Volume 10 (2009), Issue 1, Article 26, 14 pp.
SOME NEW HILBERT-PACHPATTE’S INEQUALITIES
WENGUI YANG
S CHOOL OF M ATHEMATICS AND C OMPUTATIONAL S CIENCE
X IANGTAN U NIVERSITY, X IANGTAN , 411105
H UNAN , P.R. C HINA
yangwg8088@163.com
Received 05 September, 2008; accepted 2 January, 2009
Communicated by W.S. Cheung
A BSTRACT. Some new Hilbert-Pachpatte discrete inequalities and their integral analogues are
established in this paper. Other inequalities are also given in remarks.
Key words and phrases: Hilbert-Pachpatte’s inequality; Hölder’s inequality; Jensen’s inequality; Nonnegative sequences.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
Let p ≥ 1, q ≥ 1 and {am } and {bn } be two nonnegative sequences of real numbers defined
for
and n = 1, 2, . . . , r, where k and r are natural numbers and define Am =
Pmm = 1, 2, . . . , kP
n
s=1 as and Bn =
t=1 bt . Then
( k
) 12
k X
r
p
q
X
X
Am Bn
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
unless {am } or {bn } is null, where C(p, q, k, r) = 12 pq kr.
An integral analogue of (1.1) is given in the following result.
Let p ≥ 1, q ≥ 1 and f (σ) ≥ 0,R g(τ ) ≥ 0 for σ ∈ (0, x),
R t τ ∈ (0, y), where x, y are positive
s
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
12
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
×
(y − t)(Gq−1 (t)g(t))2 dt ,
0
031-09
2
W ENGUI YANG
√
unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y) = 12 pq xy.
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].
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
definedP
for m = 1, 2, . . . , kPand n = 1, 2, . . . , r, where k and r are natural numbers and define
n
Am = m
s=1 as and Bn =
t=1 bt . Then
( k
) 21
r
k X
p
q
X
X
Am Bn
2
(k − m + 1)(Ap−1
(1.3)
1 ≤ C(p, q, k, r; α)
m am )
α + nα ) α
(m
m=1 n=1
m=1
( r
) 12
X
×
(r − n + 1)(Bnq−1 bn )2
,
n=1
√
1
α
unless {am } or {bn } is null, where C(p, q, k, r; α) = 12 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, Rg(τ ) ≥ 0 for σ ∈ (0, x),
R t τ ∈ (0, y), 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
Z
y
(1.4)
F p (s)Gq (t)dsdt
1
0
0
(sα + tα ) α
Z
x
≤ D(p, q, x, y; α)
(x − s)(F
p−1
12
(s)f (s)) ds
2
0
y
Z
×
12
(y − t)(Gq−1 (t)g(t))2 dt ,
0
√
unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y; α) = 2 pq xy.
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].
1
1 α
2. M AIN R ESULTS
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 P
for m = 1, 2, . . . , kP
and n = 1, 2, . . . , r, where k and r are
n
natural numbers and define Am = m
a
and
B
=
n
s=1 s
t=1 bt . Then
(2.1)
k X
r
X
m=1 n=1
Apm Bnq
γm
(α−1)(α+γ)
αγ
+ αn
(
×
k
X
(γ−1)(α+γ)
αγ
≤ C(p, q, k, r; α, γ)
α
(k − m + 1)(Ap−1
m am )
) α1 ( r
X
m=1
unless {am } or {bn } is null, where C(p, q, k, r; α, γ) =
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
) γ1
(r − n + 1)(Bnq−1 bn )γ
,
n=1
pq
k
α+γ
α−1
α
r
γ−1
γ
.
http://jipam.vu.edu.au/
H ILBERT-PACHPATTE ’ S I NEQUALITIES
3
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
Apm ≤ p
(2.3)
m
X
Ap−1
s as ,
m = 1, 2, . . . , k,
Btq−1 bt ,
n = 1, 2, . . . , r.
s=1
Bnq ≤ q
n
X
t=1
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
(Btq−1 bt )γ
,
n = 1, 2, . . . , r.
t=1
Using the inequality of means [12]
( n
) Ω1
(
) r1
n
n
Y
X
1
(2.6)
sωi i
≤
ωi sri
Ωn i=1
i=1
Pn
for r > 0, ωi > 0, i=1 ωi = Ωn , we observe that
(2.7)
(sω1 1 sω2 2 )r/(ω1 +ω2 ) ≤
1
(ω1 sr1 + ω2 sr2 ) .
ω1 + ω2
Let s1 = mα−1 , s2 = nγ−1 , ω1 = α1 , ω2 = γ1 and r = ω1 + ω2 , from (2.3) – (2.5) and (2.7),
we have
( m
) α1 ( n
) γ1
X
X
γ−1
α−1
α
(2.8)
(Ap−1
(Btq−1 bt )γ
Apm Bnq ≤ pqm α n γ
s as )
s=1
≤
pqαγ
α+γ
(
m
α
×
t=1
(α−1)(α+γ)
αγ
( m
X
+
n
(γ−1)(α+γ)
αγ
α
(Ap−1
s as )
s=1
)
γ
) α1 ( n
X
) γ1
(Btq−1 bt )γ
,
t=1
for m = 1, 2, . . . , k, n = 1, 2, . . . , r. From (2.8), we observe that
( m
) α1 ( n
) γ1
X
X q−1
Apm Bnq
pq
α
(2.9)
≤
(Ap−1
(Bt bt )γ
,
s as )
(α−1)(α+γ)
(γ−1)(α+γ)
α + γ s=1
+ αn αγ
γm αγ
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
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
4
W ENGUI YANG
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
(γ−1)(α+γ)
(α−1)(α+γ)
+ αn αγ
γm αγ

( m
) α1   r ( n
) γ1 
k
X
X
X
X
pq 
α

(Btq−1 bt )γ 
≤
(Ap−1
s as )
α + γ m=1 s=1
n=1
t=1
m=1 n=1
pq α−1
≤
k α
α+γ
(
k X
m
X
) α1
α
(Ap−1
s as )
r
γ−1
γ
( r n
XX
) γ1
(Btq−1 bt )γ
n=1 t=1
m=1 s=1
(
k
X
pq α−1 γ−1
α
γ
k α r
=
(k − m + 1)(Ap−1
m am )
α+γ
m=1
( r
) γ1
X
×
(r − n + 1)(Bnq−1 bn )γ
.
) α1
n=1
Remark 1. In Theorem 2.1, setting α = γ = 2, we have (1.1). In Theorem 2.1, setting
1
+ γ1 = 1, we have
α
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) =
α−1
pq
k α r
α+γ
γ−1
γ
.
Remark 2. In Theorem 2.1, setting p = q = 1, we have
(2.10)
k X
r
X
m=1 n=1
Am Bn
γm
(α−1)(α+γ)
αγ
+ αn
(γ−1)(α+γ)
αγ
(
≤ C(1, 1, k, r; α, γ)
k
X
(k − m + 1)aαm
) α1 ( r
X
m=1
unless {am } or {bn } is null, where C(1, 1, k, r; α, γ) =
) γ1
(r − n + 1)bγn
,
n=1
α−1
1
k α r
α+γ
γ−1
γ
.
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 definedP
for m = 1, 2, . . P
. , k and n = 1, 2,
P.m. . , r,
n
m
where k and r are natural numbers and define Am = s=1 as , Bn = t=1 bt , Pm = s=1 ps
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
H ILBERT-PACHPATTE ’ S I NEQUALITIES
5
P
and Qn = nt=1 qt . Let Φ and Ψ be two real-valued, nonnegative, convex, and submultiplicative
functions defined on R+ = [0, ∞). Then
(2.11)
k X
r
X
m=1 n=1
Φ(Am )Ψ(Bn )
γm
(α−1)(α+γ)
αγ
+ αn
(γ−1)(α+γ)
αγ
α ) α1
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
(
k
X
where
M (k, r; α, γ) =
1
α+γ
(
α
k X
Φ(Pm ) α−1
m=1
Pm
γ−1
) α−1
( r γ )
α
X Ψ(Qn ) γ−1 γ
n=1
Qn
.
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
Φ(Pm ) α−1 X
as
≤
m α
ps Φ
,
Pm
ps
s=1
and similarly,
(2.13)
Ψ(Bn ) ≤
Ψ(Qn )
n
Qn
γ−1
γ
Let s1 = mα−1 , s2 = nγ−1 , ω1 = α1 , ω2 =
we have
(2.14)
1
γ
( n γ ) γ1
X
bt
qt Ψ
.
qt
t=1
and r = ω1 + ω2 , from (2.7), (2.12) and (2.13),
Φ(Am )Ψ(Bn )

( m α ) α1
X
γ−1
α−1
Φ(Pm )
as

ps Φ
≤m α n γ 
Pm
p
s
s=1


( n γ ) γ1
X
Ψ(Qn )
bt

×
qt Ψ
Qn
q
t
t=1


( (α−1)(α+γ)
( m (γ−1)(α+γ) )
α ) α1
X
αγ
αγ
αγ
m
n
as
 Φ(Pm )

≤
+
ps Φ
α+γ
α
γ
Pm
p
s
s=1


( n γ ) γ1
X
Ψ(Q
)
bt
n

×
qt Ψ
Qn
q
t
t=1

J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
6
W ENGUI YANG
for m = 1, 2, . . . , k, n = 1, 2, . . . , r. From (2.14), we observe that
Φ(Am )Ψ(Bn )
(2.15)
γm
(γ−1)(α+γ)
(α−1)(α+γ)
αγ
+ αn αγ



( n ( m γ ) γ1
α ) α1
X
X
as
bt
1  Φ(Pm )

  Ψ(Qn )
qt Ψ
≤
ps Φ
Q
q
α+γ
Pm
p
n
t
s
t=1
s=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
Φ(Am )Ψ(Bn )
(α−1)(α+γ)
(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
) α−1
(
α
α ) α1
k X
m X
as
ps Φ
Pm
ps
m=1
m=1 s=1
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
γ−1
( k ) α−1
( r γ )
α
α−1
γ−1
α
γ
X
X
1
Φ(Pm )
Ψ(Qn )
=
α + γ m=1
Pm
Qn
n=1
( k
α ) α1 (X
γ ) γ1
r
X
as
bt
×
(k − m + 1) ps Φ
(r − n + 1) qt Ψ
.
ps
qt
n=1
m=1
1
≤
α+γ
(
α
k X
Φ(Pm ) α−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
(γ−1)(α+γ)
αγ
 (
) 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
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
H ILBERT-PACHPATTE ’ S I NEQUALITIES
where C(p, q, k, r; α, γ) =
k X
r
X
m=1 n=1
α−1
pq
k α r
α+γ
γ−1
γ
7
. Also,
Φ(Am )Ψ(Bn )
γm
(α−1)(α+γ)
αγ
+ αn
(γ−1)(α+γ)
αγ
 (
α ) α+γ
αγ
k
X
αγM (k, r; α, γ)  1
am
(k − m + 1) pm Φ
≤
α+γ
α m=1
pm

( r
γ ) α+γ
αγ
X
1
bn
,
+
(r − n + 1) qn Ψ
γ n=1
qn
where
1
M (k, r; α, γ) =
α+γ
(
α
k X
Φ(Pm ) α−1
γ−1
) α−1
( r γ )
α
X Ψ(Qn ) γ−1 γ
Pm
m=1
Qn
n=1
.
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 P
m = 1, 2, . . . , k and nP= 1, 2, . . . , r, where k and r are natural numbers and
n
1
1
define Am = m m
s=1 as and Bn = n
t=1 bt . Let Φ and Ψ be 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
(
×
k
X
≤ C(1, 1, k, r; α, γ)
) α1 ( r
) γ1
X
(k − m + 1)Φα (am )
(r − n + 1)Ψγ (bn )
,
m=1
where C(1, 1, k, r; α, γ) =
1
k
α+γ
α−1
α
r
n=1
γ−1
γ
.
Proof. From the hypotheses and by using Jensen’s inequality and Hölder’s inequality, it is easy
to observe that
!
m
1 X
Φ(Am ) = Φ
as
m s=1
( m
) α−1
m
α
1 X
1 α−1 X α
Φ (as )
,
≤
Φ(as ) ≤ m α
m s=1
m
s=1
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 )
.
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.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
8
W ENGUI YANG
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
= 1, 2, . . P
. , k and
Pm
m
n
n = 1, 2, . P
. . , r, where k and r are natural
numbers
and
define
P
=
p
,
Q
=
m
n
s=1 s
t=1 qt ,
Pn
m
1
1
Am = Pm s=1 pm as and Bn = Qn t=1 qn bt . Let Φ and Ψ be two 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
(
≤ 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
.
Proof. From the hypotheses and by using Jensen’s inequality and Hölder’s inequality, it is easy
to observe that
!
m
1 X
Φ(Am ) = Φ
p s as
Pm s=1
m
1 X
≤
ps Φ(as )
Pm s=1
( m
) α−1
α
1 α−1 X
α
≤
m α
[ps Φ(as )]
,
Pm
s=1
Ψ(Bn ) = Ψ
n
1 X
qt b t
Qn t=1
!
n
1 X
≤
qt Ψ(bt )
Qn t=1
( n
) γ−1
γ
X
γ−1
1
γ
γ
[qt Ψ(bt )]
.
≤
n
Qn
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.
3. I NTEGRAL A NALOGUES
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) ≥ R0, g(τ ) ≥ 0 for σ R∈ (0, x),
s
t
τ ∈ (0, y), where x, y are positive real numbers, define F (s) = 0 f (σ)dσ, G(t) = 0 g(τ )dτ
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
H ILBERT-PACHPATTE ’ S I NEQUALITIES
9
for s ∈ (0, x), t ∈ (0, y). Then
Z xZ y
F p (s)Gq (t)
dsdt
(3.1)
(γ−1)(α+γ)
(α−1)(α+γ)
αγ
0
0 γs
+ αt αγ
Z x
α1
≤ D(p, q, x, y; α, γ)
(x − s)(F p−1 (s)f (s))α ds
0
y
Z
×
(y − t)(G
q−1
γ1
(t)g(t)) dt
,
γ
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
s ∈ (0, s),
0
and
y
Z
G
(3.4)
q−1
(t)g(t)dt ≤ t
(G
0
(3.5)
q−1
γ
(τ )g(τ )) dτ
t ∈ (0, t).
,
0
1
,
α
Let s1 = sα−1 , s2 = tγ−1 , ω1 =
observe that
p
γ1
t
Z
γ−1
γ
q
F (s)G (t) ≤ pqs
α−1
α
γ−1
γ
ω1 = γ1 , r = ω1 + ω2 , from (3.2) – (3.4) and (2.7), we
Z
α1 Z
s
p−1
α
γ1
t
q−1
γ
(G (τ )g(τ )) dτ
(σ)f (σ)) dσ
0
(α−1)(α+γ)
(γ−1)(α+γ) )
pqαγ m αγ
n αγ
≤
+
α+γ
α
γ
γ1
Z s
α1 Z t
(Gq−1 (τ )g(τ ))γ dτ
×
(F p−1 (σ)f (σ))α dσ
t
(
(F
0
0
0
for s ∈ (0, x), t ∈ (0, y). From (3.5), we observe that
F p (s)Gq (t)
(3.6)
γs
(α−1)(α+γ)
αγ
+ αt
(γ−1)(α+γ)
αγ
pq
≤
α+γ
Z
s
(F
0
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,
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
10
W ENGUI YANG
α/(α − 1) and γ, γ/(γ − 1) and interchanging the order of integrals, we observe that
Z xZ y
F p (s)Gq (t)
dsdt
(γ−1)(α+γ)
(α−1)(α+γ)
αγ
0
0 γs
+ αt αγ
"Z Z
α1 # "Z y Z t
γ1 #
x
s
pq
≤
(F p−1 (σ)f (σ))α dσ
ds
(Gq−1 (τ )g(τ ))γ dτ
dt
α+γ 0
0
0
0
Z x Z s
α1
Z y Z t
γ1
γ−1
α−1
pq
p−1
α
q−1
γ
≤
x α
(F (σ)f (σ)) dσds
y γ
(G (τ )g(τ )) dτ dt
α+γ
0
0
0
0
γ1
Z x
α1 Z t
γ−1
α−1
pq
(y − t)(Gq−1 (t)g(t))γ dt
.
=
x α y γ
(x − s)(F p−1 (s)f (s))α ds
α+γ
0
0
Remark 4. In Theorem 3.1, setting α = γ = 2, we have (1.2). In Theorem 3.1, setting
1
+ γ1 = 1, we have
α
Z xZ y
F p (s)Gq (t)
dsdt ≤ D(p, q, x, y; α, γ)
α−1 + αtγ−1
0
0 γs
Z x
α1 Z y
γ1
×
(x − s)(F p−1 (s)f (s))α ds
(y − t)(Gq−1 (t)g(t))γ dt
,
0
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
.
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
≤ D(1, 1, x, y; α, γ)
(x − s)f α (s)ds
(y − t)g γ (t)dt
,
0
unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(1, 1, x, y; α, γ) =
0
α−1
1
x α y
α+γ
γ−1
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 andRq(τ ) > 0 for
s
σ ∈ (0, x), τ ∈ (0, y), where x, y are positive real numbers. Define F (s) = 0 f (σ)dσ and
Rt
Rs
Rt
G(t) = 0 g(τ )dτ , P (s) = 0 p(σ)dσ and Q(t) = 0 q(τ )dτ for 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
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
H ILBERT-PACHPATTE ’ S I NEQUALITIES
11
where
L(x, y; α, γ) =
1
α+γ
(Z
x
0
Φ(P (s))
P (s)
) α−1
(Z
α
α
α−1
ds
0
y
Ψ(Q(t))
Q(t)
γ
γ−1
) γ−1
γ
dt
.
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)
P (s) 0 p(σ) fp(σ)
dσ

Rs
(3.9)
Φ(F (s))) = Φ 
p(σ)dσ
0

R s
f (σ)
p(σ) p(σ) dσ
0

Rs
≤ Φ(P (s))Φ 
p(σ)dσ
0
Z s
f (σ)
Φ(P (s))
≤
p(σ)Φ
dσ
P (s)
p(σ)
0
Z s α α1
Φ(P (s)) α−1
f (σ)
≤
s α
p(σ)Φ
dσ
,
P (s)
p(σ)
0
and similarly,
(3.10)
Ψ(Q(t)) γ−1
Ψ(G(t)) ≤
t γ
Q(t)
Let s1 = sα−1 , s2 = tγ−1 , ω1 =
observe that
(3.11)
1
,
α
γ γ1
Z t g(τ )
q(τ )Ψ
dτ
.
q(τ )
0
ω1 = γ1 , r = ω1 + ω2 , from (3.9), (3.10) and (2.7), we
Z s α α1 #
f (σ)
Φ(P (s))
Φ(F (s))Ψ(G(t)) ≤ s t
p(σ)Φ
dσ
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)
(γ−1)(α+γ) )
αγ
m αγ
n αγ
≤
+
α+γ
α
γ
"
Z s α α1 #
Φ(P (s))
f (σ)
×
p(σ)Φ
dσ
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
"
for s ∈ (0, x), t ∈ (0, y). From (3.11), we observe that
"
Z s α α1 #
Φ(F (s))Ψ(G(t))
1
Φ(P (s))
f (σ)
(3.12)
≤
p(σ)Φ
dσ
(α−1)(α+γ)
(γ−1)(α+γ)
α+γ
P (s)
p(σ)
0
γs αγ
+ αt αγ
"
Z t γ γ1 #
Ψ(Q(t))
g(τ )
×
q(τ )Ψ
dτ
Q(t)
q(τ )
0
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
12
W ENGUI YANG
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
(γ−1)(α+γ)
(α−1)(α+γ)
αγ
0
0 γs
+ αt αγ
"Z
Z s α α1 #
x
1
f (σ)
Φ(P (s))
ds
p(σ)Φ
dσ
≤
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
α1
α Z x Z s x
Φ(P (s)) α−1
1
f (σ)
ds
dσds
≤
p(σ)Φ
α+γ
P (s)
p(σ)
0
0
0
(Z ) γ−1
γ
γ
γ1
γ Z y Z t y
Ψ(Q(t)) γ−1
g(τ )
×
dt
q(τ )Ψ
dτ dt
Q(t)
q(τ )
0
0
0
(Z ) α−1
(Z ) γ−1
α
γ
α
γ
x
y
Φ(P (s)) α−1
Ψ(Q(t)) γ−1
1
=
ds
dt
α+γ
P (s)
Q(t)
0
0
Z x
α α1 Z y
γ γ1
f (s)
g(t)
×
(x − s) p(s)Φ
ds
(y − t) q(t)Ψ
dt
.
p(s)
q(t)
0
0
Remark 6. From the inequality (2.7), we obtain
sω1 1 sω2 2 ≤
(3.13)
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 inequalities
Z xZ y
F p (s)Gq (t)
(3.14)
dsdt
(α−1)(α+γ)
(γ−1)(α+γ)
αγ
0
0 γs
+ 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
q−1
γ
+
(y − t)(G (t)g(t)) dt
,
γ
0
where D(p, q, x, y; α, γ) =
Z
0
x
Z
0
y
α−1
pq
x α y
α+γ
Φ(F (s))Ψ(G(t))
γs
(α−1)(α+γ)
αγ
+ αt
(γ−1)(α+γ)
αγ
γ−1
γ
. Also,
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
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
H ILBERT-PACHPATTE ’ S I NEQUALITIES
1
+
γ
Z
0
y
13
#
γ α+γ
αγ
g(t)
(y − t) q(t)Ψ
dt
,
q(t)
where
L(x, y; α, γ) =
 x
α
Z Φ(P (s)) α−1
1
α+γ 
P (s)
0
 α−1 (
) γ−1
γ
Z y
γ
 α
Ψ(Q(t)) γ−1
.
ds
dt

Q(t)
0
The following theorems deal with slight variants ofR(3.8) given in Theorem 3.2.
R t Before we
s
state our next theorem, we point out that “F (s) = 0 f (σ)dσ and G(t) = 0 g(τ )dτ ” are
Rs
Rt
replaced by “F (s) = 1s 0 f (σ)dσ and G(t) = 1t 0 g(τ )dτ ” in Theorem 3.4 in [9].
Theorem 3.3. Let α > 1, γ > 1 and f (σ) ≥ 0, g(τ )R ≥ 0 for σ ∈ (0, x),R τ ∈ (0, y),
s
t
where x, y are positive real numbers. Define F (s) = 1s 0 f (σ)dσ, G(t) = 1t 0 g(τ )dτ for
s ∈ (0, x), t ∈ (0, y). Let Φ and Ψ be two real-valued, nonnegative, convex functions defined
on R+ = [0, ∞). Then
Z xZ y
stΦ(F (s))Ψ(G(t))
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)Φ (f (s))ds
(y − t)Ψ (g(t))dt
,
0
where D(1, 1, x, y; α, γ) =
α−1
1
x α y
α+γ
γ−1
γ
0
.
Theorem 3.4. Let α > 1, γ > 1 and f (σ) ≥ 0, g(τ ) ≥ 0, p(σ) > R0 and q(τ ) > 0 for σR ∈ (0, x),
s
t
τ ∈ (0, y), where x, y are positive real numbers. Define P (s) = 0 p(σ)dσ, Q(t) = 0 q(τ )dτ ,
R
R
s
t
1
F (s) = P 1(s) 0 p(σ)f (σ)dσ and G(t) = Q(t)
q(τ )g(τ )dτ for s ∈ (0, x), t ∈ (0, y). Let Φ
0
and Ψ be two real-valued, 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
αγ
0
0 γs
+ αt
γ1
Z x
α1 Z y
α
γ
(y − t) [q(t)Ψ (g(t))] dt
,
≤ D(1, 1, x, y; α, γ)
(x − s) [p(s)Φ (f (s))] ds
0
0
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.
R EFERENCES
[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 AND G. POLYA, Inequalities, London: Cambridge University
Press, 1952.
[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].
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/
14
W ENGUI YANG
[4] C.J. ZHAO AND W.S. CHEUNG, Inverses of new Hilbert-Pachatte type inequalities, J. Inequ. and
Appl., 2006 (2006), 1–11.
[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].
[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.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 26, 14 pp.
http://jipam.vu.edu.au/