+
n
p
q
n
n
n=1 m=1
n=1 sin
n=1 sin
p
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
n1−λ apn
<∞
and 0 <
n=1
∞
X
n1−λ bqn < ∞,
n=1
then one has the following two equivalent inequalities:
(∞
) p1 ( ∞
) 1q
∞ X
∞
X
X
X
am b n
(1.6)
< kλ (p)
n1−λ apn
n1−λ bqn
λ
(m + n)
n=1 m=1
n=1
n=1
and
#p
∞
X
a
m
(p−1)(λ−1)
p
dy < [kλ (p)]
n1−λ apn ,
(1.7)
n
λ
(m
+
n)
n=1
m=1
n=1
, q+λ−2
and [kλ (p)]p are the best possible (B(u, v)
where the constant factors kλ (p) = B p+λ−2
p
q
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
n=1
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
<∞
and 0 <
∞
X
n(q−1)(1−α) bqn < ∞,
n=1
http://jipam.vu.edu.au/
O N B EST E XTENSIONS OF H ARDY-H ILBERT ’ S I NEQUALITY WITH T WO PARAMETERS
3
then one has two equivalent inequalities as:
(∞
) p1 ( ∞
) 1q
∞ X
∞
X
X
X
am b n
π
n(p−1)(1−α) apn
(1.8)
<
n(q−1)(1−α) bqn
α + nα
π
m
α sin
n=1
n=1 m=1
n=1
p
and
p

#p
∞
X
π

(1.9)
n
n(p−1)(1−α) apn ,
π
α sin p
n=1
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
1
of (1.3) by introducing two parameters λ and α as: If α ≥ 1, 1 − αr
< λ ≤ 1 (r = p, q), then
(∞
) p1 ( ∞
) 1q
∞ X
∞
X
X
X
am b n
(1.10)
< Hλ,α (p)
nα(1−λ) apn
nα(1−λ) bpn
,
α + nα )λ
(m
n=1 m=1
n=1
n=1
∞
X
"
α−1
∞
X
am
α
m + nα
m=1
dy < 
where
p1 1q
1
1
1
1
Hλ,α (p) = B 1 −
,λ +
−1
,λ +
−1
.
B 1−
αq
αq
αp
αp
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 < λ ≤ min{p, q}, satisfy
0<
∞
X
np−1−λ apn
<∞
0<
and
n=1
∞
X
nq−1−λ apn < ∞,
n=1
then one has the following two equivalent inequalities:
(∞
) p1 ( ∞
) 1q
∞ X
∞
X
X
X
am b n
(1.11)
< Kλ (p)
np−1−λ apn
nq−1−λ bpn
λ
(m + n)
n=1 m=1
n=1
n=1
and
#p
∞
X
a
m
p
(1.12)
n(p−1)λ−1
dy
<
[K
(p)]
np−1−λ apn ,
λ
λ
(m + n)
n=1
m=1
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:
(∞
) p1 ( ∞
) 1q
∞ X
∞
X
X
X
am b n
π
(1.13)
<
np−2 apn
nq−2 bqn
π
m+n
sin
n=1 m=1
n=1
n=1
"
∞
X
∞
X
p
and
(1.14)
∞
X
n=1
n
p−2
∞
X
am
m+n
m=1
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
!p
p

<
sin
∞
π  X p−2 p
n an .
π
p
n=1
http://jipam.vu.edu.au/
4
B ICHENG YANG
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 λ and α and using the Beta function, related to the
P∞ P
am bn
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.
2. S OME L EMMAS
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
1−φr
Z ∞
1
xλα−φr
dy
(x > 0; r = p, q).
(2.2)
ωr (x) :=
α
α
λ
(x + y )
y
0
Then for x > 0, each ωr (x) is constant, that is
1
φp φq
(2.3)
ωr (x) = B
,
(x > 0; r = p, q).
α
α α
α
1
Proof. Setting u = xy in the integral (2.2), one has dy = αx u α −1 du and
1−φr
Z ∞
1
1
x 1 −1
λα−φr
ωr (x) = x
u α du
α
α
λ
1/α
(x + x u)
xu
α
0
Z
φr
1 ∞
1
−1
α
u
du
(r = p, q).
=
α 0 (1 + u)λ
By (2.1), since φp + φq = λα, one has (2.3). The lemma is proved.
Lemma 2.2. If p > 1, p1 +
0 < ε < qφp , then one has
1
q
= 1, λ, α > 0, φr > 0 (r = p, q), satisfy φp + φq = λα, and
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
(2.4)
∞
Z
∞
If 0 < p < 1 and 0 < ε < −qφq , with the above assumption, then one has
ε
∞ Z ∞
X
m−1+φq − p −1+φp − qε
I2 :=
y
dy
α + y α )λ
(m
0
m=1
∞
1
φp
ε φq
ε X 1
(2.5)
= B
−
,
+
.
α
α
qα α
qα m=1 m1+ε
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
http://jipam.vu.edu.au/
O N B EST E XTENSIONS OF H ARDY-H ILBERT ’ S I NEQUALITY WITH T WO PARAMETERS
Proof. Setting u =
I1 =
=
=
>
=
(2.6)
y α
in the integral I1 ,
x
Z ∞
Z ∞
−1+φq − pε
5
one has
1
−1+φp − qε
y
x
dy dx
(xα + y α )λ
1
1
Z
Z
φp
ε
1 ∞ −1−ε ∞
1
− qα
−1
α
u
dudx
x
λ
1
(1 + u)
α 1
xα
Z ∞ φαp − qαε −1
Z
Z 1 φp ε
1 ∞ −1−ε xα u α − qα −1
1
u
dudx
du −
x
(1 + u)λ
εα 0 (1 + u)λ
α 1
0
Z ∞ φαp − qαε −1
Z
Z 1
1 ∞ −1 xα φαp − qαε −1
1
u
u
du −
x
dudx
εα 0 (1 + u)λ
α 1
0
−2
Z ∞ φαp − qαε −1
1
u
ε
du − φp −
.
εα 0 (1 + u)λ
q
y α
By (2.1), it follows that (2.4) is valid. For 0 < p < 1, setting u = ( m
) in the integral of I2 , in
the same manner, one has (2.5). The lemma is thus proved.
3. M AIN R ESULTS
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
∞
∞
X
X
0<
np(1−φq )−1 apn < ∞ and 0 <
nq(1−φp )−1 bqn < ∞,
n=1
n=1
then one has
) p1 ( ∞
) 1q
(X
∞
X
am b n
1
φp φq
(3.1)
< B
,
np(1−φq )−1 apn
nq(1−φp )−1 bqn
,
α + nα )λ
(m
α
α
α
n=1 m=1
n=1
n=1
where the constant factor α1 B φαp , φαq is the best possible.
∞ X
∞
X
Proof. By Hölder’s inequality with weight (see [15]), one has
H(am , bn ) :=
∞ X
∞
X
am b n
(mα + nα )λ
n=1 m=1
∞ X
∞
X
(3.2)
(1−φq )/q (1−φp )/p 1
m
n
=
a
b
m
α
α
λ
(1−φ
)/p
(1−φq )/q n
p
(m
+
n
)
n
m
n=1 m=1
( ∞ "∞
#
) p1
X X mλα−φp
1
≤
·
mp(1−φq )−1 apm
α + nα )λ n1−φp
(m
m=1 n=1
(∞ " ∞
#
) 1q
X X nλα−φq
1
×
·
nq(1−φp )−1 bqn
.
α + nα )λ m1−φq
(m
n=1 m=1
Since λ, α > 0, and 1 − φr ≥ 0 (r = p, q), in view of (2.2), we rewrite (3.2) as
( ∞
) p1 ( ∞
) 1q
X
X
ωq (n)nq(1−φp )−1 bqn
,
H(am , bn ) <
ωp (m)mp(1−φq )−1 apm
m=1
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
n=1
http://jipam.vu.edu.au/
6
B ICHENG YANG
ε
and then by (2.3), one has (3.1). For 0 < ε < qφp , setting a0n and b0n as: a0n = n−1+φq − p ,
ε
b0n = n−1+φp − q , n ∈ N, then we find
(∞
X
(3.3)
np(1−φq )−1 a0p
n
) p1 ( ∞
X
n=1
φp φq
,
α α
=1+
constant k (with k < α1 B φαp , φαq
k. In particular, by (2.4) and (3.3),
∞
X
1
1+ε
n
n=2
Z
∞
1
<1+
dt
1+ε
t
1
1
= (1 + ε).
ε
nq(1−φp )−1 b0q
n
n=1
If the constant factor α1 B
1
B
α
) 1q
φp
ε φq
ε
−
,
+
α
qα α
qα
in (3.1) is not the best possible, then there exists a positive
), such that (3.1) is still valid if one replaces α1 B φαp , φαq by
− εO(1) < εI1
< εH(a0m , b0n )
(∞
) p1 ( ∞
) 1q
X
X
< εk
np(1−φq )−1 a0p
nq(1−φp )−1 b0q
n
n
n=1
n=1
= k(1 + ε),
1
B
α
φp φq
,
α α
1
B
α
+
φp φq
,
α α
≤ k (ε → 0 ). This contradicts the fact that k <
the constant factor α1 B φαp , φαq in (3.1) is the best possible. The theorem is proved.
and then
Theorem 3.2. If p > 1,
satisfy
1
p
. Hence
+ 1q = 1, λ, α > 0, 0 < φr ≤ 1 (r = p, q), φp + φq = λα and an ≥ 0
0<
∞
X
np(1−φq )−1 apn < ∞,
n=1
then one has
(3.4)
∞
X
n=1
"
n
pφp −1
∞
X
am
α
(m + nα )λ
m=1
where the constant factor
(3.1).
h
1
B
α
φp φq
,
α α
#p
ip
1
<
B
α
φp φq
,
α α
p X
∞
np(1−φq )−1 apn ,
n=1
is the best possible. Inequality (3.4) is equivalent to
Proof. Set
"
bn := npφp −1
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
∞
X
am
α
(m + nα )λ
m=1
#p−1
,
http://jipam.vu.edu.au/
O N B EST E XTENSIONS OF H ARDY-H ILBERT ’ S I NEQUALITY WITH T WO PARAMETERS
7
and use (3.1) to obtain
(3.5)
0<
=
∞
X
n=1
∞
X
nq(1−φp )−1 bqn
"
npφp −1
n=1
∞ X
∞
X
∞
X
am
α
(m + nα )λ
m=1
#p
am b n
+ nα )λ
n=1 m=1
) p1 ( ∞
) 1q
(X
∞
X
1
φp φq
nq(1−φp )−1 bqn
≤ B
,
np(1−φq )−1 apn
α
α α
n=1
n=1
=
(mα
and
0<
(∞
X
) p1
nq(1−φp )−1 bqn
n=1
# ) p1
a
m
p
=
npφp −1
α
α
λ
(m + n )
n=1
m=1
(
) p1
X
∞
1
φp φq
≤ 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
" ∞
#
∞ X
∞
∞
X
X
X n 1q −1+φp am h 1−φ − 1 i
am b n
(3.7)
=
n p q bn
α + nα )λ
α + nα )λ
(m
(m
n=1 m=1
n=1 m=1
(∞
" ∞
#p ) p1 ( ∞
) 1q
X
X
X
a
m
.
≤
npφp −1
nq(1−φp )−1 bqn
α + nα )λ
(m
n=1
m=1
n=1
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). Hence the constant factor in (3.4) is still the best
possible. Thus the theorem is proved.
Theorem 3.3. 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
np(1−φq )−1 apn
n=1
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
<∞
and
0<
∞
X
nq(1−φp )−1 bqn < ∞,
n=1
http://jipam.vu.edu.au/
8
B ICHENG YANG
then for (λ, α) ∈ A, one has
(3.8)
∞
∞ X
X
n=1 m=1
am b n
+ nα )λ
(mα
1
> B
α
where 0 < θp (n) = O
1
nφp
φp φq
,
α α
(X
∞
[1 − θp (n)]np(1−φq )−1 apn
) p1 ( ∞
X
n=1
) 1q
nq(1−φp )−1 bqn
,
n=1
< 1; the constant α1 B
φp φq
,
α α
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
$p (n)np(1−φq )−1 apn
) p1 ( ∞
X
n=1
) 1q
ωq (n)nq(1−φp )−1 bqn
,
n=1
where ωq (n) is defined as in (2.2) and
(3.10)
$p (n) :=
∞
X
nλα−φp
(nα + k α )λ
k=1
1−φp
1
k
(n ∈ N).
Define θp (n) as
(3.11)
nλα−φp
θp (n) :=
ωp (n)
Z
1
1
α
(n + y α )λ
0
1−φp
1
dy
y
(n ∈ N).
Since
Z
ωp (n) >
0
1
nλα−φp
(nα + y α )λ
1−φp
1
dy,
y
then we find 0 < θp (n) < 1, and
Z
(3.12)
$p (n) >
1
∞
nλα−φp
(nα + y α )λ
1−φp
1
dy = ωp (n) [1 − θp (n)] .
y
By (3.12), (2.3) and (3.9), one has (3.8). Since
1−φp
1
1
1
(3.13)
dy =
· φp ,
y
ωp (n)φp n
0
and ωp (n) is a constant, we have θp (n) = O n1φp (n → ∞).
ε
ε
For 0 < ε < min{q(φp − 1), −qφq }, setting a0n and b0n as: a0n = n−1+φq − p , b0n = n−1+φp − q ,
n ∈ N, since φp > 0, then
nλα−φp
0 < θp (n) <
ωp (n)
∞
X
O
n=1
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
Z
1
1
nλα
1
nφp+1+ε
= O(1)(ε → 0+ ),
http://jipam.vu.edu.au/
O N B EST E XTENSIONS OF H ARDY-H ILBERT ’ S I NEQUALITY WITH T WO PARAMETERS
9
and
(∞
X
(3.14)
[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
∞


X 1
n=1 O nφp+1+ε
P∞
=
1−
1

n1+ε 
n=1 n1+ε
n=1
∞
X
1
1
p.
=
(1
−
o(1))
n1+ε
n=1
If the constant
1
B
α
φp φq
,
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 replaces α1 B φαp , φαq by K. In
particular, by (3.14) and (2.5), one has
K
∞
X
1
1
{1 − o(1)} p
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
0
H(am , b0n )
n=1
∞
1
φp
ε φq
ε X 1
< I2 = B
−
,
+
,
α
α
qα α
qα n=1 n1+ε
and then K ≤ α1 B φαp , φαq (ε → 0+ ). By this contradiction we can conclude that the constant
φp φq
1
B α , α in (3.8) is the best possible. Thus the theorem is proved.
α
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
np(1−φq )−1 apn < ∞,
n=1
for (λ, α) ∈ A, one has
" ∞
∞
X
X
(3.15)
npφp −1
#p
p X
∞
φp φq
,
[1 − θp (n)]np(1−φq )−1 apn ,
α α
n=1
n=1
h
ip
is the best possible.
where 0 < θp (n) = O n1φp < 1, and the constant factor α1 B φαp , φαq
Inequality (3.15) is equivalent to (3.8).
am
α
(m + nα )λ
m=1
1
>
B
α
Proof. Still setting
"
bn := npφp −1
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
∞
X
am
α
(m + nα )λ
m=1
#p−1
,
http://jipam.vu.edu.au/
10
B ICHENG YANG
by (3.8), one has
(3.16)
0<
=
∞
X
n=1
∞
X
nq(1−φp )−1 bqn
"
npφp −1
n=1
∞
X
am
α
(m + nα )λ
m=1
#p
∞ 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
=
(mα
and
0<
(∞
X
) p1
nq(1−φp )−1 bqn
n=1
=
(∞
X
"
npφp −1
n=1
(3.17)
1
≥ B
α
∞
X
m=1
φp φq
,
α α
(mα
(X
∞
am
+ nα )λ
#p ) p1
) p1
[1 − θp (n)]np(1−φq )−1 apn
.
n=1
P
q(1−φp )−1 q
If ∞
bn < ∞, by using (3.8), (3.16) takes the form of strict inequality; so does
n=1 n P
∞
q(1−φp )−1 q
(3.17). If n=1 n
bn = ∞, (3.17) takes naturally strict inequality. Hence we have
(3.15).
On the other hand, if (3.15) is valid, by the reverse of Hölder’s inequality,
" ∞
#
∞ X
∞
∞
X
X
X n 1q −1+φp am
am b n
1−φp − 1q
(3.18)
=
[n
bn ]
α + nα )λ
α + nα )λ
(m
(m
n=1 m=1
n=1 m=1
(∞
" ∞
#p ) p1 ( ∞
) 1q
X
X
X
a
m
.
≥
npφp −1
nq(1−φp )−1 bqn
α
α
λ
(m
+
n
)
n=1
m=1
n=1
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 B(φp (1, 1), φq (1, 1)) =
π
(or π2 ), and rφr (1, 1) 6= 1 (r = p, q), one can get a best extension of (1.1) but not a best
sin(π/p)
extension of (1.3). For example, setting φr = 1r (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:
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
http://jipam.vu.edu.au/
O N B EST E XTENSIONS OF H ARDY-H ILBERT ’ S I NEQUALITY
WITH
T WO PARAMETERS
11
Corollary 3.5. If p > 1, p1 + 1q = 1, λ > 0, α > 2−min{p, q}, 1r (α − 2) + 1 λ ≤ 1 (r = p, q),
an, bn ≥ 0, satisfy
∞
X
0<
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
one has equivalent inequalities as:
(3.19)
∞
∞ X
X
am b n
+ nα )λ
n=1 m=1
(∞
) p1 ( ∞
) 1q
X
X
< Kλ,α (p) ×
np[1−λ(α−1)]+(α−2)λ−1 apn
nq[1−λ(α−1)]+(α−2)λ−1 bqn
(mα
n=1
n=1
and
(3.20)
∞
X
"
n
(p+α−2)λ−1
n=1
∞
X
am
α
(m + nα )λ
m=1
#p
p
< [Kλ,α (p)]
∞
X
np[1−λ(α−1)]+(α−2)λ−1 apn ,
n=1
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
(∞
) p1 ( ∞
) 1q
∞ X
∞
X
X
X
am b n
< K1,α (p)
n(p−1)(2−α)−1 apn
n(q−1)(2−α)−1 bqn
(3.21)
α + nα
m
n=1 m=1
n=1
n=1
and
∞
X
(3.22)
"
n
p+α−3
n=1
∞
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
(∞ ) 1q
) p1 (X
∞ X
∞
∞
X
am b n
π X
2 apn
bqn
(3.23)
>
1−
m2 + n2
2 n=1
πn n
n
n=1 m=1
n=1
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.
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
http://jipam.vu.edu.au/
12
B ICHENG YANG
4. S OME B EST E XTENSIONS OF (1.3)
Setting φr =
λα
r
(r = p, q), by Theorems 3.1 – 3.4, one has
1
p
Corollary 4.1. If p > 1,
∞
X
0<
+
1
q
= 1, λ, α > 0, λα ≤ min{p, q}, an , bn ≥ 0, satisfy
n(p−1)(1−λα) apn
<∞
0<
and
n=1
∞
X
n(q−1)(1−λα) bqn < ∞,
n=1
then one has the following equivalent inequalities:
(∞
) p1 ( ∞
) 1q
∞
∞ X
X
X
Kλ (p) X (p−1)(1−λα) p
am b n
n(q−1)(1−λα) bqn
<
n
an
(4.1)
α + nα )λ
(m
α
n=1
n=1 m=1
n=1
and
∞
X
(4.2)
"
n
λα−1
n=1
where Kλ (p) = B
λ λ
,
p q
∞
X
am
(mα + nα )λ
m=1
#p
Kλ (p)
<
α
p X
∞
n(p−1)(1−λα) apn ,
n=1
. In particular,
(i) for α = 1, one has 0 < λ ≤ min{p, q} and the following two equivalent inequalities:
(∞
) p1 ( ∞
) 1q
∞ X
∞
X
X
X
am b n
(4.3)
< Kλ (p)
n(p−1)(1−λ) apn
n(q−1)(1−λ) bqn
λ
(m
+
n)
n=1 m=1
n=1
n=1
and
∞
X
(4.4)
"
n
λ−1
n=1
∞
X
am
(m + n)λ
m=1
#p
p
< [Kλ (p)]
∞
X
n(p−1)(1−λ) apn ;
n=1
(ii) for λ = 1, 0 < α ≤ min{p, q} one has two equivalent inequalities as:
(∞
) p1 ( ∞
) 1q
∞ X
∞
X
X
X
am b n
π
n(p−1)(1−α) apn
n(q−1)(1−α) bqn
(4.5)
<
α + nα
π
m
α sin
n=1 m=1
n=1
n=1
p
and
(4.6)
∞
X
"
n
α−1
n=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.
< 0, then A = Φ. It follows that both (4.1) and (4.2) do
Note: Since for 0 < p < 1, φq = λα
q
λα−1
not possess reversions. Setting φr = 2 + 1r (r = p, q), by Theorems 3.1 – 3.4, one has
n
o
n
o
Corollary 4.2. If p > 1, p1 + 1q = 1, λ, α > 0, 1 − 2 min p1 , 1q < λα ≤ 1 + 2 min p1 , 1q ,
an , bn ≥ 0, satisfy
0<
∞
X
n
p
(1−λα)
2
apn
n=1
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
<∞
and
0<
∞
X
q
n 2 (1−λα) bqn < ∞,
n=1
http://jipam.vu.edu.au/
O N B EST E XTENSIONS OF H ARDY-H ILBERT ’ S I NEQUALITY
WITH
T WO PARAMETERS
13
then one has the following equivalent inequalities:
(4.7)
∞ X
∞
X
n=1 m=1
am b n
+ nα )λ
(mα
1
< B
α
pλα − p + 2 qλα − q + 2
,
2pα
2qα
(X
∞
n
p
(1−λα)
2
apn
) p1 ( ∞
X
n=1
) 1q
n
q
(1−λα)
2
bqn
n=1
and
(4.8)
∞
X
"
n
p
(λα−1)
2
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
∞
X
In particular,
(i) for α = 1, one has 1 − 2 min
equivalent inequalities:
(4.9)
n
1 1
,
p q
o
< λ ≤ 1 + 2 min
n
1 1
,
p q
o
, and the following two
∞ X
∞
X
am b n
(m + n)λ
n=1 m=1
<B
pλ − p + 2 qλ − q + 2
,
2p
2q
(X
∞
n
p
(1−λ)
2
apn
) p1 ( ∞
X
n=1
) 1q
n
q
(1−λ)
2
bqn
n=1
and
#p
am
(4.10)
n
dy
λ
(m
+
n)
n=1
m=1
p X
∞
p
pλ − p + 2 qλ − q + 2
< B
,
n 2 (1−λα) apn ;
2p
2q
n=1
o
n
o
n
1 1
1 1
(ii) for λ = 1, one has 1 − 2 min p , q < α ≤ 1 + 2 min p , q and two equivalent
inequalities as:
∞
X
"
p
(λ−1)
2
∞
X
∞ X
∞
X
am b n
mα + nα
n=1 m=1
(4.11)
<
1
B
α
pα − p + 2 qα − q + 2
,
2pα
2qα
(X
∞
n=1
p
n 2 (1−α) apn
) p1 ( ∞
X
) 1q
q
n 2 (1−α) bqn
n=1
and
(4.12)
∞
X
n=1
"
n
p
(α−1)
2
#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
∞
X
where the constant factors in the above inequalities are the best possible.
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
http://jipam.vu.edu.au/
14
B ICHENG YANG
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 −
(r = p, q), by Theorems 3.1 – 3.4, one has
Corollary 4.3. If p > 1,
0<
1
p
1
q
+
∞
X
1
r
(λα − 2) + 1
= 1, λ, α > 0, 2 − min{p, q} < λα ≤ 2, an , bn ≥ 0, satisfy
n1−λα apn
<∞
0<
and
n=1
∞
X
n1−λα bqn < ∞,
n=1
then one has the following equivalent inequalities:
(4.13)
∞ X
∞
X
n=1 m=1
am b n
+ nα )λ
(mα
1
< B
α
p + λα − 2 q + λα − 2
,
pα
qα
(X
∞
n1−λα apn
) p1 ( ∞
X
n=1
) 1q
n1−λα bqn
n=1
and
(4.14)
∞
X
"
n(p−1)(λα−1)
n=1
#p
am
(mα + nα )λ
m=1
p X
∞
1
p + λα − 2 q + λα − 2
<
B
,
n1−λα apn ,
α
pα
qα
n=1
∞
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:
(∞ ) p1 ( ∞
) 1q
∞ X
∞
X
X
X1
am b n
1
1 p
(4.15)
> kα
1−
an
bqn
α + nα )2/α
(m
k
n
n
n
α
n=1 m=1
n=1
n=1
and
(4.16)
∞
X
"
n
p−1
n=1
where kα = α1 B
1
1
,
α α
∞
X
am
(mα + nα )2/α
m=1
#p
>
kαp
∞ X
n=1
1
1−
kα n
1 p
a ,
n n
, and the constant factors are the best possible.
Proof. For 0 < p < 1,
φr =
1
1−
r
(λα − 2) + 1 ≤ 1
(r = p, q),
we obtain λα = 2 and φr = 1. By (3.13), we find
0 < θp (n) <
1
1
1
· φp =
.
ωp (n)φp n
kα n
Hence by (3.8) and (3.15), one has (4.15) and (4.16). The corollary is proved.
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
http://jipam.vu.edu.au/
O N B EST E XTENSIONS OF H ARDY-H ILBERT ’ S I NEQUALITY
WITH
T WO PARAMETERS
15
Remark 4.4.
(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.
(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.
R EFERENCES
[1] G.H. HARDY, J.E. LITTLEWOOD
Cambridge, 1952.
AND
G. POLYA, Inequalities, Cambridge University Press,
[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.
[4] BICHENG YANG AND MINGZHE GAO, On a best value of hardy-Hilbert’s inequality, Advances
in Math., 26(2) (1997), 159–164.
[5] MINGZHE GAO AND BICHENG YANG, On the extended Hilbert’s inequality, Proc. Amer. Math.
Soc., 126(3) (1998), 751–759.
[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.
[8] BICHENG YANG, On generalizations of Hardy-Hilbert’s inequality and their equivalent forms, J.
of Math., 24(1) (2004), 24–30.
[9] JICHANG KUANG, On a new extension of Hilbert’s integral inequality, J. Math. Anal. Appl., 235
(1999), 608–614.
[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.
J. Inequal. Pure and Appl. Math., 6(3) Art. 81, 2005
http://jipam.vu.edu.au/