SOME NEW HILBERT-PACHPATTE’S
INEQUALITIES
Hilbert-Pachpatte’s Inequalities
WENGUI YANG
School of Mathematics and Computational Science
Xiangtan University, Xiangtan, 411105
Hunan, P.R. China
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
EMail: yangwg8088@163.com
Contents
Received:
05 September, 2008
Accepted:
2 January, 2009
Communicated by:
W.S. Cheung
2000 AMS Sub. Class.:
26D15.
Key words:
Hilbert-Pachpatte’s inequality; Hölder’s inequality; Jensen’s inequality; Nonnegative sequences.
Abstract:
Some new Hilbert-Pachpatte discrete inequalities and their integral analogues are
established in this paper. Other inequalities are also given in remarks.
JJ
II
J
I
Page 1 of 28
Go Back
Full Screen
Close
Contents
1
Introduction
3
2
Main Results
6
3
Integral Analogues
17
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 2 of 28
Go Back
Full Screen
Close
1.
Introduction
Let p ≥ 1, q ≥ 1 and {am } and {bn } be two nonnegative sequences of real numbers
defined for m = 1,
1, 2, . . . , r, where k and r are natural numbers
P2, . . . , k and n = P
n
a
and
B
=
and define Am = m
n
s=1 s
t=1 bt . Then
( k
) 12
k X
r
X
X
Apm Bnq
2
(1.1)
≤ C(p, q, k, r)
(k − m + 1)(Ap−1
m am )
m
+
n
m=1 n=1
m=1
( r
) 12
X
×
,
(r − n + 1)(Bnq−1 bn )2
n=1
1
pq
2
√
unless {am } or {bn } is null, where C(p, q, k, r) =
kr.
An integral analogue of (1.1) is given in the following result.
Let p ≥ 1, q ≥ 1 and f (σ) ≥ 0, g(τ ) ≥ 0Rfor σ ∈ (0, x), τ ∈ (0, y),
R t where x, y
s
are positive real numbers and define F (s) = 0 f (σ)dσ and G(t) = 0 g(τ )dτ, for
s ∈ (0, x), t ∈ (0, y). Then
Z x
21
Z xZ y p
F (s)Gq (t)dsdt
p−1
2
(1.2)
≤ D(p, q, x, y)
(x − s)(F (s)f (s)) ds
s+t
0
0
0
Z y
12
q−1
2
×
(y − t)(G (t)g(t)) dt ,
0
√
1
pq xy.
2
unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y) =
Inequalities (1.1) and (1.2) are the well known Hilbert-Pachpatte inequalities [1],
which gave new estimates on Hilbert type inequalities [2]. It is well known that the
Hilbert-Pachpatte inequalities play a dominant role in analysis, so the literature on
such inequalities and their applications is vast [3] – [8].
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 3 of 28
Go Back
Full Screen
Close
Young-Ho Kim [9] gave new inequalities similar to the Hilbert-Pachpatte inequalities as follows.
Let p ≥ 1, q ≥ 1, α > 0, and {am } and {bn } be two nonnegative sequences of
real numbers defined for m = 1, 2,
P.m. . , k and n = 1,P2,n . . . , r, where k and r are
natural numbers and define Am = s=1 as and Bn = t=1 bt . Then
(1.3)
k X
r
X
m=1 n=1
(
Apm Bnq
(mα
+
1
nα ) α
≤ C(p, q, k, r; α)
k
X
) 12
2
(k − m + 1)(Ap−1
m am )
Hilbert-Pachpatte’s Inequalities
Wengui Yang
m=1
×
( r
X
vol. 10, iss. 1, art. 26, 2009
) 12
(r − n + 1)(Bnq−1 bn )2
,
n=1
Title Page
1 √
unless {am } or {bn } is null, where C(p, q, k, r; α) = 12 α pq kr.
An integral analogue of (1.3) is given in the following result.
Let p ≥ 1, q ≥ 1, α > 0 and f (σ) ≥ 0, g(τ ) ≥ R0 for σ ∈ (0, x), τ ∈ (0,Ry), where
s
t
x, y are positive real numbers and define F (s) = 0 f (σ)dσ and G(t) = 0 g(τ )dτ,
for s ∈ (0, x), t ∈ (0, y). Then
Z xZ y p
F (s)Gq (t)dsdt
(1.4)
1
(sα + tα ) α
0
0
Z x
12
p−1
2
≤ D(p, q, x, y; α)
(x − s)(F (s)f (s)) ds
0
Z
×
y
(y − t)(G
q−1
12
(t)g(t)) dt ,
0
unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y; α) =
1
2
α1
√
pq xy.
2
Contents
JJ
II
J
I
Page 4 of 28
Go Back
Full Screen
Close
The purpose of the present paper is to derive some new generalized inequalities
(1.1) and (1.2) that are similar to (1.3) and (1.4). By applying an elementary inequality, we also obtain some new inequalities similar to some results in [1, 9].
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 5 of 28
Go Back
Full Screen
Close
2.
Main Results
Now we give our results as follows in this paper.
Theorem 2.1. Let p ≥ 1, q ≥ 1, α > 1, γ > 1 and {am } and {bn } be two nonnegative sequences of real numbers defined for m = 1, P
2, . . . , k and n = 1, 2,
P.n. . , r,
m
where k and r are natural numbers and define Am = s=1 as and Bn = t=1 bt .
Then
k X
r
X
Apm Bnq
(2.1)
≤ C(p, q, k, r; α, γ)
(γ−1)(α+γ)
(α−1)(α+γ)
αγ
αγ
+ αn
m=1 n=1 γm
( k
) α1 ( r
) γ1
X
X
α
,
×
(k − m + 1)(Ap−1
(r − n + 1)(Bnq−1 bn )γ
m am )
m=1
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
n=1
unless {am } or {bn } is null, where C(p, q, k, r; α, γ) =
Contents
α−1
pq
k α r
α+γ
γ−1
γ
.
Proof. The idea for the proof Theorem 2.1 comes from Theorem 1 of [1] and Theorem 2.1 of [9]. From the hypotheses of Theorem 2.1 and using the following inequality (see [10, 11]),
( n
)β
( m )β−1
n
X
X
X
(2.2)
zm
≤β
zm
zk
,
m=1
m=1
k=1
where β ≥ 1 is a constant and zm ≥ 0, (m = 1, 2, . . . , n), it is easy to observe that
m
X
p
(2.3)
Am ≤ p
Ap−1
m = 1, 2, . . . , k,
s as ,
Bnq ≤ q
s=1
n
X
t=1
Btq−1 bt ,
n = 1, 2, . . . , r.
JJ
II
J
I
Page 6 of 28
Go Back
Full Screen
Close
From (2.3) and Hölder’s inequality, we have
( m
) α1
m
X
X
α−1
α
α
(2.4)
Ap−1
,
(Ap−1
s as ≤ m
s as )
s=1
m = 1, 2, . . . , k,
s=1
and
(2.5)
n
X
Btq−1 bt
≤n
γ−1
γ
( n
X
t=1
) γ1
Hilbert-Pachpatte’s Inequalities
(Btq−1 bt )γ
,
n = 1, 2, . . . , r.
t=1
Using the inequality of means [12]
) r1
( n
(
) Ω1
n
n
X
Y
1
≤
ωi sri
(2.6)
sωi i
Ωn i=1
i=1
P
for r > 0, ωi > 0, ni=1 ωi = Ωn , we observe that
(sω1 1 sω2 2 )r/(ω1 +ω2 ) ≤
(2.7)
α−1
Let s1 = m , s2 = n
and (2.7), we have
(2.8)
Apm Bnq ≤ pqm
γ−1
α−1
α
, ω1 =
n
γ−1
γ
1
,
α
( m
X
≤
pqαγ
α+γ
m
ω2 =
1
γ
Contents
) α1 ( n
X
t=1
+
n
(γ−1)(α+γ)
αγ
γ
)
II
J
I
Go Back
) γ1
(Btq−1 bt )γ
JJ
Page 7 of 28
and r = ω1 + ω2 , from (2.3) – (2.5)
α
(Ap−1
s as )
(α−1)(α+γ)
αγ
α
Title Page
1
(ω1 sr1 + ω2 sr2 ) .
ω1 + ω2
s=1
(
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Full Screen
Close
×
( m
X
α
(Ap−1
s as )
) α1 ( n
X
s=1
) γ1
(Btq−1 bt )γ
,
t=1
for m = 1, 2, . . . , k, n = 1, 2, . . . , r. From (2.8), we observe that
Apm Bnq
(2.9)
γm
(α−1)(α+γ)
αγ
+ αn
(γ−1)(α+γ)
αγ
pq
≤
α+γ
Hilbert-Pachpatte’s Inequalities
( m
X
α
(Ap−1
s as )
s=1
) α1 ( n
X
Wengui Yang
) γ1
(Btq−1 bt )γ
,
t=1
for m = 1, 2, . . . , k, n = 1, 2, . . . , r. Taking the sum on both sides of (2.9) first
over n from 1 to r and then over m from 1 to k of the resulting inequality and using
Hölder’s inequality with indices α, α/(α − 1) and γ, γ/(γ − 1) and interchanging
the order of summations, we observe that
k X
r
X
Apm Bnq
(α−1)(α+γ)
vol. 10, iss. 1, art. 26, 2009
(γ−1)(α+γ)
γm αγ
+ αn αγ

( m
) α1   r ( n
) γ1 
k
X
X
X
X
pq 
α

(Ap−1
≤
(Btq−1 bt )γ 
s as )
α + γ m=1 s=1
n=1
t=1
Title Page
Contents
JJ
II
J
I
Page 8 of 28
m=1 n=1
≤
pq α−1
k α
α+γ
(
k X
m
X
m=1 s=1
) α1
α
(Ap−1
s as )
r
γ−1
γ
( r n
XX
n=1 t=1
) γ1
(Btq−1 bt )γ
Go Back
Full Screen
Close
=
pq
k
α+γ
α−1
α
×
r
γ−1
γ
(
k
X
) α1
α
(k − m + 1)(Ap−1
m am )
m=1
( r
X
) γ1
(r − n + 1)(Bnq−1 bn )γ
.
n=1
Hilbert-Pachpatte’s Inequalities
Remark 1. In Theorem 2.1, setting α = γ = 2, we have (1.1). In Theorem 2.1,
setting α1 + γ1 = 1, we have
k X
r
X
m=1 n=1
Apm Bnq
≤ C(p, q, k, r; α, γ)
γmα−1 + αnγ−1
( k
) α1 ( r
) γ1
X
X
α
,
×
(k − m + 1)(Ap−1
(r − n + 1)(Bnq−1 bn )γ
m am )
m=1
n=1
unless {am } or {bn } is null, where C(p, q, k, r; α, γ) =
Remark 2. In Theorem 2.1, setting p = q = 1, we have
(2.10)
k X
r
X
α−1
pq
k α r
α+γ
γ−1
γ
.
(α−1)(α+γ)
αγ
Contents
JJ
II
J
I
Page 9 of 28
Full Screen
Am Bn
γm
Title Page
Go Back
(γ−1)(α+γ)
+ αn αγ
( k
) α1 ( r
) γ1
X
X
≤ C(1, 1, k, r; α, γ)
(k − m + 1)aαm
(r − n + 1)bγn
,
m=1 n=1
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
m=1
unless {am } or {bn } is null, where C(1, 1, k, r; α, γ) =
n=1
α−1
1
k α r
α+γ
γ−1
γ
.
Close
In the following theorem we give a further generalization of the inequality (2.10)
obtained in Remark 2. Before we give our result, we point out that {pm } and {qn }
should be two positive sequences for m = 1, 2, . . . , k and n = 1, 2, . . . , r in Theorem
2.3 of [9].
Theorem 2.2. Let α > 1, γ > 1 and {am } and {bn } be two nonnegative sequences of
real numbers and {pm } and {qn } be positive sequences defined for m = 1,P
2, . . . , k
and n P
= 1, 2, . . . , r, where
k and r are natural
numbers and define Am = m
s=1 as ,
P
P
n
Bn = nt=1 bt , Pm = m
p
and
Q
=
q
.
Let
Φ
and
Ψ
be
two
real-valued,
s
n
t
s=1
t=1
nonnegative, convex, and submultiplicative functions defined on R+ = [0, ∞). Then
(2.11)
k X
r
X
Φ(Am )Ψ(Bn )
γm
Title Page
where
M (k, r; α, γ) =
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
(γ−1)(α+γ)
αγ
(α−1)(α+γ)
αγ
+ αn
( k
α ) α1
X
am
≤ M (k, r; α, γ)
(k − m + 1) pm Φ
pm
m=1
( r
γ ) γ1
X
bn
×
(r − n + 1) qn Ψ
,
q
n
n=1
m=1 n=1
Hilbert-Pachpatte’s Inequalities
Contents
JJ
II
J
I
Page 10 of 28
Go Back
Full Screen
1
α+γ
(
γ−1
( r γ )
α ) α−1
α
k X Ψ(Qn ) γ−1 γ
X
Φ(Pm ) α−1
m=1
Pm
n=1
Qn
Close
.
Proof. From the hypotheses of Φ and Ψ and by using Jensen’s inequality and Hölder’s
inequality, it is easy to observe that
P
Pm m
ps as /ps
s=1
Pm
(2.12)
Φ(Am ) = Φ
s=1 ps
Pm
m
ps as /ps
Φ(Pm ) X
as
s=1
Pm
≤ Φ(Pm )Φ
≤
ps Φ
Pm s=1
ps
s=1 ps
( m α ) α1
as
Φ(Pm ) α−1 X
m α
ps Φ
,
≤
Pm
p
s
s=1
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
and similarly,
Title Page
(2.13)
Ψ(Bn ) ≤
Ψ(Qn ) γ−1
n γ
Qn
( n γ ) γ1
X
bt
.
qt Ψ
q
t
t=1
Let s1 = mα−1 , s2 = nγ−1 , ω1 = α1 , ω2 = γ1 and r = ω1 + ω2 , from (2.7), (2.12)
and (2.13), we have


( m α ) α1
X
γ−1
α−1
Φ(Pm )
as

Φ(Am )Ψ(Bn ) ≤ m α n γ 
ps Φ
(2.14)
Pm
p
s
s=1


( n γ ) γ1
X
Ψ(Qn )
bt

×
qt Ψ
Qn
q
t
t=1
( (α−1)(α+γ)
(γ−1)(α+γ) )
αγ
m αγ
n αγ
+
≤
α+γ
α
γ
Contents
JJ
II
J
I
Page 11 of 28
Go Back
Full Screen
Close
) α1  
( m ( n ) γ1 
α
γ
X
Φ(Pm ) X
as
bt
  Ψ(Qn )

×
ps Φ
qt Ψ
Pm
p
Q
q
s
n
t
s=1
t=1

for m = 1, 2, . . . , k, n = 1, 2, . . . , r. From (2.14), we observe that
Φ(Am )Ψ(Bn )
(2.15)
γm
(α−1)(α+γ)
αγ
+ αn
(γ−1)(α+γ)
αγ
Hilbert-Pachpatte’s Inequalities

( m α ) α1
X
as
1  Φ(Pm )

≤
ps Φ
α+γ
Pm
p
s
s=1


( n γ ) γ1
X
Ψ(Qn )
bt

×
qt Ψ
Qn
q
t
t=1

for m = 1, 2, . . . , k, n = 1, 2, . . . , r. Taking the sum on both sides of (2.15) first
over n from 1 to r and then over m from 1 to k of the resulting inequality and using
Hölder’s inequality with indices α, α/(α − 1) and γ, γ/(γ − 1) and interchanging
the order of summations, we observe that
k X
r
X
(γ−1)(α+γ)
γm αγ
+ αn αγ



( m ( n α ) α1
γ ) γ1
k
r
X
X
X
X
1 
Φ(Pm )
as
Ψ(Qn )
bt


≤
ps Φ
qt Ψ
α + γ m=1 Pm
p
Q
q
s
n
t
s=1
n=1
t=1
m=1 n=1
≤
1
α+γ
(
α
k X
Φ(Pm ) α−1
m=1
Pm
Title Page
Contents
JJ
II
J
I
Page 12 of 28
Go Back
Φ(Am )Ψ(Bn )
(α−1)(α+γ)
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
) α−1
(
α
α
k X
m X
as
ps Φ
ps
m=1 s=1
) α1
Full Screen
Close
γ−1 (
( r γ )
γ ) γ1
r X
n X Ψ(Qn ) γ−1 γ
X
bt
×
qt Ψ
Qn
qt
n=1
n=1 t=1
α−1
γ−1
( k ) α ( r γ )
α
α−1
γ−1
γ
X
X
1
Φ(Pm )
Ψ(Qn )
=
α + γ m=1
Pm
Qn
n=1
1
( k
γ ) γ1
α ) α (X
r
X
bt
as
×
(r − n + 1) qt Ψ
.
(k − m + 1) ps Φ
ps
qt
n=1
m=1
Remark 3. From the inequality (2.7), we obtain
1
(2.16)
sω1 1 sω2 2 ≤
ω1 sω1 1 +ω2 + ω2 sω2 1 +ω2
ω1 + ω2
for ω1 > 0, ω2 > 0. If we apply the elementary inequality (2.16) on the right-hand
sides of (2.1) in Theorem 2.1 and (2.11) in Theorem 2.2, then we get the following
inequalities
k
X
r
X
m=1 n=1
Apm Bnq
γm
(α−1)(α+γ)
αγ
+ αn
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 13 of 28
Go Back
(γ−1)(α+γ)
αγ
Full Screen
 (
) α+γ
αγ
k
X
αγC(p, q, k, r; α, γ)  1
p−1
α
≤
(k − m + 1)(Am am )
α+γ
α m=1

( r
) α+γ
αγ
X
1
,
+
(r − n + 1)(Bnq−1 bn )γ
γ n=1
Close
where C(p, q, k, r; α, γ) =
k X
r
X
α−1
pq
k α r
α+γ
γ−1
γ
. Also,
Φ(Am )Ψ(Bn )
(γ−1)(α+γ)
+ αn αγ
 (
α ) α+γ
αγ
k
X
am
αγM (k, r; α, γ)  1
≤
(k − m + 1) pm Φ
α+γ
α m=1
pm
m=1 n=1
γm
(α−1)(α+γ)
αγ
Hilbert-Pachpatte’s Inequalities
Wengui Yang

( r
γ ) α+γ
αγ
X
bn
1
,
+
(r − n + 1) qn Ψ
γ n=1
qn
vol. 10, iss. 1, art. 26, 2009
Title Page
where
Contents
1
M (k, r; α, γ) =
α+γ
(
γ−1
) α−1
( r γ )
α
α−1
γ−1
α
k γ
X
X
Φ(Pm )
Ψ(Qn )
m=1
Pm
n=1
Qn
.
The following theorems deal with slight variants of the inequality (2.11) given in
Theorem 2.2.
Theorem 2.3. Let α > 1, γ > 1 and {am } and {bn } be two nonnegative sequences
of real numbers defined for m = 1, 2,P
. . . , k and n = 1, 2, .P
. . , r, where k and r are
n
1
natural numbers and define Am = m1 m
a
and
B
=
s
n
s=1
t=1 bt . Let Φ and Ψ be
n
two real-valued, nonnegative, convex functions defined on R+ = [0, ∞). Then
k X
r
X
m=1 n=1
mnΦ(Am )Ψ(Bn )
γm
(α−1)(α+γ)
αγ
+ αn
(γ−1)(α+γ)
αγ
≤ C(1, 1, k, r; α, γ)
JJ
II
J
I
Page 14 of 28
Go Back
Full Screen
Close
(
×
k
X
) α1 ( r
) γ1
X
,
(k − m + 1)Φα (am )
(r − n + 1)Ψγ (bn )
m=1
where C(1, 1, k, r; α, γ) =
n=1
α−1
1
k α r
α+γ
γ−1
γ
.
Proof. From the hypotheses and by using Jensen’s inequality and Hölder’s inequality, it is easy to observe that
!
( m
) α−1
m
m
α
X
X
X
1
1 α−1
1
α
α
,
Φ(Am ) = Φ
as ≤
Φ(as ) ≤ m
Φ (as )
m s=1
m s=1
m
s=1
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
n
Ψ(Bn ) = Ψ
1X
bt
n t=1
!
n
1X
1 γ−1
≤
Ψ(bt ) ≤ n γ
n t=1
n
( n
X
) γ−1
γ
Ψγ (bt )
Contents
.
t=1
The rest of the proof can be completed by following the same steps as in the
proofs of Theorems 2.1 and 2.2 with suitable changes and hence we omit the details.
Theorem 2.4. Let α > 1, γ > 1 and {am } and {bn } be two nonnegative sequences of
real numbers and {pm } and {qn } be positive sequences defined for m = 1,P
2, . . . , k
and n P
= 1, 2, . . . , r, whereP
k and r are natural numbers
and define Pm = m
s=1 ps ,
P
n
1
Qn = nt=1 qt , Am = P1m m
p
a
and
B
=
q
b
.
Let
Φ
and
Ψ
be
two
n
s=1 m s
t=1 n t
Qn
real-valued, nonnegative, convex functions defined on R+ = [0, ∞). Then
k X
r
X
m=1 n=1
Pm Qn Φ(Am )Ψ(Bn )
γm
(α−1)(α+γ)
αγ
+ αn
(γ−1)(α+γ)
αγ
JJ
II
J
I
Page 15 of 28
Go Back
Full Screen
Close
(
≤ C(1, 1, k, r; α, γ)
k
X
) α1
(k − m + 1) [pm Φ (am )]α
m=1
×
( r
X
) γ1
(r − n + 1) [qn Ψ (bn )]γ
,
n=1
where C(1, 1, k, r; α, γ) =
α−1
1
k α r
α+γ
γ−1
γ
Hilbert-Pachpatte’s Inequalities
.
Wengui Yang
Proof. From the hypotheses and by using Jensen’s inequality and Hölder’s inequality, it is easy to observe that
!
m
1 X
p s as
Φ(Am ) = Φ
Pm s=1
( m
) α−1
m
α
1 X
1 α−1 X
,
≤
ps Φ(as ) ≤
m α
[ps Φ(as )]α
Pm s=1
Pm
s=1
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 16 of 28
Ψ(Bn ) = Ψ
n
1 X
qt b t
Qn t=1
!
n
1 X
1 γ−1
≤
qt Ψ(bt ) ≤
n γ
Qn t=1
Qn
Go Back
( n
X
Full Screen
) γ−1
γ
[qt Ψ(bt )]γ
.
t=1
The rest of the proof can be completed by following the same steps as in the
proofs of Theorems 2.1 and 2.2 with suitable changes and hence we omit the details.
Close
3.
Integral Analogues
Now we give the integral analogues of the inequalities in Theorems 2.1 – 2.4.
An integral analogue of Theorem 2.1 is given in the following theorem.
Theorem 3.1. Let p ≥ 0, q ≥ 0, α > 1, γ > 1 and f (σ) ≥ 0, g(τ ) ≥R 0 for σ ∈
s
(0, x), τ ∈ (0, y), where x, y are positive real numbers, define F (s) = 0 f (σ)dσ,
Rt
G(t) = 0 g(τ )dτ for s ∈ (0, x), t ∈ (0, y). Then
Z xZ y
F p (s)Gq (t)
(3.1)
dsdt
(γ−1)(α+γ)
(α−1)(α+γ)
αγ
0
0 γs
+ αt αγ
Z x
α1
p−1
α
≤ D(p, q, x, y; α, γ)
(x − s)(F (s)f (s)) ds
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
0
Contents
Z
×
y
γ1
q−1
γ
,
(y − t)(G (t)g(t)) dt
0
unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y; α, γ) =
α−1
pq
x α y
α+γ
γ−1
γ
.
Proof. From the hypotheses of F (s) and G(t), it is easy to observe that
Z s
p
(3.2)
F (s) = p
F p−1 (σ)f (σ)dσ,
s ∈ (0, x),
0
Z t
q
G (t) = q
Gq−1 (τ )g(τ )dτ,
t ∈ (0, y).
0
From (3.2) and Hölder’s inequality, we have
Z s
α1
Z x
α−1
(3.3)
F p−1 (σ)f (σ)dσ ≤ s α
(F p−1 (σ)f (σ))α dσ
,
0
0
s ∈ (0, s),
JJ
II
J
I
Page 17 of 28
Go Back
Full Screen
Close
and
Z
(3.4)
y
Gq−1 (t)g(t)dt ≤ t
γ−1
γ
Z
0
t
(Gq−1 (τ )g(τ ))γ dτ
γ1
,
t ∈ (0, t).
0
Let s1 = sα−1 , s2 = tγ−1 , ω1 = α1 , ω1 = γ1 , r = ω1 + ω2 , from (3.2) – (3.4) and
(2.7), we observe that
Hilbert-Pachpatte’s Inequalities
(3.5)
F p (s)Gq (t)
α−1
α
γ−1
Wengui Yang
Z
α1 Z
s
γ1
t
(Gq−1 (τ )g(τ ))γ dτ
t γ
(F p−1 (σ)f (σ))α dσ
0
0
( (α−1)(α+γ)
(γ−1)(α+γ) )
pqαγ m αγ
n αγ
≤
+
α+γ
α
γ
Z s
α1 Z t
γ1
p−1
α
q−1
γ
×
(F (σ)f (σ)) dσ
(G (τ )g(τ )) dτ
≤ pqs
0
Title Page
Contents
0
for s ∈ (0, x), t ∈ (0, y). From (3.5), we observe that
γs
(α−1)(α+γ)
αγ
+ αt
II
J
I
Go Back
(γ−1)(α+γ)
αγ
pq
≤
α+γ
JJ
Page 18 of 28
F p (s)Gq (t)
(3.6)
vol. 10, iss. 1, art. 26, 2009
Z
s
(F
0
Full Screen
p−1
α
α1 Z
(σ)f (σ)) dσ
t
(G
q−1
γ
γ1
(τ )g(τ )) dτ
0
for s ∈ (0, x), t ∈ (0, y). Taking the integral on both sides of (3.6) first over t from
0 to y and then over s from 0 to x of the resulting inequality and using Hölder’s
inequality with indices α, α/(α − 1) and γ, γ/(γ − 1) and interchanging the order
Close
of integrals, we observe that
Z xZ y
F p (s)Gq (t)
0
≤
≤
(α−1)(α+γ)
αγ
γs
"Z
0
pq
α+γ
x
α−1
pq
x α
α+γ
(γ−1)(α+γ)
αγ
dsdt
s
Z
0
+ αt
(F p−1 (σ)f (σ))α dσ
α1
# "Z
0
x
s
Z
(F p−1 (σ)f (σ))α dσds
0
0
γ−1
α−1
pq
=
x α y γ
α+γ
Z
t
Z
(Gq−1 (τ )g(τ ))γ dτ
ds
0
Z
y
α1
y
γ−1
γ
Z
y
0
(x − s)(F
p−1
α
#
dt
0
α1 Z
x
γ1
Z
t
γ1
(Gq−1 (τ )g(τ ))γ dτ dt
0
t
(y − t)(G
(s)f (s)) ds
0
q−1
γ1
.
(t)g(t)) dt
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
γ
0
Title Page
Contents
Remark 4. In Theorem 3.1, setting α = γ = 2, we have (1.2). In Theorem 3.1,
setting α1 + γ1 = 1, we have
Z
0
x
Z
0
y
JJ
II
J
I
Page 19 of 28
F p (s)Gq (t)
dsdt
γsα−1 + αtγ−1
Go Back
Z
≤ D(p, q, x, y; α, γ)
x
(x − s)(F p−1 (s)f (s))α ds
α1
Full Screen
0
Z
×
y
γ1
q−1
γ
(y − t)(G (t)g(t)) dt
,
0
unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y; α, γ) =
α−1
pq
x α y
α+γ
γ−1
γ
.
Close
Remark 5. In Theorem 3.1, setting p = q = 1, we have
Z xZ y
F (s)G(t)
(3.7)
dsdt
(α−1)(α+γ)
(γ−1)(α+γ)
αγ
0
0 γs
+ αt αγ
Z x
α1 Z y
γ1
α
γ
≤ D(1, 1, x, y; α, γ)
(x − s)f (s)ds
(y − t)g (t)dt
,
0
0
α−1
1
x α y
α+γ
γ−1
γ .
unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(1, 1, x, y; α, γ) =
In the following theorem we give a further generalization of the inequality (3.7)
obtained in Remark 5.
Theorem 3.2. Let α > 1, γ > 1 and f (σ) ≥ 0, g(τ ) ≥ 0, p(σ) > 0 and q(τ ) > 0
for
real numbers. Define
R s σ ∈ (0, x), τ ∈ (0, y),
R t where x, y are positive
Rs
R t F (s) =
f (σ)dσ and G(t) = 0 g(τ )dτ , P (s) = 0 p(σ)dσ and Q(t) = 0 q(τ )dτ for
0
s ∈ (0, x), t ∈ (0, y). Let Φ and Ψ be two real-valued, nonnegative, convex, and
submultiplicative functions defined on R+ = [0, ∞). Then
Z xZ y
Φ(F (s))Ψ(G(t))
(3.8)
dsdt
(α−1)(α+γ)
(γ−1)(α+γ)
αγ
0
0 γs
+ αt αγ
Z x
α α1
f (s)
≤ L(x, y; α, γ)
(x − s) p(s)Φ
ds
p(s)
0
Z y
γ γ1
g(t)
×
(y − t) q(t)Ψ
dt
,
q(t)
0
where
(Z ) α−1
(Z ) γ−1
γ
α
α−1
γ−1
α
γ
x
y
1
Φ(P (s))
Ψ(Q(t))
L(x, y; α, γ) =
ds
dt
.
α+γ
P (s)
Q(t)
0
0
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 20 of 28
Go Back
Full Screen
Close
Proof. From the hypotheses of Φ and Ψ and by using Jensen’s inequality and Hölder’s
inequality, it is easy to see that


Rs
(σ)
dσ
P (s) 0 p(σ) fp(σ)

Rs
(3.9)
Φ(F (s))) = Φ 
p(σ)dσ
0

R s
f (σ)
p(σ)
dσ
p(σ)
0

Rs
≤ Φ(P (s))Φ 
p(σ)dσ
0
Z
Φ(P (s)) s
f (σ)
≤
p(σ)Φ
dσ
P (s)
p(σ)
0
Z s α α1
Φ(P (s)) α−1
f (σ)
dσ
,
≤
s α
p(σ)Φ
P (s)
p(σ)
0
and similarly,
(3.10)
Ψ(G(t)) ≤
α−1
Ψ(Q(t)) γ−1
t γ
Q(t)
γ−1
Let s1 = s , s2 = t
(2.7), we observe that
, ω1 =
1
,
α
Z t γ γ1
g(τ )
dτ
.
q(τ )Ψ
q(τ )
0
ω1 =
1
,
γ
r = ω1 + ω2 , from (3.9), (3.10) and
Z s α α1 #
f (σ)
Φ(P (s))
p(σ)Φ
dσ
(3.11) Φ(F (s))Ψ(G(t)) ≤ s t
P (s)
p(σ)
0
"
Z t γ γ1 #
Ψ(Q(t))
g(τ )
×
q(τ )Ψ
dτ
Q(t)
q(τ )
0
α−1
α
γ−1
γ
"
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 21 of 28
Go Back
Full Screen
Close
αγ
≤
α+γ
"
(
m
(α−1)(α+γ)
αγ
α
+
n
(γ−1)(α+γ)
αγ
)
γ
Z s α α1 #
Φ(P (s))
f (σ)
×
p(σ)Φ
dσ
P (s)
p(σ)
0
"
Z t γ γ1 #
g(τ )
Ψ(Q(t))
×
q(τ )Ψ
dτ
Q(t)
q(τ )
0
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
for s ∈ (0, x), t ∈ (0, y). From (3.11), we observe that
Φ(F (s))Ψ(G(t))
(3.12)
γs
(α−1)(α+γ)
αγ
Title Page
(γ−1)(α+γ)
αγ
+ αt
"
Z s α α1 #
Φ(P (s))
f (σ)
1
dσ
≤
p(σ)Φ
α+γ
P (s)
p(σ)
0
"
Z t γ γ1 #
g(τ )
Ψ(Q(t))
×
q(τ )Ψ
dτ
Q(t)
q(τ )
0
for s ∈ (0, x), t ∈ (0, y). Taking the integral on both sides of (3.12) first over t from
0 to y and then over s from 0 to x of the resulting inequality and using Hölder’s
inequality with indices α, α/(α − 1) and γ, γ/(γ − 1) and interchanging the order
of integrals, we observe that
Z xZ y
Φ(F (s))Ψ(G(t))
dsdt
(α−1)(α+γ)
(γ−1)(α+γ)
αγ
αγ
0
0 γs
+ αt
Contents
JJ
II
J
I
Page 22 of 28
Go Back
Full Screen
Close
Z s α α1 #
1
Φ(P (s))
f (σ)
≤
p(σ)Φ
dσ
ds
α+γ 0
P (s)
p(σ)
0
"Z
Z t γ γ1 #
y
g(τ )
Ψ(Q(t))
×
q(τ )Ψ
dτ
dt
Q(t)
q(τ )
0
0
(Z ) α−1
α
α−1
α1
α
α Z x Z s x
Φ(P (s))
1
f (σ)
≤
ds
p(σ)Φ
dσds
α+γ
P (s)
p(σ)
0
0
0
(Z ) γ−1
γ
γ
γ1
γ−1
γ Z y Z t y
Ψ(Q(t))
g(τ )
dτ dt
×
dt
q(τ )Ψ
q(τ )
Q(t)
0
0
0
) γ−1
) α−1
(Z (Z γ
α
γ
α
y
x
Ψ(Q(t)) γ−1
1
Φ(P (s)) α−1
dt
ds
=
Q(t)
α+γ
P (s)
0
0
Z x
α α1 Z y
γ γ1
g(t)
f (s)
×
ds
(y − t) q(t)Ψ
dt
.
(x − s) p(s)Φ
p(s)
q(t)
0
0
"Z
x
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 23 of 28
Go Back
Remark 6. From the inequality (2.7), we obtain
(3.13)
sω1 1 sω2 2
1
ω1 sω1 1 +ω2 + ω2 sω2 1 +ω2
≤
ω1 + ω2
for ω1 > 0, ω2 > 0. If we apply the elementary inequality (3.13) on the right-hand
sides of (3.1) in Theorem 3.1 and (3.8) in Theorem 3.2, then we get the following
Full Screen
Close
inequalities
Z xZ
(3.14)
y
F p (s)Gq (t)
dsdt
(γ−1)(α+γ)
+ αt αγ
" Z
α+γ
x
αγ
αγD(p, q, x, y; α, γ) 1
p−1
α
≤
(x − s)(F (s)f (s)) ds
α+γ
α
0
#
Z y
α+γ
αγ
1
,
+
(y − t)(Gq−1 (t)g(t))γ dt
γ
0
0
0
γs
(α−1)(α+γ)
αγ
pq
where D(p, q, x, y; α, γ) = α+γ
x
Z xZ y
Φ(F (s))Ψ(G(t))
0
0
γs
(α−1)(α+γ)
αγ
+ αt
α−1
α
y
(γ−1)(α+γ)
αγ
γ−1
γ
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
. Also,
Title Page
Contents
dsdt
" Z
α α+γ
x
αγ
αγL(x, y; α, γ) 1
f (s)
≤
(x − s) p(s)Φ
ds
α+γ
α
p(s)
0
#
Z y
γ α+γ
αγ
1
g(t)
+
(y − t) q(t)Ψ
dt
,
γ
q(t)
0
II
J
I
Page 24 of 28
Go Back
Full Screen
where
L(x, y; α, γ) =
JJ
 x
α
Z Φ(P (s)) α−1
1
α+γ 
P (s)
0
ds
 α−1 (
Z
 α

0
y
Ψ(Q(t))
Q(t)
γ
γ−1
Close
) γ−1
γ
dt
.
The following theorems deal with slight variants of (3.8) Rgiven in Theorem 3.2.
s
Before we state our next theorem, we point out that “F (s) = 0 f (σ)dσ and G(t) =
Rt
g(τ )dτ ” are replaced by “F (s) =
orem 3.4 in [9].
0
1
s
Rs
0
f (σ)dσ and G(t) =
1
t
Rt
0
g(τ )dτ ” in The-
Theorem 3.3. Let α > 1, γ > 1 and f (σ) ≥ 0, g(τ ) ≥ 0 for
R s σ ∈ (0, x), τ ∈
(0, y), where x, y are positive real numbers. Define F (s) = 1s 0 f (σ)dσ, G(t) =
R
1 t
g(τ )dτ for s ∈ (0, x), t ∈ (0, y). Let Φ and Ψ be two real-valued, nonnegative,
t 0
convex functions defined on R+ = [0, ∞). Then
Z xZ y
stΦ(F (s))Ψ(G(t))
dsdt
(γ−1)(α+γ)
(α−1)(α+γ)
αγ
αγ
0
0 γs
+ αt
γ1
Z x
α1 Z y
γ
α
(y − t)Ψ (g(t))dt
,
≤ D(1, 1, x, y; α, γ)
(x − s)Φ (f (s))ds
0
0
where D(1, 1, x, y; α, γ) =
α−1
1
x α y
α+γ
γ−1
γ
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
.
Theorem 3.4. Let α > 1, γ > 1 and f (σ) ≥ 0, g(τ ) ≥ 0, p(σ) > 0 and
q(τ ) > 0 forR σ ∈ (0, x), τ ∈ R(0, y), where x, y are positive
real numbers. DeRs
s
t
fine P (s) = 0 p(σ)dσ, Q(t) = 0 q(τ )dτ , F (s) = P 1(s) 0 p(σ)f (σ)dσ and G(t) =
Rt
1
q(τ )g(τ )dτ for s ∈ (0, x), t ∈ (0, y). Let Φ and Ψ be two real-valued,
Q(t) 0
nonnegative, convex functions defined on R+ = [0, ∞). Then
Z xZ y
P (s)Q(t)Φ(F (s))Ψ(G(t))
dsdt
(α−1)(α+γ)
(γ−1)(α+γ)
αγ
0
0 γs
+ αt αγ
Z x
α1
α
≤ D(1, 1, x, y; α, γ)
(x − s) [p(s)Φ (f (s))] ds
0
Z
×
0
y
γ1
(y − t) [q(t)Ψ (g(t))] dt
,
γ
JJ
II
J
I
Page 25 of 28
Go Back
Full Screen
Close
where D(1, 1, k, r; α, γ) =
α−1
1
x α y
α+γ
γ−1
γ
.
The proofs of Theorems 3.3 and 3.4 are similar to the proof of Theorem 3.2 and
similar to the proofs of Theorems 2.3 and 2.4. Hence, we leave out the details.
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 26 of 28
Go Back
Full Screen
Close
References
[1] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequality, J.
Math. Anal. and Appl., 226 (1998), 116–179.
[2] G.H. HARDY, J.E. LITTLEWOOD
Cambridge University Press, 1952.
AND
G. POLYA, Inequalities, London:
[3] G.D. HANDLEY, J.J. KOLIHA AND J. PEČARIĆ, Hilbert-Pachpatte type multidimensional integral inequalities, J. Ineq. Pure and Appl. Math., 5(2) (2004),
Art. 34. [ONLINE http://jipam.vu.edu.au/article.php?sid=
389].
Hilbert-Pachpatte’s Inequalities
[4] C.J. ZHAO AND W.S. CHEUNG, Inverses of new Hilbert-Pachatte type inequalities, J. Inequ. and Appl., 2006 (2006), 1–11.
Title Page
[5] B.G. PACHPATTE, Inequalities similar to certain extensions of Hilbert’s inequality, J. Math. Anal. and Appl., 243 (2000), 217–227.
[6] Z.X. LÜ, Some new inequalities similar to Hilbert-Pachpatte type inequalities, J. Ineq. Pure and Appl. Math., 4(2) (2003), Art. 33. [ONLINE http:
//jipam.vu.edu.au/article.php?sid=271].
[7] J. PEČARIĆ, I. PERIĆ AND P. VUKVIĆ, Hilbert-Pachpatte type inequalities
from Bonsall’s form of Hilbert’s inequality, J. Ineq. Pure and Appl. Math., 9(1)
(2008), Art. 9. [ONLINE http://jipam.vu.edu.au/article.php?
sid=949].
[8] S.S. DRAGOMIR AND YOUNG-HO KIM, Hilbert-Pachpatte type integral inequalities and their improvement, J. Ineq. Pure and Appl. Math., 4(1) (2003),
Art. 16. [ONLINE http://jipam.vu.edu.au/article.php?sid=
252].
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Contents
JJ
II
J
I
Page 27 of 28
Go Back
Full Screen
Close
[9] YOUNG-HO KIM, An improvement of some inequalities similar to Hilbert’s
inequality, Int. J. Math. and Math. Sci., 28(4) (2001), 211–221.
[10] G.S. DAVIS AND G.M. PETERSON, On an inequality of Hardy’s (II), Quart.
J. Math. Oxford Ser., 15 (1964), 35–40.
[11] J. NÉMETH, Generalizations of the Hardy-Littlewood inequality, Acta Sci.
Math. (Szeged), 32 (1971), 295–299.
[12] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic, Dordrecht, 1993.
[13] YOUNG-HO KIM AND BYUNG-IL KIM, An analogue of Hilbert’s inequality
and its extensions, Bull. Korean Math. Soc., 39(3) (2002), 377–388.
Hilbert-Pachpatte’s Inequalities
Wengui Yang
vol. 10, iss. 1, art. 26, 2009
Title Page
Contents
JJ
II
J
I
Page 28 of 28
Go Back
Full Screen
Close