Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 2, Article 33, 2003
SOME NEW INEQUALITIES SIMILAR TO HILBERT-PACHPATTE TYPE
INEQUALITIES
ZHONGXUE LÜ
D EPARTMENT OF BASIC S CIENCE OF T ECHNOLOGY C OLLEGE ,
X UZHOU N ORMAL U NIVERSITY, 221011,
P EOPLE ’ S R EPUBLIC OF C HINA
lvzx1@163.net
Received 25 May, 2002; accepted 7 April, 2003
Communicated by S.S. Dragomir
A BSTRACT. In this paper, some new inequalities similar to Hilbert-Pachpatte type inequalities
are given.
Key words and phrases: Inequalities, Hilbert-Pachpatte inequalities, Hölder inequality.
1991 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In [1, Chap. 9], the well-known Hardy-Hilbert inequality is given as follows.
P
P∞ q
p
Theorem 1.1. Let p > 1, q > 1, p1 + 1q = 1, am , bn ≥ 0, 0 < ∞
n=1 an < ∞, 0 <
n=1 bn <
∞. Then
! p1 ∞ ! 1q
∞ X
∞
∞
X
X
X
am b n
π
p
(1.1)
≤
a
bqn
m
λ
(m
+
n)
sin(π/p)
m=1 n=1
m=1
n=1
where
π
sin(π/p)
is best possible.
The integral analogue of the Hardy-Hilbert inequality can be stated as follows
R∞
Theorem 1.2. Let p > 1, q > 1, p1 + 1q = 1, f (x), g(y) ≥ 0, 0 < 0 f p (x)dx < ∞, 0 <
R∞ q
g (y)dy < ∞. Then
0
1q
Z ∞
p1 Z ∞
Z ∞Z ∞
f (x)g(y)
π
q
p
g (y)dy ,
(1.2)
dxdy ≤
f (x)dx
x+y
sin(π/p)
0
0
0
0
π
is best possible.
where sin(π/p)
In [1, Chap. 9] the following extension of Hardy-Hilbert’s double-series theorem is given.
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
059-02
2
Z HONGXUE L Ü
1
p
1
q
≥ 1, 0 < λ = 2 −
! p1
∞ X
∞
∞
X
X
am b n
≤K
apm
λ
(m + n)
m=1 n=1
m=1
Theorem 1.3. Let p > 1, q > 1,
+
1
p
−
1
q
∞
X
=
1
p
+
1
q
≤ 1. Then
! 1q
bqn
,
n=1
where K = K(p, q) depends on p and q only.
The following integral analogue of Theorem 1.3 is also given in [1, Chap. 9].
Theorem 1.4. Under the same conditions as in Theorem 1.1 we have
Z ∞
p1 Z ∞
1q
Z ∞Z ∞
f (x)g(y)
p
q
dxdy ≤ K
f dx
g dy ,
(x + y)λ
0
0
0
0
where K = K(p, q) depends on p and q only.
The inequalities in Theorems 1.1 and 1.2 were studied by Yang and Kuang (see [2, 3]). In
[4, 5], some new inequalities similar to the inequalities given in Theorems 1.1, 1.2, 1.3 and 1.4
were established.
In this paper, we establish some new inequalities similar to the Hilbert-Pachpatte inequality.
2. M AIN R ESULTS
In what follows we denote by R the set of real numbers. Let N = {1, 2, . . . },N0 = {0, 1, 2, . . . }.
We define the operator ∇ by ∇u(t) = u(t) − u(t − 1) for any function u defined on N . For any
function u(t) : [0, ∞) → R, we denote by u0 the derivatives of u.
First we introduce some Lemmas.
Lemma 2.1. (see [2]). Let p > 1, q > 1, p1 + 1q = 1, λ > 2 − min{p, q}, define the weight
function ω1 (q, x) as
2−λ
Z ∞
x q
1
dy, x ∈ [0, ∞).
ω1 (q, x) :=
(x + y)λ y
0
Then
(2.1)
ω1 (q, x) = B
q+λ−2 p+λ−2
,
q
p
x1−λ ,
where B(p, q) is β-function.
Lemma 2.2. (see [3]). Let p > 1, q > 1, p1 + 1q = 1, λ > 2 − min{p, q}, define the weight
function ω2 (q, x) as
2−λ
Z ∞
1
x q
ω2 (q, x) :=
dy, x ∈ [0, ∞).
xλ + y λ y
0
Then
(2.2)
1
ω2 (q, x) = B
λ
q+λ−2 p+λ−2
,
qλ
pλ
x1−λ .
Lemma 2.3. Let p > 1, q > 1, p1 + 1q = 1, λ > 2 − min{p, q}, define the weight function
ω3 (q, m) as
∞
m 2−λ
X
1
q
ω3 (q, m) :=
, m ∈ {1, 2, . . . }.
λ
(m + n)
n
n=1
J. Inequal. Pure and Appl. Math., 4(2) Art. 33, 2003
http://jipam.vu.edu.au/
S OME N EW I NEQUALITIES S IMILAR TO H ILBERT-PACHPATTE T YPE I NEQUALITIES
3
Then
ω3 (q, m) < B
(2.3)
q+λ−2 p+λ−2
,
q
p
m1−λ ,
where B(p, q) is β-function.
Proof. By Lemma 2.1,we have
2−λ
Z ∞
1
m q
q+λ−2 p+λ−2
ω3 (q, m) <
dy = B
,
m1−λ .
λ
q
p
(m + y)
y
0
The proof is completed.
Lemma 2.4. Let p > 1, q > 1, p1 + 1q = 1, λ > 2 − min{p, q}, define the weight function
ω4 (q, m) as
∞
m 2−λ
X
1
q
ω4 (q, m) :=
, m ∈ {1, 2, . . . }.
λ
λ
m
+
n
n
n=1
Then
1
ω4 (q, m) < B
λ
(2.4)
q+λ−2 p+λ−2
,
qλ
pλ
m1−λ .
Proof. By Lemma 2.2, we have
2−λ
Z ∞
m q
1
q+λ−2 p+λ−2
1
dy = B
,
m1−λ .
ω4 (q, m) <
λ + yλ
m
y
λ
qλ
pλ
0
The proof is completed.
Our main result is given in the following theorem.
Theorem 2.5. Let p > 1, p1 + 1q = 1, and f (x), g(y) be real-valued continuous functions
defined on [0, ∞), respectively, and let f (0) = g(0) = 0, and
Z ∞Z x
Z ∞Z y
p
q
0
0<
|f (τ )| dτ dx < ∞, 0 <
|g 0 (δ)| dδdy < ∞.
0
0
0
0
Then
Z
∞
Z
(2.5)
0
0
∞
|f (x)| |g(y)|
dxdy
(qxp−1 + py q−1 )(x+y)
Z ∞ Z x
p1 Z ∞ Z y
1q
π
p
q
0
0
≤
|f (τ )| dτ dx
|g (δ)| dδdy .
sin(π/p)pq
0
0
0
0
In particular, when p = q = 2, we have
Z ∞Z ∞
|f (x)| |g(y)|
(2.6)
dxdy
(x + y)2
0
0
Z ∞ Z x
12 Z ∞ Z y
12
π
2
2
0
0
≤
|f (τ )| dτ dx
|g (δ)| dδdy .
2
0
0
0
0
Proof. From the hypotheses, we have the following identities
Z x
(2.7)
f (x) =
f 0 (τ )dτ,
0
J. Inequal. Pure and Appl. Math., 4(2) Art. 33, 2003
http://jipam.vu.edu.au/
4
Z HONGXUE L Ü
and
y
Z
g 0 (δ)dδ
g(y) =
(2.8)
0
for x, y ∈ (0, ∞). From (2.7) and (2.8) and using Hölder’s integral inequality, respectively, we
have
Z x
p1
1
p
0
q
(2.9)
|f (x)| ≤ x
|f (τ )| dτ
0
and
|g(y)| ≤ y
(2.10)
1
p
y
Z
1q
q
0
|g (δ)| dδ
0
for x, y ∈ (0, ∞). From (2.9) and (2.10) and using the elementary inequality
z1 z2 ≤
(2.11)
z1p z2q
1 1
+ , z1 ≥ 0, z2 ≥ 0, + = 1, p > 1,
p
q
p q
we observe that
1
q
|f (x)| |g(y)| ≤ x y
1
p
Z
x
0
p1 Z
p
(2.12)
≤
0
|f (τ )| dτ
xp−1 y q−1
+
p
q
1q
q
|g (δ)| dδ
0
y
0
Z
x
0
p
p1 Z
|f (τ )| dτ
0
y
0
q
1q
|g (δ)| dδ
0
for x, y ∈ (0, ∞). From (2.12) we observe that
Z x
p1 Z y
1q
1
|f (x)| |g(y)|
p
q
0
0
(2.13)
≤
|f (τ )| dτ
|g (δ)| dδ .
qxp−1 + py q−1
pq
0
0
Hence
Z ∞Z ∞
|f (x)| |g(y)|
(2.14)
dxdy
(qxp−1 + py q−1 )(x + y)
0
0
1q
p1 R y 0
Z ∞Z ∞ Rx 0
p
q
|g
(δ)|
dδ
|f
(τ
)|
dτ
1
0
0
dxdy.
≤
x
+
y
pq 0
0
By Hölder’s integral inequality and (2.1), we have
1 R y
1
Z ∞Z ∞ Rx 0
|f (τ )|p dτ p 0 |g 0 (δ)|q dδ q
0
dxdy
x+y
0
0
p1 1 R y 0
1q 1
Z ∞Z ∞ Rx 0
p
q
pq
|f
(τ
)|
dτ
|g
(δ)|
dδ
x
y pq
0
0
=
dxdy
1
1
y
x
0
0
(x + y) p
(x + y) q
! p1
1q
Z ∞Z ∞ Rx 0
p
|f
(τ
)|
dτ
x
0
≤
dxdy
x+y
y
0
0
Z ∞ Z ∞ R y 0
1q
|g (δ)|q dδ y p1
0
×
dxdy
x+y
x
0
0
Z ∞ Z x
p1 Z ∞ Z y
1q
π
p
q
0
0
(2.15)
≤
|f (τ )| dτ dx
|g (δ)| dδdy
sin(π/p)
0
0
0
0
J. Inequal. Pure and Appl. Math., 4(2) Art. 33, 2003
http://jipam.vu.edu.au/
S OME N EW I NEQUALITIES S IMILAR TO H ILBERT-PACHPATTE T YPE I NEQUALITIES
5
by (2.14) and (2.15), we get (2.5). The proof of Theorem 2.5 is complete.
In a similar way to the proof of Theorem 2.5, we can prove the following theorems.
Theorem 2.6. Let p > 1, p1 + 1q = 1, λ > 2 − min{p, q}, and f (x), g(y) be real-valued
continuous functions defined on [0, ∞),respectively, and let f (0) = g(0) = 0, and
Z ∞Z x
Z ∞Z y
p
q
1−λ
0
0<
x
|f (τ )| dτ dx < ∞, 0 <
y 1−λ |g 0 (δ)| dδdy < ∞,
0
0
0
0
then
Z
∞
Z
∞
(2.16)
0
|f (x)| |g(y)|
dxdy
+ py q−1 )(x + y)λ
p+λ−2 Z
∞Z
B q+λ−2
,
q
p
(qxp−1
0
≤
pq
0
x
x
1−λ
p1
|f (τ )| dτ dx
p
0
0
Z
∞
Z
y
×
y
0
1−λ
0
q
1q
|g (δ)| dδdy
.
0
In particular, when p = q = 2,
Z ∞Z ∞
|f (x)| |g(y)|
(2.17)
dxdy
(x + y)1+λ
0
0
Z ∞ Z x
12 Z ∞ Z y
21
B λ2 , λ2
2
2
≤
x1−λ |f 0 (τ )| dτ dx
y 1−λ |g 0 (δ)| dδdy .
2
0
0
0
0
Theorem 2.7. Let p > 1, p1 + 1q = 1, λ > 2 − min{p, q}, and f (x), g(y) be real-valued
continuous functions defined on [0, ∞), respectively, and let f (0) = g(0) = 0, and
Z ∞Z x
Z ∞Z y
p
q
1−λ
0
0<
x
|f (τ )| dτ dx < ∞, 0 <
y 1−λ |g 0 (δ)| dδdy < ∞.
0
0
0
0
Then
Z
∞
Z
(2.18)
0
0
∞
|f (x)| |g(y)|
dxdy
+ py q−1 )(xλ + y λ )
p+λ−2 Z
∞Z
,
B q+λ−2
qλ
pλ
(qxp−1
≤
λpq
0
x
x
1−λ
p1
|f (τ )| dτ dx
p
0
0
Z
∞
Z
×
y
y
0
1−λ
0
q
|g (δ)| dδdy
1q
.
0
In particular, when p = q = 2,
Z ∞Z ∞
|f (x)| |g(y)|
(2.19)
dxdy
λ
(x + y λ )(x + y)
0
0
Z ∞ Z x
12 Z ∞ Z y
21
π
2
2
1−λ
0
1−λ 0
≤
x
|f (τ )| dτ dx
y
|g (δ)| dδdy .
2λ
0
0
0
0
J. Inequal. Pure and Appl. Math., 4(2) Art. 33, 2003
http://jipam.vu.edu.au/
6
Z HONGXUE L Ü
3. D ISCRETE A NALOGUES
Theorem 3.1. Let p > 1, p1 + 1q = 1, and {a(m)} and {b(n)} be two sequences of real
P∞ Pm
p
numbers
where
m,
n
∈
N
,
and
a(0)
=
b(0)
=
0,
and
0
<
0
τ =1 |∇a(τ )| < ∞,
m=1
P∞ Pn
q
0 < n=1 δ=1 |∇b(δ)| < ∞, then
! p1 ∞ n ! 1q
∞ X
∞
∞ X
m
XX
X
X
|am | |bn |
π
p
a
bqr
.
(3.1)
≤
k
p−1 + pnq−1 )(m + n)
(qm
sin(π/p)pq
m=1 n=1
m=1 k=1
n=1 r=1
In particular, when p = q = 2, we have
(3.2)
∞ X
∞
X
|am | |bn |
π
≤
2
(m + n)
2
m=1 n=1
∞ X
m
X
! 12
∞ X
n
X
a2k
m=1 k=1
! 12
b2r
.
n=1 r=1
Proof. From the hypotheses, it is easy to observe that the following identities hold
m
X
(3.3)
am =
∇a(τ ),
τ =1
and
bn =
(3.4)
n
X
∇b(δ)
δ=1
for m, n ∈ N. From (3.3) and (3.4) and using Hölder’s inequality, we have
! p1
m
X
1
,
(3.5)
|am | ≤ m q
|∇a(τ )|p
τ =1
and
|bn | ≤ n
(3.6)
1
p
n
X
! 1q
|∇b(δ)|q
δ=1
for m, n ∈ N. From (3.5) and (3.6) and using the elementary inequality (2.11), we observe that
! p1
! 1q
m
n
X
X
1
1
|am | |bn | ≤ m q n p
|∇a(τ )|p
|∇b(δ)|q
τ =1
≤
(3.7)
mp−1
+
p
δ=1
X
m
nq−1
q
! p1
|∇a(τ )|p
τ =1
n
X
! 1q
|∇b(δ)|q
δ=1
for m, n ∈ N. From (3.7), we observe that
(3.8)
|am | |bn |
1
≤
p−1
q−1
qm
+ pn
pq
m
X
! p1
|∇a(τ )|p
τ =1
n
X
! 1q
|∇b(δ)|q
.
δ=1
Hence
(3.9)
∞ X
∞
X
m=1 n=1
(qmp−1
|am | |bn |
+ pnq−1 )(m + n)
P
∞ ∞
p p1 Pn
q 1q
1 XX( m
τ =1 |∇a(τ )| ) (
δ=1 |∇b(δ)| )
≤
.
pq m=1 n=1
m+n
J. Inequal. Pure and Appl. Math., 4(2) Art. 33, 2003
http://jipam.vu.edu.au/
S OME N EW I NEQUALITIES S IMILAR TO H ILBERT-PACHPATTE T YPE I NEQUALITIES
7
By the Hölder inequality and (2.3)
P
∞ X
∞
p p1 Pn
q 1q
X
( m
τ =1 |∇a(τ )| ) (
δ=1 |∇b(δ)| )
m+n
m=1 n=1
1
P
P
∞ X
∞
p p1 X
( m
m pq1 ( nδ=1 |∇b(δ)|q ) q n pq1
τ =1 |∇a(τ )| )
=
1
1
m
n
(m + n) q
(m + n) p
m=1 n=1
!1
!1
∞ X
∞ Pn
∞ X
∞ Pm
q p p1 q
1q p X
X
|∇b(δ)|
|∇a(τ
)|
m
n
δ=1
τ =1
≤
m+n
m
m
+
n
n
m=1 n=1
m=1 n=1
! 1q
! p1 ∞ n
∞ X
m
X
X
X
π
(3.10)
<
|∇b(δ)|q
|∇a(τ )|p
sin(π/p) m=1 τ =1
n=1 δ=1
by (3.9) and (3.10), we get (3.1). The proof of Theorem 3.1 is complete.
In a similar manner to the proof of Theorem 3.1, we can prove the following theorems.
Theorem 3.2. Let p > 1, p1 + 1q = 1, λ > 2 − min{p, q}, and {a(m)} and {b(n)} be two
sequences of real numbers where m, n ∈ N0 ,and a(0) = b(0) = 0, and
0<
0<
∞ X
m
X
m=1 τ =1
n
∞ X
X
m1−λ |∇a(τ )|p < ∞,
n1−λ |∇b(δ)|q < ∞,
n=1 δ=1
then
(3.11)
∞ X
∞
X
|am | |bn |
+ pnq−1 )(m + n)λ
m=1 n=1
! p1
p+λ−2
∞ X
m
B q+λ−2
,
X
q
p
≤
m1−λ |∇a(τ )|p
pq
m=1 τ =1
(qmp−1
∞ X
n
X
! 1q
n1−λ |∇b(δ)|q
.
n=1 δ=1
In particular, when p = q = 2, we have
(3.12)
∞ X
∞
X
|am | |bn |
(m + n)1+λ
m=1 n=1
≤
B
λ λ
,
2 2
∞ X
m
X
2
! 12
m1−λ |∇a(τ )|2
m=1 τ =1
∞ X
n
X
! 21
n1−λ |∇b(δ)|2
.
n=1 δ=1
Theorem 3.3. Let p > 1, p1 + 1q = 1, λ > 2 − min{p, q}, and {a(m)} and {b(n)} be two
sequences of real numbers where m, n ∈ N0 , and a(0) = b(0) = 0, and
0<
0<
∞ X
m
X
m=1 τ =1
∞ X
n
X
m1−λ |∇a(τ )|p < ∞,
n1−λ |∇b(δ)|q < ∞,
n=1 δ=1
J. Inequal. Pure and Appl. Math., 4(2) Art. 33, 2003
http://jipam.vu.edu.au/
8
Z HONGXUE L Ü
then
(3.13)
∞ X
∞
X
|am | |bn |
(qmp−1 + pnq−1 )(mλ + nλ )
m=1 n=1
! p1
q+λ−2 p+λ−2
∞ X
m
B
, pλ
X
qλ
p
≤
m1−λ |∇a(τ )|
λpq
m=1 τ =1
∞ X
n
X
! 1q
q
n1−λ |∇b(δ)|
.
n=1 δ=1
In particular, when p = q = 2, we have
(3.14)
∞
∞ X
X
m=1 n=1
|am | |bn |
(m + n)(mλ + nλ )
π
≤
2λ
∞ X
m
X
! 12
m1−λ |∇a(τ )|2
m=1 τ =1
∞ X
n
X
! 21
n1−λ |∇b(δ)|2
.
n=1 δ=1
R EFERENCES
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, Cambridge Univ. Press, London,
1952.
[2] BICHENG YANG, A generalized Hilbert’s integral inequality with the best constant, Ann. of
Math.(Chinese), 21A:4 (2000), 401–408.
[3] JICHANG KUANG,On new extensions of Hilbert’s integral inequality, J. Math. Anal. Appl., 235
(1999), 608–614.
[4] B.G. PACHPATTE, Inequalities similar to certain extensions of Hilbert’s inequality, J.Math. Anal.
Appl., 243 (2000), 217–227.
[5] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequality, J. Math. Anal. Appl.,
226 (1998), 166–179.
[6] D.S. MITRINOVIĆ, Analytic Inequalities, Springer-Verlag, Berlin/New York,1970.
[7] E.F. BECKENBACH AND R. BELLMAN, Inequalities, Springer-Verlag, Berlin, 1983.
J. Inequal. Pure and Appl. Math., 4(2) Art. 33, 2003
http://jipam.vu.edu.au/