Journal of Inequalities in Pure and
Applied Mathematics
ON SEVERAL NEW INEQUALITIES CLOSE TO
HILBERT-PACHPATTE’S INEQUALITY
volume 7, issue 4, article 154,
2006.
BING HE AND YONGJIN LI
Department of Mathematics
Sun Yat-sen University
Guangzhou, 510275 P. R. China
Received 27 April, 2006;
accepted 09 October, 2006.
Communicated by: B. Yang
EMail: hzs314@163.com
EMail: stslyj@mail.sysu.edu.cn
Abstract
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Abstract
In this paper we establish several new inequalities similar to Hilbert-Pachpatte’s
inequality. Moreover, some new generalizations of Hilbert-Pachpatte’s inequality are presented.
2000 Mathematics Subject Classification: 26D15.
Key words: Hilbert’s inequality, Hölder’s inequality, Jensen’s inequality, Power mean
inequality.
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
The work was partially supported by the Foundation of Sun Yat-sen University Advanced Research Centre.
Bing He and Yongjin Li
Contents
Title Page
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
6
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1.
Introduction
The well known Hardy-Hilbert’s inequality is (see [1]):
P∞ p0
P
p
Theorem 1.1. If p > 1, p0 = p/(p − 1) and ∞
a
≤
A,
m
m=1
n=1 bn ≤ B,
then
(1.1)
∞ X
∞
X
1
1
am b n
π
(A) p (B) p0 ,
<
m+n
sin πp
n=1 m=1
unless the sequence {am } or {bn } is null.
The integral analogue can be stated as follows:
R∞
R∞ 0
Theorem 1.2. If p > 1, p0 = p/(p − 1) and 0 f p (x)dx ≤ F, 0 g p (y)dy ≤
G, then
Z ∞Z ∞
1
1
f (x)g(y)
π
F p G p0
(1.2)
dxdy <
x+y
0
0
sin πp
unless f ≡ 0 or g ≡ 0.
The following two theorems were studied by Pachpatte (see [2])
Theorem 1.3. Let p ≥ 1, q ≥ 1 and f (σ) ≥ 0, g(τ ) ≥ 0 for σR ∈ (0, x),
s
τ ∈ (0, y), where x, y are positive real numbers and define F (s) = 0 f (σ)dσ,
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
Bing He and Yongjin Li
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G(t) =
Z xZ
0
Rt
0
0
y
g(τ )dτ , for s ∈ (0, x), t ∈ (0, y). Then
F p (s)Gq (t)
dsdt
s+t
Z
≤ D(p, q, x, y)
x
(x − s)(F
p−1
2
12
(s)f (s)) ds
0
Z
×
y
(y − t)(G
q−1
12
(t)g(t)) dt ,
2
0
unless f ≡ 0 or g ≡ 0, where
1 √
D(p, q, x, y) = pq xy.
2
Theorem 1.4. Let f, g, F, G be as in the above theorem, let p(σ) and q(τ ) be
two
R s positive functions
R t defined for σ ∈ (0, x), τ ∈ (0, y) and define P (s) =
p(σ)dσ, Q(t) = 0 q(τ )dτ for s ∈ (0, x), t ∈ (0, y), where x, y are posi0
tive real numbers. Let φ and ψ be real-valued, nonnegative, convex, and submultiplicative functions defined on R+ = [0, ∞). Then
Z xZ y
φ(F (s))ψ(G(t))
(1.3)
dsdt
s+t
0
0
(Z
2 ) 12
x
f (s)
ds
≤ L(x, y)
(x − s) p(s)φ
p(s)
0
(Z
2 ) 12
y
g(t)
×
(y − t) q(t)ψ
dt
q(t)
0
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
Bing He and Yongjin Li
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where
L(x, y) =
1
2
Z
0
x
φ(P (s))
P (s)
! 12
2
Z
ds
0
y
ψ(Q(t))
Q(t)
2
! 12
dt
.
The inequalities in these theorems were studied extensively and numerous
variants, generalizations, and extensions have appeared in the literature (see [3]
– [9]).
The main purpose of the present article is to establish some new inequalities
similar to the Hilbert-Pachpatte inequalities.
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
Bing He and Yongjin Li
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2.
Main Results
Lemma 2.1. Suppose p > 1, p1 + 1q = 1, r > 1, 1r + w1 = 1, λ > 0, define the
weight functions ω
e1 (w, p, x), ω
e2 (w, p, x), ω
e3 (w, p, x) as
Z ∞
λ
x(p−1)(1− r )
1
(2.1)
ω
e1 (w, p, x) :=
·
dy,
x ∈ (0, ∞),
λ
xλ + y λ
y 1− w
0
Z
(2.2)
ω
e2 (w, p, x) :=
∞
0
λ
1
x(p−1)(1− r )
dy,
·
λ
max {xλ , y λ }
y 1− w
x ∈ (0, ∞),
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
Bing He and Yongjin Li
Z
(2.3)
Then
ω
e3 (w, p, x) :=
0
∞
)
(p−1)(1− λ
r
ln x/y x
·
λ
xλ − y λ
y 1− w
dy,
λ
π
xp(1− r )−1 ,
π
λ sin r
rw p(1− λ )−1
r
x
ω
e2 (w, p, x) =
,
λ
#2
"
λ
π
xp(1− r )−1 .
ω
e3 (w, p, x) =
π
λ sin r
ω
e1 (w, p, x) =
Proof. For fixed x, let u = y λ /xλ , then (2.1) turns into
Z
1
1 p(1− λ )−1 ∞ 1
r
ω
e1 (w, p, x) = x
u−1+ w du
λ
1+u
0
x ∈ (0, ∞).
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λ
1
= xp(1− r )−1 B
λ
1 1
,
w r
λ
π
xp(1− r )−1
λ sin πp
=
and similarly, we can prove the others.
Theorem 2.2. Let m ≥ 1, n ≥ 1, pi > 1, p1i + q1i = 1 for i = 0, 1, 2, 3, 4,
p0 = p, pR3 = k, p4 = r; q0 = q, qR3 = l, q4 = w and f (σ) ≥ 0, g(τ ) ≥ 0. Define
s
t
F (s) = 0 f (σ)dσ and G(t) = 0 g(τ )dτ such that
Z ∞
Z ∞
λ
p(1− λ
)−1 p
r
0<
s
Ff (s)ds < ∞,
0<
tq(1− w )−1 Gqg (t)dt < ∞
0
0
for σ, τ, s, t ∈ (0, ∞). Then
Z ∞Z ∞
F m (s)Gn (t)
dsdt
(2.4)
(lsk/p1 + ktl/p2 )(sλ + tλ )
0
0
Z ∞
p1
p
)−1
p(1− λ
r
Ff (s)ds
≤ E1 (m, n, k, r, λ)
s
Bing He and Yongjin Li
Title Page
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0
∞
Z
×
λ
)−1
q(1− w
t
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
1q
q
Gg (t)dt
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0
unless f ≡ 0 or g ≡ 0, where E1 (m, n, k, r, λ) = λkl πmn
,
sin(π/r)
Z s
q1
1
m−1
q1
Ff (s) =
(F
(σ)f (σ)) dσ
and
0
Z
Gg (t) =
t
(G
0
n−1
q2
(τ )g(τ )) dτ
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2
.
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Proof. From the hypotheses, we can easily observe that
Z s
m
(2.5)
F (s) = m
F m−1 (σ)f (σ)dσ,
s ∈ (0, ∞),
0
t
Z
n
Gn−1 (τ )g(τ )dτ,
G (t) = n
(2.6)
t ∈ (0, ∞).
0
From (2.5) and (2.6), applying Hölder’s inequality, equality (2.1) and Young’s
inequality:
ap b q
ab ≤
+ ,
p
q
where a ≥ 0, b ≥ 0, p > 1,
1
p
+
1
q
= 1, we obtain
≤ mns
0
1/p1 1/p2
Z
t
(F
0
≤ mn
s
m−1
q1 Z t
q1
2
1
n−1
q2
(σ)f (σ)) dσ)
(G (τ )g(τ )) dτ )
q1
0
Z s
q1
1
sk/p1 tl/p2
m−1
q1
+
(F
(σ)f (σ)) dσ)
k
l
0
Z t
q1
2
n−1
q2
×
(G (τ )g(τ )) dτ )
.
0
Bing He and Yongjin Li
Title Page
F m (s)Gn (t)
Z s
Z t
m−1
n−1
= mn
F
(σ)f (σ)dσ
G (τ )g(τ )dτ
0
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
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Note that
Z
Ff (s) =
0
Hence
Z ∞Z
0
≤
=
≤
=
=
s
(F m−1 (σ)f (σ))q1 dσ
q1
1
Z
,
Gg (t) =
t
(Gn−1 (τ )g(τ ))q2 dτ
q1
2
.
0
∞
F m (s)Gn (t)
dsdt
(lsk/p1 + ktl/p2 )(sλ + tλ )
0
n
o1
1
Z ∞ Z ∞ R s (F m−1 (σ)f (σ))q1 dσ) q1 R t (Gn−1 (τ )g(τ ))q2 dτ ) q2
0
0
mn
dsdt
λ
λ
s +t
kl 0
0
#"
#
"
Z Z
λ
λ
t(1− w )/p
mn ∞ ∞
1
s(1− r )/q
Gg (t)
dsdt
Ff (s)
λ
λ
kl 0
sλ + tλ
t(1− w )/p
s(1− r )/q
0
(Z Z
) p1
∞
∞ F p (s)
)
(p−1)(1− λ
r
mn
s
f
·
dsdt
λ
λ
λ
kl
s +t
t(1− w )
0
0
(Z Z
) 1q
λ
∞
∞ Gq (t)
(q−1)(1− w
)
t
g
×
·
dsdt
λ
λ
λ
s +t
s(1− r )
0
0
Z ∞
p1 Z ∞
1q
mn
p
q
ω
e1 (w, p, s)Ff (s)ds
ω
e1 (r, q, t)Gg (t)dt
kl
0
0
Z ∞
p1 Z ∞
1q
λ
λ
πmn
p
sp(1− r )−1 Ff (s)ds
tq(1− w )−1 Gqg (t)dt
λkl sin(π/r)
0
0
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
Bing He and Yongjin Li
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Z
∞
= E1 (m, n, k, r, λ)
s
p(1− λ
)−1
r
p1 Z
∞
Ffp (s)ds
λ
q(1− w
)−1
t
0
1q
Gqg (t)dt
.
0
This completes the proof.
Theorem 2.3. Let m ≥ 1, n ≥ 1, pi > 1, p1i + q1i = 1 for i = 0, 1, 2, 3, 4, p0 = p,
p3 = k, p4 = r; q0 = q, q3 = l, q4 = w and f (σ) ≥ 0, g(τ ) ≥ 0. Define F (s) =
Rs
Rt
R∞
λ
f (σ)dσ and G(t) = 0 g(τ )dτ such that 0 < 0 sp(1− r )−1 Ffp (s)ds < ∞,
0 R
λ
∞
0 < 0 tq(1− w )−1 Gqg (t)dt < ∞ for σ, τ, s, t ∈ (0, ∞). Then
Z
0
∞
Z
0
∞
F m (s)Gn (t)
dsdt
(lsk/p1 + ktl/p2 ) max{sλ , tλ }
p1
Z ∞
p
)−1
p(1− λ
r
s
Ff (s)ds
≤ E2 (m, n, k, r, λ)
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
Bing He and Yongjin Li
Title Page
0
∞
Z
×
λ
)−1
q(1− w
t
0
unless f ≡ 0 or g ≡ 0, where E2 (m, n, k, r, λ) =
Z
mnrw
λkl
s
(F m−1 (σ)f (σ))q1 dσ
Ff (s) =
and
0
Z
Gg (t) =
t
(Gn−1 (τ )g(τ ))q2 dτ
,
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1
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1q
Gqg (t)dt
,
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q1
2
.
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Proof. By (2.5) and (2.6), using Hölder’s inequality, equality (2.2) and Young’s
p
q
inequality: ab ≤ ap + bq , where a ≥ 0, b ≥ 0, p > 1, p1 + 1q = 1. We obtain
F m (s)Gn (t)
Z s
Z t
m−1
n−1
= mn
F
(σ)f (σ)dσ
G (τ )g(τ )dτ
0
≤ mns
0
Z
1/p1 1/p2
s
t
(F
m−1
q1 Z t
q1
2
1
n−1
q2
(σ)f (σ)) dσ)
(G (τ )g(τ )) dτ )
q1
0
≤ mn
s
k/p1
k
0
l/p2
+
Z
×
t
Z
l
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
q1
1
m−1
q1
(F
(σ)f (σ)) dσ)
0
t
(G
s
n−1
Bing He and Yongjin Li
q1
2
(τ )g(τ )) dτ )
.
q2
0
Title Page
Note
Z
Ff (s) =
(F
m−1
(G
n−1
q1
Contents
q1
s
1
(σ)f (σ)) dσ
,
0
Z
Gg (t) =
t
q2
(τ )g(τ )) dτ
q1
2
.
0
Hence
Z ∞Z
∞
F m (s)Gn (t)
dsdt
(lsk/p1 + ktl/p2 ) max{sλ , tλ }
0
0
n
o1
1
Z ∞ Z ∞ R s (F m−1 (σ)f (σ))q1 dσ) q1 R t (Gn−1 (τ )g(τ ))q2 dτ ) q2
0
0
mn
≤
dsdt
λ
kl 0
max{s , tλ }
0
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#"
#
"
λ
λ
s(1− r )/q
t(1− w )/p
1
Gg (t)
dsdt
Ff (s)
λ
λ
max{sλ , tλ }
t(1− w )/p
s(1− r )/q
0
0
(Z Z
) p1
p
∞
∞
(p−1)(1− λ
)
F
(s)
r
mn
s
f
≤
dsdt
λ
λ
kl
max{s , t } t(1− wλ )
0
0
(Z Z
) 1q
λ
∞
∞
)
(q−1)(1− w
Gqg (t)
t
×
dsdt
λ
λ
max{s , t } s(1− λr )
0
0
Z ∞
p1 Z ∞
1q
mn
p
=
ω
e2 (w, p, s)Ff (s)ds
ω
e2 (r, q, t)Gqg (t)dt
kl
0
0
Z ∞
p1 Z ∞
1q
λ
mnrw
p(1− λ
)−1 p
q(1− w
)−1 q
r
=
s
Ff (s)ds
t
Gg (t)dt
λkl
0
0
Z ∞
p1 Z ∞
1q
λ
p
)−1
q(1−
)−1
q
p(1− λ
r
w
= E2 (m, n, k, r, λ)
s
Ff (s)ds
t
Gg (t)dt .
mn
=
kl
Z
∞
Z
∞
0
0
This completes the proof.
Theorem 2.4. Let m ≥ 1, n ≥ 1, pi > 1, p1i + q1i = 1 for i = 0, 1, 2, 3, 4,
p0 = p, pR3 = k, p4 = r; q0 = q, Rq3 = l, q4 = w and f (σ) ≥ 0, g(τ ) ≥ 0. Define
s
t
F (s) = 0 f (σ)dσ and G(t) = 0 g(τ )dτ such that
Z ∞
λ
0<
sp(1− r )−1 Ffp (s)ds < ∞,
Z0 ∞
λ
0<
tq(1− w )−1 Gqg (t)dt < ∞
0
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
Bing He and Yongjin Li
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for σ, τ, s, t ∈ (0, ∞). Then
Z
∞
Z
0
∞
ln(s/t)
(sλ
0
−
tλ )(lsk/p1
Z
+ ktl/p2 )
∞
×
s
p(1− λ
)−1
r
F m (s)Gn (t)dsdt ≤ E3 (m, n, k, r, λ)
Ffp (s)ds
p1 Z
∞
λ
q(1− w
)−1
t
0
1q
Gqg (t)dt
,
0
unless f ≡ 0 or g ≡ 0, where E3 (m, n, k, r, λ) =
Z
π 2 mn
,
λ2 kl(sin(π/r))2
q1
s
Ff (s) =
(F
m−1
q1
1
(σ)f (σ)) dσ
,
0
Z
Gg (t) =
t
(Gn−1 (τ )g(τ ))q2 dτ
2
.
Proof. From (2.5) and (2.6), by Hölder’s inequality, equality (2.3) and Young’s
p
q
inequality: ab ≤ ap + bq , where a ≥ 0, b ≥ 0, p > 1, p1 + 1q = 1, and
using a similar method of proof to that of Theorem 2.2, the result can be clearly
seen.
P
Theorem 2.5. Let mi ≥ 1, pi > 1, ni=1 p1i = 1, p1i + q1i = 1, and fi (σi ) ≥
0R for σi ∈ (0, xi ), where xi are positive real numbers and define Fi (ti ) =
ti
fi (σi )dσi , for ti ∈ (0, xi ), i = 1, 2, . . . , n. Then
0
x1
Z
x2
Z
···
(2.7)
0
0
0
xn
Qn
mi
i=1 Fi (ti )
Pn ti dt1 dt2 · · · dtn
i=1 pi
Bing He and Yongjin Li
q1
0
Z
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Close to Hilbert-Pachpatte’s
Inequality
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≤
n
Y
xi
Z
D(mi , xi , pi )
qi q1i
e
(xi − ti ) Fi (ti ) dti
,
0
i=1
1
p
unless f ≡ 0 or g ≡ 0, where D(mi , xi , pi ) = mi xi i , Fei (ti ) = F mi −1 (ti )fi (ti ).
Proof. From the hypotheses, we know that
Z ti
(m −1)
mi
(2.8)
Fi (ti ) = mi
Fi i (σi )fi (σi )dσi ,
ti ∈ (0, xi ).
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
0
Then
n
Y
(2.9)
Fimi (ti ) =
i=1
n
Y
Bing He and Yongjin Li
Z
ti
Fimi −1 (σi )fi (σi )dσi .
mi
0
i=1
Title Page
Using Hölder’s inequality, we have
Z
ti
(2.10)
Fimi −1 (σi )dσi
Contents
1
pi
Z
1
pi
Z
ti
≤ ti
0
(Fimi −1 (σi )fi (σi ))qi dσi
0
ti
, ti
q1
qi
i
Fei (σi ) dσi
.
0
i=1
Fimi (ti )
≤
n
Y
i=1
i
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By (2.9) and (2.10) it follows that
n
Y
q1
1
pi
Z
mi ti
0
ti
q1
qi
i
Fei (σi ) dσi
.
Page 14 of 20
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Applying Young’s inequality
n
X
1
|ak | ≤
|ak |pk ,
p
k
k=1
k=1
n
Y
where 1 < pk < ∞,
Pn
1
k=1 pk
= 1, we observe that
Qn
Z
n
mi
Y
i=1 Fi (ti )
Pn ti
≤
mi
i=1 pi
ti
q1
qi
i
Fei (σi ) dσi
.
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
0
i=1
Integrating over ti from 0 to xi where i runs from 1 to n, and using the Hölder’s
inequality, we get
Z x1 Z x2
Z xn Q n
mi
i=1 Fi (ti )
Pn ti dt1 dt2 · · · dtn
···
0
0
0
i=1 pi
≤
n
Y
"Z
≤
=
i=1
n
Y
i=1
ti
Z
mi
0
i=1
n
Y
xi
1
pi
q1
qi
i
Fei (σi ) dσi
dti
#
0
Z
xi
ti
Z
mi xi
0
q1
qi
i
Fei (σi ) dσi dti
0
Z
D(mi , xi , pi )
0
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xi
qi q1i
(xi − ti ) Fei (ti ) dti
Quit
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This completes the proof.
J. Ineq. Pure and Appl. Math. 7(4) Art. 154, 2006
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Remark 1. In the special case when p1 = p2 = · · · = pn = n, the inequality
(2.7) reduces to the following inequality,
Z x1 Z x2
Z xn Q n
F mi (ti )
i=1
Pn i
(2.11)
···
dt1 dt2 · · · dtn
0
0
0
i=1 ti
Z xi
n−1
n
n
n−1
n
1Y
D(mi , xi )
(xi − ti ) Fei (ti )
dti
≤
n i=1
0
1
where D(mi , xi ) = mi xin . Moreover, (i) (2.11) reduces to Theorem 1.3 which
belongs to Pachpatte for n = 2; (ii) when m1 = m2 = · · · = mn = 1, (2.11)
turns into
Z x1 Z x2
Z xn Q n
i=1 Fi (ti )
P
(2.12)
···
dt1 dt2 · · · dtn
n
0
0
0
i=1 ti
n−1
Z xi
n
n
n
1Y
≤
D(mi , xi )
(xi − ti )(fi (ti )) n−1 dti
n i=1
0
Theorem 2.6. Let fi , Fi be as in the above theorem, p > 1, p1i + q1i = 1,
let
R ti pi (σi ) be positive function defined for σi ∈ (0, xi ), and define Pi (ti ) =
pi (σi )dσi for ti ∈ (0, xi ),where xi are positive real numbers. Let φi be
0
real-valued, nonnegative, convex, and sub-multiplicative function defined on
R+ = [0, ∞), i = 1, 2, . . . , n. Then
Z x1 Z x2
Z xn Q n
i=1 φi (Fi (ti ))
Pn ti
dt1 dt2 · · · dtn
(2.13)
···
0
0
0
i=1 pi
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
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≤
n
Y
xi
Z
L(xi , pi )
0
i=1
qi q1
i
fi (ti )
(xi − ti ) pi (ti )φi
dti
,
pi (ti )
where
Z
xi
L(xi , pi ) =
0
φi (Pi (ti ))
Pi (ti )
p1
pi
dti
i
.
Proof. Applying Jensen’s inequality and Hölder’s inequality, it is clear to observe that
!
Rt
i)
Pi (ti ) 0 i pi (σi ) pfii (σ
dσi
(σi )
φi (Fi (ti )) = φi
R ti
pi (σi )dσi
0
Z ti
fi (σi )
φi (Pi (ti ))
≤
pi (σi )φi
dσi
Pi (ti )
pi (σi )
0
Z ti qi
q1
i
φi (Pi (ti )) p1i
fi (σi )
≤
ti
pi (σi )φi
dσi
.
Pi (ti )
pi (σi )
0
Using Young’s inequality, we obtain that
Qn
n
Y φi (Pi (ti ))
i=1 φi (Fi (ti ))
Pn ti
≤
Pi (ti )
i=1 pi
i=1
Z
0
ti
qi
q1
i
fi (σi )
pi (σi )φi
dσi
.
pi (σi )
Integrating both sides of the above inequality over ti from 0 to xi with i running
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
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from i to n, and applying the Hölder inequality, we get
Z x1 Z x2
Z xn Qn
i=1 φi (Fi (ti ))
Pn ti
dt1 dt2 · · · dtn
···
0
0
0
i=1 pi
#
qi
q1
ti i
φi (Pi (ti ))
fi (σi )
≤
pi (σi )φi
dσi
dti
P
(t
)
p
(σ
)
i
i
i
i
0
0
i=1
qi
q1
p
p1 Z xi Z ti n Z xi Y
i
i
fi (σi )
φi (Pi (ti )) i
pi (σi )φi
dσi dti
≤
dti
pi (σi )
Pi (ti )
0
0
0
i=1
1
p
p Z xi
qi q1
n Z xi Y
i
i
fi (ti )
φi (Pi (ti )) i
dti
(xi − ti ) pi (ti )φi
=
dti
pi (ti )
Pi (ti )
0
0
i=1
1
Z xi
qi q
n
Y
i
fi (ti )
L(xi , pi )
dti
=
(xi − ti ) pi (ti )φi
pi (ti )
0
i=1
n
Y
"Z
xi
Z
This completes the theorem.
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
Bing He and Yongjin Li
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References
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge Univ. Press, Cambridge, 1952.
[2] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequality, J. Math. Anal. Appl., 226(1) (1998), 166–179.
[3] G.S. DAVIES AND G.M. PETERSEN, On an inequality of Hardy’s, (II),
Quart. J. Math. Oxford Ser., 15(2) (1964), 35–40.
[4] D.S. MITRINOVIĆ, Analytic inequalities. In cooperation with P. M. Vasić.
Die Grundlehren der mathematischen Wisenschaften, Band 1965, SpringerVerlag, New York-Berlin, 1970.
[5] D.S. MITRINOVIĆ AND J.E. PEČARIĆ, On inequalities of Hilbert and
Widder, Proc. Edinburgh Math. Soc. (2), 34(3) (1991), 411–414.
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
Bing He and Yongjin Li
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[6] Z. LÜ, Some new inequalities similar to Hilbert-Pachpatte type inequalities,
J. Inequal. Pure Appl. Math., 4(2) (2003), Art. 33. [ONLINE: http://
jipam.vu.edu.au/article.php?sid=271].
JJ
J
II
I
[7] Y.H. KIM, Some new inverse-type Hilbert-Pachpatte integral inequalities,
Acta Math. Sin. (Engl. Ser.), 20(1) (2004), 57–62.
Go Back
[8] B. YANG, On an extension of Hilbert’s integral inequality with some
parameters, Aust. J. Math. Anal. Appl., 1(1) (2004), Art. 11. [ONLINE:
http://ajmaa.org].
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[9] E.F. BECKENBACH AND R. BELLMAN, Inequalities. Second revised
printing. Ergebnisse der Mathematik und ihrer Grenzgebiete. Neue Folge,
Band 30, Springer-Verlag, New York, Inc. 1965.
On Several New Inequalities
Close to Hilbert-Pachpatte’s
Inequality
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