Volume 10 (2009), Issue 1, Article 25, 8 pp.
ON A DECOMPOSITION OF HILBERT’S INEQUALITY
BICHENG YANG
D EPARTMENT OF M ATHEMATICS
G UANGDONG E DUCATION I NSTITUTE
G UANGZHOU , G UANGDONG 510303, P.R. CHINA
bcyang@pub.guangzhou.gd.cn
Received 07 October, 2008; accepted 20 March, 2009
Communicated by B. Opic
A BSTRACT. 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.
Key words and phrases: Hilbert’s inequality; Weight coefficient; Equivalent form; Hilbert-type inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
InP
1908, H. Weyl publishedPthe following Hilbert inequality: If {an }, {bn } are real sequences,
∞
2
2
0< ∞
n=1 an < ∞ and 0 <
n=1 bn < ∞, then [1]
(1.1)
∞
∞
∞ X
∞
X
X
X
am b n
<π
a2n
b2n
m
+
n
n=1 m=1
n=1
n=1
! 12
,
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) as [2]: If p > 1, an , bn ≥ 0,
P
P∞ q
p
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
n=1
! p1
apn
∞
X
! 1q
bqn
,
n=1
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 ≥
The author expresses his sincerest thanks to the referee for his thoughtful suggestions in improving this work.
277-08
2
B ICHENG YANG
1, 0 < λ = 2 − ( p1 + 1q ) ≤ 1, then
(1.3)
∞ X
∞
X
∞
X
am b n
≤
K(p,
q)
apn
λ
(m
+
n)
n=1 m=1
n=1
! p1
∞
X
! 1q
bqn
,
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.
In 1997-1998, by using weight coefficients, Yang and Gao [6], [7] gave a strengthened version
of (1.2) as:
(∞ "
# ) p1 ( ∞ "
# ) 1q
∞ X
∞
X
X
X
1−γ p
1−γ q
am b n
π
π
− 1/p an
− 1/q bn
,
<
(1.4)
m+n
sin( πp )
n
sin( πp )
n
n=1 m=1
n=1
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
! 12
X
∞ X
∞
∞
∞
X
X
am b n
λ λ
(1.5)
<B
,
n1−λ a2n
n1−λ b2n
,
λ
(m
+
n)
2
2
n=1 m=1
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
! 1q
! p1 ∞
∞ X
∞
∞
X
X
X
am b n
π
nq−2 bqn
.
(1.6)
<
np−2 apn
π
m
+
n
sin(
)
p
n=1
n=1 m=1
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 > 1) and parameters α, λ >
P
P
−1 p
q (1− αλ
−1 q
p(1− αλ
r )
s )
0 (αλ ≤ min{r, s}) as: If 0 < ∞
an < ∞ and 0 < ∞
bn < ∞,
n=1 n
n=1 n
then
) p1 ( ∞
) 1q
(∞
∞ X
∞
X
X
X
αλ
αλ
am bn
(1.7)
< kαλ (r)
np(1− r )−1 apn
nq(1− s )−1 bqn
,
α + nα )λ
(m
n=1
n=1 m=1
n=1
where the constant factor kαλ (r) = α1 B( λr , λs ) is the best possible. T. K. Pogány [11] also
1
considered a best extension of (1.2) with the non-homogeneous kernel as (λm +ρ
µ (µ, λm , ρn >
n)
0).
We have a non-negative decomposition of kernel in (1.1):
1
max{m, n} min{m, n}
=
+
(m, n ∈ N)
m+n
(m + n)2
(m + n)2
(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}
(1.8)
am b n <
+1
a2n
b2n
;
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 m=1
n=1
n=1
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp.
http://jipam.vu.edu.au/
H ILBERT ’ S I NEQUALITY
3
where the sum of two best constant factors is π. The equivalent forms and extended inequalities
with a pair of conjugate exponents are considered.
2. S OME L EMMAS
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
where P1 (x) = x − [x] − 12 is the Bernoulli function of the first order; if g ∈ C 3 [1, ∞),
(−1)i g (i) (x) > 0, g (i) (∞) = 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
∞
X
max{m, n}mα
ω(α, m) :=
(2.3)
(m ∈ N),
(m + n)2 nα
n=1
then we have
(2.4)
k(α) = Aα (m) + ω(α, m);
ω
1
,m
2
1
π
<k
= + 1,
2
2
where
1
k(α) :=
α
Z
0
∞
max u1/α , 1 1 −2
u α du
(u1/α + 1)2
and Aα (m) = O(mα−1 ), (m → ∞).
Proof. For fixed
follows that
(2.5)
1
2
≤ α < 1, m ∈ N, setting f (x) :=
ω(α, m) = m
α
∞
X
max{m,x}
,
(m+x)2 xα
x ∈ (0, ∞), then by (2.1), it
f (n)
n=1
=m
=m
α
α
Z
∞
1
∞
Z
1
f (x)dx + f (1) +
2
Z
∞
0
P1 (x)f (x)dx
1
f (x)dx − mα ρ(α, m), P1 (x)f 0 (x)dx.
0
Z
(2.6)
ρ(α, m) :=
0
1
1
f (x)dx − f (1) −
2
Z
∞
P1 (x)f 0 (x)dx.
1
We find
1
−m
−1
1
− f (1) =
=
+
,
2
2
2(m + 1)
2(m + 1) 2(m + 1)2
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp.
http://jipam.vu.edu.au/
4
B ICHENG YANG
and
1
1
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
1
m
f (x)dx ≤
dx =
.
2
α
(1 − α)m
0
0 m x
Z
Z
For x ∈ (0, m), f (x) =
f (x) =
x1−α
,
(m+x)2
m
,
(m+x)2 xα
it follows f 0 (x) =
−2m
(m+x)3 xα
−
αm
;
(m+x)2 xα+1
for x ∈ (m, ∞),
we find
1−α
−2x1−α
+
(m + x)3 (m + x)2 xα
−2(x + m − m)
1−α
=
+
(m + x)3 xα
(m + x)2 xα
−2
2m
1−α
=
+
+
.
(m + x)2 xα (m + x)3 xα (m + x)2 xα
f 0 (x) =
1
1
In the following, it is obvious that g1 (x) = (m+x)
3 xα , g2 (x) = (m+x)2 xα+1 and g3 (x) =
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
2mP1 (x)dx
αmP1 (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
2−α
1
α+1
=
−
−
+
;
2+α
2
48m
12(m + 1) 12(m + 1)
6(m + 1)3
Z
(2.8)
1
(m+x)2 xα
∞
−
m
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−1
α
2−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
−1
2 − 12
1
2
+ 2 2+1/2 −
−
+
2
48m
12(m + 1) 12(m + 1)
6(m + 1)3
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp.
http://jipam.vu.edu.au/
H ILBERT ’ S I NEQUALITY
5
11
9
−1
1
+
+
+
2
2+1/2
24(m + 1) 24(m + 1)
96m
6(m + 1)3
9
−1
1
11
≥
+
+
+
> 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
m
Z
0
∞
P1 (x)f (x)dx −
1
P1 (x)f 0 (x)dx
m
1
1−α
<
+
2+α
48m
48m2+α
2−α
=
.
48m2+α
Then by (2.6), it follows
0 < m1−α [mα ρ(α, m)]
−m
m
1
2−α
<
+
+
+
2
2(m + 1) 2(m + 1)
1 − α 48m1+α
1
1
→
−
(m → ∞).
1−α 2
(2.10)
Setting u = (x/m)α , we find
Z ∞
Z ∞
max{m, x}
α
α
(2.11)
dx
m
f (x)dx = m
(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)
=2
du
=
4
k
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.
Similar to Lemma 2.2, we still have
Lemma 2.3. If
1
2
≤ α < 1, setting the weight coefficient $(α, m) as
$(α, m) :=
(2.13)
∞
X
min{m, n}mα
(m ∈ N),
(m + n)2 nα
n=1
then we have
(2.14)
e
k(α) = Bα (m) + $(α, m);
$
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 → ∞).
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp.
http://jipam.vu.edu.au/
6
B ICHENG YANG
3. M AIN R ESULTS AND THEIR E QUIVALENT F ORMS
P
P
p
q
−1 p
−1 q
2
2
Theorem 3.1. If p > 1, p1 + 1q = 1, an , bn ≥ 0, 0 < ∞
an < ∞ and 0 < ∞
bn <
n=1 n
n=1 n
∞, then we have the following equivalent inequalities
(3.1)
(3.2)
! 1q
! p1 ∞
∞ X
∞
∞
π
X
X q
X
p
max{m, n}
;
n 2 −1 bqn
I :=
am b n <
+1
n 2 −1 apn
2
(m
+
n)
2
n=1 m=1
n=1
n=1
" ∞
#p
∞
∞
π
p X
X
X max{m, n}
q
p
−1
J :=
am <
+1
n 2 −1 apn ,
n2
2
(m + n)
2
m=1
n=1
n=1
where the constant factors
π
2
+ 1 and
π
2
p
+ 1 are the best possible.
Proof. By Hölder’s inequality and (2.3) – (2.4), we find
" ∞
#p ( ∞
)p
X max{m, n}
X max{m, n} m1/(2q) n1/(2p) am =
am
(3.3)
(m + n)2
(m + n)2
n1/(2p)
m1/(2q)
m=1
m=1
" ∞
#
X max{m, n} mp/(2q)
p
≤
a
(m + n)2 n1/2 m
m=1
" ∞
#p−1
X max{m, n} nq/(2p)
×
(m + n)2 m1/2
m=1
∞
X
1
max{m, n} mp/(2q) p
p−1
1− p2
,n n
a
=ω
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
J≤
=
=
π
2
π
2
π
2
∞ X
∞
p−1 X
max{m, n} mp/(2q) p
+1
am
2
1/2
(m
+
n)
n
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
∞
∞
π
p−1 X
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
(3.4)
I=
∞
X
n=1
"
1
1
n2−p
#
∞
h −1 1 i
X
1
max{m, n}
am n 2 + p b n ≤ J p
2
(m + n)
m=1
∞
X
! 1q
q
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
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp.
http://jipam.vu.edu.au/
H ILBERT ’ S I NEQUALITY
7
P
q
−1 q
2
then it follows ∞
bn = J. By (3.3), we confirm that J < ∞. If J = 0, then (3.2) is
n=1 n
naturally valid; if 0 < J < ∞, then by (3.1), we find
! p1 ∞
! 1q
∞
∞
π
X
X q
X
q
p
+1
n 2 −1 apn
n 2 −1 bqn
;
n 2 −1 bqn = J = I <
2
n=1
n=1
n=1
! p1
! p1
∞
∞
π
X p
X q
1
n 2 −1 bqn
= Jp <
,
+1
n 2 −1 apn
2
n=1
n=1
and inequality (3.2) is valid, which is equivalent to (3.1).
−1
−1
− pε
− qε
2
2
e
eb = {ebn }∞ as e
For 0 < ε < 2q , setting e
,
b
, for n ∈ N, if there exists
a = {e
an } ∞
,
a
n
n
n=1
n=1
π
a constant 0 < k ≤ 2 + 1, such that (3.1) is still valid when we replace π2 + 1 by k, then we find
! p1 ∞
! 1q
∞ X
∞
∞
∞
X
X
X
X
e
p
q
max{m, n}e
am bn
1
−1 p
−1eq
e
2
2
I :=
<k
n e
an
n bn
=k
;
2
1+ε
(m + n)
n
n=1 m=1
n=1
n=1
n=1
#
" ∞
∞
X
X max{m, n} −1 − ε
−1
ε
m 2 p n 2 −q
Ie =
2
(m + n)
n=1 m=1
1
ε
∞
∞
X
1 X max{m, n}m 2 + q
1
1 ε
=
=
ω
+ ,m .
1
ε
1+ε
1+ε
2n 2 + q
m
m
2
q
(m
+
n)
m=1
m=1
n=1
And then by (2.4) and the above results, it follows that


X
∞
∞
X
1
1 ε
1 
1
A 1 + ε (m)
(3.5)
k
>k
+
1− 1+ε
1+ε
2
q
1
ε
n
2
q
m
k 2+q
n=1
m=1


X
∞
∞
X
1
1 ε 
1
1
=k
+
− A 1 ε (m)
1+ε
1+ε 2 + q
1
ε
2 q
m
m
k
+
m=1
2
q m=1


P
∞
X
1
∞
1
ε
(m)]
A
m=1 m1+ε 2 + q
1 ε
1 
;
P
=k
+
1
−
∞
1
ε
1
2 q m=1 m1+ε
k 2+q
m=1 m1+ε

1 ε 
P∞
1
1 2−q
O
m=1 m1+ε
m
1 ε 
.
P
k>k
+
1−
∞
1
2 q
k 1+ε
∞
X
2
q
m=1 m1+ε
+ qε , by Fatou’s Lemma, it follows that
Z
1 ε
1 ∞ max{u1/α , 1} 1 −2
lim k
+
= lim+
u α du
ε→0+
2 q
(u1/α + 1)2
α→ 12 α 0
Z ∞
1
max{u1/α , 1} 1 −2
π
≥2
lim+
u α du = k
= + 1.
1/α
2
(u + 1)
2
2
α→ 21
0
Setting α =
1
2
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.
In the same manner, by Lemma 2.3, we have:
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp.
http://jipam.vu.edu.au/
8
B ICHENG YANG
P
P
p
q
−1 p
−1 q
2
2
Theorem 3.2. If p > 1, p1 + 1q = 1, an , bn ≥ 0, 0 < ∞
an < ∞ and 0 < ∞
bn <
n=1 n
n=1 n
∞, then we have the following equivalent inequalities
! p1 ∞
! 1q
∞ X
∞
∞
π
X
X
X q
p
min{m, n}
(3.6)
am b n <
−1
n 2 −1 apn
;
n 2 −1 bqn
2
(m
+
n)
2
n=1 m=1
n=1
n=1
" ∞
#p
∞
∞
X p
X min{m, n}
pX q
π
n 2 −1
a
<
−
1
n 2 −1 apn ,
m
2
(m
+
n)
2
n=1
m=1
n=1
where the constant factors
π
2
− 1 and ( π2 − 1)p are the best possible.
Remark 1. For p = q = 2, (3.1) reduces to (1.8) and (3.6) reduces to (1.9).
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