# SOME NEW HILBERT-PACHPATTE’S INEQUALITIES

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## SOME NEW HILBERT-PACHPATTE’S INEQUALITIES

WENGUI YANG

School of Mathematics and Computational Science Xiangtan University, Xiangtan, 411105

Hunan, P.R. China

EMail:yangwg8088@163.com

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; Nonneg- ative sequences.

Abstract: Some new Hilbert-Pachpatte discrete inequalities and their integral analogues are established in this paper. Other inequalities are also given in remarks.

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### Contents

1 Introduction 3

2 Main Results 6

3 Integral Analogues 17

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### 1.Introduction

Letp≥1, q≥1and{am}and{bn}be two nonnegative sequences of real numbers defined form = 1,2, . . . , kandn = 1,2, . . . , r, wherek andr are natural numbers and defineAm =Pm

s=1asandBn=Pn

t=1bt. Then (1.1)

k

X

m=1 r

X

n=1

ApmBnq

m+n ≤C(p, q, k, r) ( k

X

m=1

(k−m+ 1)(Ap−1m am)2 )12

× ( r

X

n=1

(r−n+ 1)(Bnq−1bn)2 )12

, unless{am}or{bn}is null, whereC(p, q, k, r) = 12pq√

kr.

An integral analogue of (1.1) is given in the following result.

Letp ≥1, q ≥ 1andf(σ) ≥ 0, g(τ)≥ 0forσ ∈(0, x), τ ∈ (0, y),wherex, y are positive real numbers and defineF(s) = Rs

0 f(σ)dσ andG(t) =Rt

0 g(τ)dτ,for s∈(0, x), t∈(0, y). Then

(1.2) Z x

0

Z y 0

Fp(s)Gq(t)dsdt

s+t ≤D(p, q, x, y) Z x

0

(x−s)(Fp−1(s)f(s))2ds 12

× Z y

0

(y−t)(Gq−1(t)g(t))2dt 12

, unlessf(σ)≡0org(τ)≡0, whereD(p, q, x, y) = 12pq√

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].

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Young-Ho Kim [9] gave new inequalities similar to the Hilbert-Pachpatte inequal- ities as follows.

Letp ≥ 1, q ≥ 1, α > 0, and{am} and{bn}be two nonnegative sequences of real numbers defined for m = 1,2, . . . , k and n = 1,2, . . . , r, where k and r are natural numbers and defineAm =Pm

s=1asandBn=Pn

t=1bt. Then (1.3)

k

X

m=1 r

X

n=1

ApmBnq

(mα+nα)α1 ≤C(p, q, k, r;α) ( k

X

m=1

(k−m+ 1)(Ap−1m am)2 )12

× ( r

X

n=1

(r−n+ 1)(Bnq−1bn)2 )12

,

unless{am}or{bn}is null, whereC(p, q, k, r;α) = 121α pq√

kr.

An integral analogue of (1.3) is given in the following result.

Letp≥1, q ≥1, α >0andf(σ)≥0, g(τ)≥0forσ∈(0, x), τ ∈(0, y),where x, y are positive real numbers and defineF(s) =Rs

0 f(σ)dσ andG(t) =Rt

0 g(τ)dτ, fors∈(0, x), t∈(0, y). Then

(1.4) Z x

0

Z y 0

Fp(s)Gq(t)dsdt (sα+tα)α1

≤D(p, q, x, y;α) Z x

0

(x−s)(Fp−1(s)f(s))2ds 12

× Z y

0

(y−t)(Gq−1(t)g(t))2dt 12

,

unlessf(σ)≡0org(τ)≡0, whereD(p, q, x, y;α) = 121α pq√

xy.

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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 inequal- ity, we also obtain some new inequalities similar to some results in [1,9].

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### 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 non- negative sequences of real numbers defined form= 1,2, . . . , k andn= 1,2, . . . , r, wherek andr are natural numbers and defineAm = Pm

s=1as andBn = Pn t=1bt. Then

(2.1)

k

X

m=1 r

X

n=1

ApmBnq γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

≤C(p, q, k, r;α, γ)

× ( k

X

m=1

(k−m+ 1)(Ap−1m am)α

)α1 ( r X

n=1

(r−n+ 1)(Bnq−1bn)γ )1γ

,

unless{am}or{bn}is null, whereC(p, q, k, r;α, γ) = α+γpq kα−1α rγ−1γ .

Proof. The idea for the proof Theorem2.1comes from Theorem 1 of [1] and The- orem 2.1 of [9]. From the hypotheses of Theorem 2.1 and using the following in- equality (see [10,11]),

(2.2)

( n X

m=1

zm )β

≤β

n

X

m=1

zm ( m

X

k=1

zk )β−1

,

whereβ ≥1is a constant andzm ≥0,(m= 1,2, . . . , n), it is easy to observe that Apm ≤p

m

X

s=1

Ap−1s as, m= 1,2, . . . , k, (2.3)

Bnq ≤q

n

X

t=1

Btq−1bt, n = 1,2, . . . , r.

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From (2.3) and Hölder’s inequality, we have (2.4)

m

X

s=1

Ap−1s as ≤mα−1α ( m

X

s=1

(Ap−1s as)α )α1

, m= 1,2, . . . , k,

and (2.5)

n

X

t=1

Btq−1bt≤nγ−1γ ( n

X

t=1

(Btq−1bt)γ )1γ

, n = 1,2, . . . , r.

Using the inequality of means [12]

(2.6)

( n Y

i=1

sωii )Ωn1

≤ ( 1

n

n

X

i=1

ωisri )r1

forr >0, ωi >0,Pn

i=1ωi = Ωn,we observe that (2.7) (sω11sω22)r/(ω12) ≤ 1

ω121sr12sr2).

Lets1 =mα−1, s2 =nγ−1, ω1 = α1, ω2 = 1γ andr=ω12,from (2.3) – (2.5) and (2.7), we have

ApmBnq ≤pqmα−1α nγ−1γ ( m

X

s=1

(Ap−1s as)α

)α1 ( n X

t=1

(Btq−1bt)γ )1γ (2.8)

≤ pqαγ α+γ

(m

(α−1)(α+γ) αγ

α + n

(γ−1)(α+γ) αγ

γ )

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× ( m

X

s=1

(Ap−1s as)α

)α1 ( n X

t=1

(Btq−1bt)γ )1γ

,

form= 1,2, . . . , k, n= 1,2, . . . , r.From (2.8), we observe that (2.9) ApmBnq

γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

≤ pq α+γ

( m X

s=1

(Ap−1s as)α

)α1 ( n X

t=1

(Btq−1bt)γ )1γ

, for m = 1,2, . . . , k, n = 1,2, . . . , r. Taking the sum on both sides of (2.9) first overnfrom1torand then overmfrom1tokof the resulting inequality and using Hölder’s inequality with indicesα, α/(α−1)andγ, γ/(γ −1)and interchanging the order of summations, we observe that

k

X

m=1 r

X

n=1

ApmBnq γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

≤ pq α+γ

k

X

m=1

( m X

s=1

(Ap−1s as)α )1α

r

X

n=1

( n X

t=1

(Btq−1bt)γ )1γ

≤ pq α+γkα−1α

( k X

m=1 m

X

s=1

(Ap−1s as)α )α1

r

γ−1 γ

( r X

n=1 n

X

t=1

(Btq−1bt)γ )γ1

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= pq

α+γkα−1α rγ−1γ ( k

X

m=1

(k−m+ 1)(Ap−1m am)α )1α

× ( r

X

n=1

(r−n+ 1)(Bnq−1bn)γ )γ1

.

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

m=1 r

X

n=1

ApmBnq

γmα−1+αnγ−1 ≤C(p, q, k, r;α, γ)

× ( k

X

m=1

(k−m+ 1)(Ap−1m am)α

)α1 ( r X

n=1

(r−n+ 1)(Bnq−1bn)γ )1γ

,

unless{am}or{bn}is null, whereC(p, q, k, r;α, γ) = α+γpq kα−1α rγ−1γ . Remark 2. In Theorem2.1, settingp=q= 1, we have

(2.10)

k

X

m=1 r

X

n=1

AmBn

γm(α−1)(α+γ)αγ +αn(γ−1)(α+γ)αγ

≤C(1,1, k, r;α, γ) ( k

X

m=1

(k−m+ 1)aαm

)α1 ( r X

n=1

(r−n+ 1)bγn )1γ

,

unless{am}or{bn}is null, whereC(1,1, k, r;α, γ) = α+γ1 kα−1α r

γ−1 γ .

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In the following theorem we give a further generalization of the inequality (2.10) obtained in Remark2. Before we give our result, we point out that{pm}and {qn} should be two positive sequences form= 1,2, . . . , kandn= 1,2, . . . , rin Theorem 2.3 of [9].

Theorem 2.2. Letα >1, γ >1and{am}and{bn}be two nonnegative sequences of real numbers and{pm}and{qn}be positive sequences defined form = 1,2, . . . , k andn = 1,2, . . . , r, wherek andr are natural numbers and defineAm =Pm

s=1as, Bn =Pn

t=1bt,Pm =Pm

s=1psandQn =Pn

t=1qt. LetΦandΨbe two real-valued, nonnegative, convex, and submultiplicative functions defined onR+ = [0,∞).Then

(2.11)

k

X

m=1 r

X

n=1

Φ(Am)Ψ(Bn) γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

≤M(k, r;α, γ) ( k

X

m=1

(k−m+ 1)

pmΦ am

pm

α)α1

× ( r

X

n=1

(r−n+ 1)

qnΨ bn

qn

γ)1γ , where

M(k, r;α, γ) = 1 α+γ

( k X

m=1

Φ(Pm) Pm

α−1α )α−1α ( r X

n=1

Ψ(Qn) Qn

γ−1γ )γ−1γ . Proof. From the hypotheses ofΦandΨand by using Jensen’s inequality and Hölder’s

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inequality, it is easy to observe that Φ(Am) = Φ

PmPm

s=1psas/ps

Pm s=1ps

(2.12)

≤Φ(Pm)Φ Pm

s=1psas/ps Pm

s=1ps

≤ Φ(Pm) Pm

m

X

s=1

psΦ as

ps

≤ Φ(Pm) Pm mα−1α

( m X

s=1

psΦ

as ps

α)α1 ,

and similarly,

(2.13) Ψ(Bn)≤ Ψ(Qn) Qn nγ−1γ

( n X

t=1

qtΨ

bt qt

γ)1γ .

Lets1 =mα−1, s2 =nγ−1, ω1 = α1, ω2 = 1γ andr=ω12,from (2.7), (2.12) and (2.13), we have

Φ(Am)Ψ(Bn)≤mα−1α n

γ−1 γ

Φ(Pm) Pm

( m X

s=1

psΦ

as ps

α)α1 (2.14) 

×

Ψ(Qn) Qn

( n X

t=1

qtΨ

bt qt

γ)1γ

≤ αγ α+γ

(m

(α−1)(α+γ) αγ

α + n

(γ−1)(α+γ) αγ

γ )

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×

Φ(Pm) Pm

( m X

s=1

psΦ

as

ps

α)α1

Ψ(Qn) Qn

( n X

t=1

qtΨ

bt

qt

γ)1γ

form = 1,2, . . . , k, n= 1,2, . . . , r.From (2.14), we observe that (2.15) Φ(Am)Ψ(Bn)

γm(α−1)(α+γ)αγ +αn(γ−1)(α+γ)αγ

≤ 1 α+γ

Φ(Pm) Pm

( m X

s=1

psΦ

as ps

α)α1

×

Ψ(Qn) Qn

( n X

t=1

qtΨ

bt qt

γ)1γ

for m = 1,2, . . . , k, n = 1,2, . . . , r. Taking the sum on both sides of (2.15) first overnfrom1torand then overmfrom1tokof the resulting inequality and using Hölder’s inequality with indicesα, α/(α−1)andγ, γ/(γ −1)and interchanging the order of summations, we observe that

k

X

m=1 r

X

n=1

Φ(Am)Ψ(Bn) γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

≤ 1 α+γ

k

X

m=1

Φ(Pm) Pm

( m X

s=1

psΦ

as ps

α)α1

r

X

n=1

Ψ(Qn) Qn

( n X

t=1

qtΨ

bt qt

γ)γ1

≤ 1 α+γ

( k X

m=1

Φ(Pm) Pm

α−1α )α−1α ( k X

m=1 m

X

s=1

psΦ

as

ps

α)α1

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× ( r

X

n=1

Ψ(Qn) Qn

γ−1γ )γ−1γ ( r X

n=1 n

X

t=1

qtΨ

bt qt

γ)1γ

= 1

α+γ ( k

X

m=1

Φ(Pm) Pm

α−1α )α−1α ( r X

n=1

Ψ(Qn) Qn

γ−1γ )γ−1γ

× ( k

X

m=1

(k−m+ 1)

psΦ as

ps

α)1α( r X

n=1

(r−n+ 1)

qtΨ bt

qt

γ)1γ .

Remark 3. From the inequality (2.7), we obtain (2.16) sω11sω22 ≤ 1

ω12

ω1sω1122sω212

forω1 >0, ω2 >0. If we apply the elementary inequality (2.16) on the right-hand sides of (2.1) in Theorem2.1and (2.11) in Theorem2.2, then we get the following inequalities

k

X

m=1 r

X

n=1

ApmBnq γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

≤ αγC(p, q, k, r;α, γ) α+γ

 1 α

( k X

m=1

(k−m+ 1)(Ap−1m am)α )α+γαγ

+1 γ

( r X

n=1

(r−n+ 1)(Bnq−1bn)γ

)α+γαγ

,

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whereC(p, q, k, r;α, γ) = α+γpq kα−1α rγ−1γ .Also,

k

X

m=1 r

X

n=1

Φ(Am)Ψ(Bn) γm(α−1)(α+γ)αγ +αn(γ−1)(α+γ)αγ

≤ αγM(k, r;α, γ) α+γ

 1 α

( k X

m=1

(k−m+ 1)

pmΦ am

pm

α)α+γαγ

+ 1 γ

( r X

n=1

(r−n+ 1)

qnΨ bn

qn

γ)α+γαγ

,

where

M(k, r;α, γ) = 1 α+γ

( k X

m=1

Φ(Pm) Pm

α−1α )α−1α ( r X

n=1

Ψ(Qn) Qn

γ−1γ )γ−1γ . The following theorems deal with slight variants of the inequality (2.11) given in Theorem2.2.

Theorem 2.3. Letα > 1, γ >1and{am}and{bn}be two nonnegative sequences of real numbers defined form = 1,2, . . . , k andn = 1,2, . . . , r, wherekandrare natural numbers and defineAm = m1 Pm

s=1asandBn = n1 Pn

t=1bt. LetΦandΨbe two real-valued, nonnegative, convex functions defined onR+ = [0,∞).Then

k

X

m=1 r

X

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)

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× ( k

X

m=1

(k−m+ 1)Φα(am)

)α1 ( r X

n=1

(r−n+ 1)Ψγ(bn) )1γ

,

whereC(1,1, k, r;α, γ) = α+γ1 kα−1α rγ−1γ .

Proof. From the hypotheses and by using Jensen’s inequality and Hölder’s inequal- ity, it is easy to observe that

Φ(Am) = Φ 1 m

m

X

s=1

as

!

≤ 1 m

m

X

s=1

Φ(as)≤ 1 mmα−1α

( m X

s=1

Φα(as) )α−1α

,

Ψ(Bn) = Ψ 1 n

n

X

t=1

bt

!

≤ 1 n

n

X

t=1

Ψ(bt)≤ 1 nnγ−1γ

( n X

t=1

Ψγ(bt) )γ−1γ

. The rest of the proof can be completed by following the same steps as in the proofs of Theorems2.1and2.2with suitable changes and hence we omit the details.

Theorem 2.4. Letα >1, γ >1and{am}and{bn}be two nonnegative sequences of real numbers and{pm}and{qn}be positive sequences defined form = 1,2, . . . , k andn = 1,2, . . . , r, wherek andr are natural numbers and definePm = Pm

s=1ps, Qn=Pn

t=1qt, Am = P1

m

Pm

s=1pmasandBn = Q1

n

Pn

t=1qnbt. LetΦandΨbe two real-valued, nonnegative, convex functions defined onR+= [0,∞).Then

k

X

m=1 r

X

n=1

PmQnΦ(Am)Ψ(Bn) γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

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≤C(1,1, k, r;α, γ) ( k

X

m=1

(k−m+ 1) [pmΦ (am)]α )α1

× ( r

X

n=1

(r−n+ 1) [qnΨ (bn)]γ )1γ

,

whereC(1,1, k, r;α, γ) = α+γ1 kα−1α rγ−1γ .

Proof. From the hypotheses and by using Jensen’s inequality and Hölder’s inequal- ity, it is easy to observe that

Φ(Am) = Φ 1 Pm

m

X

s=1

psas

!

≤ 1 Pm

m

X

s=1

psΦ(as)≤ 1 Pmmα−1α

( m X

s=1

[psΦ(as)]α )α−1α

,

Ψ(Bn) = Ψ 1 Qn

n

X

t=1

qtbt

!

≤ 1 Qn

n

X

t=1

qtΨ(bt)≤ 1 Qn

nγ−1γ ( n

X

t=1

[qtΨ(bt)]γ )γ−1γ

.

The rest of the proof can be completed by following the same steps as in the proofs of Theorems2.1and2.2with suitable changes and hence we omit the details.

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### 3.Integral Analogues

Now we give the integral analogues of the inequalities in Theorems2.1–2.4.

An integral analogue of Theorem2.1is given in the following theorem.

Theorem 3.1. Letp ≥ 0, q ≥ 0, α > 1, γ > 1and f(σ) ≥ 0, g(τ) ≥ 0forσ ∈ (0, x), τ ∈ (0, y), wherex, yare positive real numbers, defineF(s) = Rs

0 f(σ)dσ, G(t) =Rt

0 g(τ)dτ fors∈(0, x), t∈(0, y). Then (3.1)

Z x 0

Z y 0

Fp(s)Gq(t) γs(α−1)(α+γ)αγ +αt(γ−1)(α+γ)αγ

dsdt

≤D(p, q, x, y;α, γ) Z x

0

(x−s)(Fp−1(s)f(s))αds α1

× Z y

0

(y−t)(Gq−1(t)g(t))γdt 1γ

,

unlessf(σ)≡0org(τ)≡0, whereD(p, q, x, y;α, γ) = α+γpq xα−1α yγ−1γ . Proof. From the hypotheses ofF(s)andG(t), it is easy to observe that

Fp(s) = p Z s

0

Fp−1(σ)f(σ)dσ, s∈(0, x), (3.2)

Gq(t) = q Z t

0

Gq−1(τ)g(τ)dτ, t ∈(0, y).

From (3.2) and Hölder’s inequality, we have (3.3)

Z x 0

Fp−1(σ)f(σ)dσ≤sα−1α Z s

0

(Fp−1(σ)f(σ))αα1

, s∈(0, s),

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and (3.4)

Z y 0

Gq−1(t)g(t)dt≤tγ−1γ Z t

0

(Gq−1(τ)g(τ))γγ1

, t∈(0, t).

Lets1 =sα−1, s2 =tγ−1, ω1 = α1, ω1 = γ1, r =ω12,from (3.2) – (3.4) and (2.7), we observe that

Fp(s)Gq(t) (3.5)

≤pqsα−1α tγ−1γ Z s

0

(Fp−1(σ)f(σ))α

α1 Z t 0

(Gq−1(τ)g(τ))γγ1

≤ pqαγ α+γ

(m(α−1)(α+γ)αγ

α + n(γ−1)(α+γ)αγ γ

)

× Z s

0

(Fp−1(σ)f(σ))α

α1 Z t 0

(Gq−1(τ)g(τ))γ1γ

fors∈(0, x), t∈(0, y).From (3.5), we observe that (3.6) Fp(s)Gq(t)

γs

(α−1)(α+γ)

αγ +αt

(γ−1)(α+γ) αγ

≤ pq α+γ

Z s 0

(Fp−1(σ)f(σ))α

α1 Z t 0

(Gq−1(τ)g(τ))γ1γ

fors ∈(0, x), t ∈ (0, y).Taking the integral on both sides of (3.6) first overtfrom 0 to y and then over s from 0 to x of the resulting inequality and using Hölder’s inequality with indicesα,α/(α−1)andγ,γ/(γ−1)and interchanging the order

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of integrals, we observe that Z x

0

Z y 0

Fp(s)Gq(t) γs(α−1)(α+γ)αγ +αt(γ−1)(α+γ)αγ

dsdt

≤ pq α+γ

"

Z x 0

Z s 0

(Fp−1(σ)f(σ))αα1

ds

# "

Z y 0

Z t 0

(Gq−1(τ)g(τ))γ1γ

dt

#

≤ pq α+γxα−1α

Z x 0

Z s 0

(Fp−1(σ)f(σ))αdσds α1

yγ−1γ Z y

0

Z t 0

(Gq−1(τ)g(τ))γdτ dt 1γ

= pq

α+γxα−1α yγ−1γ Z x

0

(x−s)(Fp−1(s)f(s))αds

1αZ t 0

(y−t)(Gq−1(t)g(t))γdt 1γ

.

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 x 0

Z y 0

Fp(s)Gq(t) γsα−1+αtγ−1dsdt

≤D(p, q, x, y;α, γ) Z x

0

(x−s)(Fp−1(s)f(s))αds α1

× Z y

0

(y−t)(Gq−1(t)g(t))γdt 1γ

,

unlessf(σ)≡0org(τ)≡0, whereD(p, q, x, y;α, γ) = α+γpq xα−1α yγ−1γ .

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Remark 5. In Theorem3.1, settingp=q= 1, we have (3.7)

Z x 0

Z y 0

F(s)G(t) γs

(α−1)(α+γ)

αγ +αt

(γ−1)(α+γ) αγ

dsdt

≤D(1,1, x, y;α, γ) Z x

0

(x−s)fα(s)ds

α1 Z y 0

(y−t)gγ(t)dt 1γ

, unlessf(σ)≡0org(τ)≡0, whereD(1,1, x, y;α, γ) = α+γ1 xα−1α yγ−1γ .

In the following theorem we give a further generalization of the inequality (3.7) obtained in Remark5.

Theorem 3.2. Letα > 1, γ > 1andf(σ) ≥ 0, g(τ) ≥0, p(σ) > 0andq(τ) > 0 for σ ∈ (0, x), τ ∈ (0, y), where x, y are positive real numbers. Define F(s) = Rs

0 f(σ)dσ and G(t) = Rt

0 g(τ)dτ, P(s) = Rs

0 p(σ)dσ and Q(t) = Rt

0 q(τ)dτ for s ∈ (0, x), t ∈ (0, y). LetΦ andΨ be two real-valued, nonnegative, convex, and submultiplicative functions defined onR+= [0,∞).Then

(3.8) Z x

0

Z y 0

Φ(F(s))Ψ(G(t)) γs

(α−1)(α+γ)

αγ +αt

(γ−1)(α+γ) αγ

dsdt

≤L(x, y;α, γ) Z x

0

(x−s)

p(s)Φ f(s)

p(s) α

ds α1

× Z y

0

(y−t)

q(t)Ψ g(t)

q(t) γ

dt 1γ

, where

L(x, y;α, γ) = 1 α+γ

(Z x 0

Φ(P(s)) P(s)

α−1α ds

)α−1α ( Z y

0

Ψ(Q(t)) Q(t)

γ−1γ dt

)γ−1γ .

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Proof. From the hypotheses ofΦandΨand by using Jensen’s inequality and Hölder’s inequality, it is easy to see that

Φ(F(s))) = Φ

P(s)Rs

0 p(σ)

f(σ) p(σ)

dσ Rs

0 p(σ)dσ

 (3.9) 

≤Φ(P(s))Φ

 Rs

0 p(σ)

f(σ) p(σ)

dσ Rs

0 p(σ)dσ

≤ Φ(P(s)) P(s)

Z s 0

p(σ)Φ

f(σ) p(σ)

≤ Φ(P(s)) P(s) sα−1α

Z s 0

p(σ)Φ

f(σ) p(σ)

α

α1

,

and similarly,

(3.10) Ψ(G(t))≤ Ψ(Q(t)) Q(t) tγ−1γ

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

1γ

.

Lets1 =sα−1, s2 =tγ−1, ω1 = α1, ω1 = γ1, r=ω12,from (3.9), (3.10) and (2.7), we observe that

Φ(F(s))Ψ(G(t))≤sα−1α tγ−1γ

"

Φ(P(s)) P(s)

Z s 0

p(σ)Φ

f(σ) p(σ)

α

α1# (3.11)

×

"

Ψ(Q(t)) Q(t)

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

1γ#

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≤ αγ α+γ

(m

(α−1)(α+γ) αγ

α + n

(γ−1)(α+γ) αγ

γ )

×

"

Φ(P(s)) P(s)

Z s 0

p(σ)Φ

f(σ) p(σ)

α

α1#

×

"

Ψ(Q(t)) Q(t)

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

1γ#

fors∈(0, x), t∈(0, y).From (3.11), we observe that (3.12) Φ(F(s))Ψ(G(t))

γs(α−1)(α+γ)αγ +αt(γ−1)(α+γ)αγ

≤ 1 α+γ

"

Φ(P(s)) P(s)

Z s 0

p(σ)Φ

f(σ) p(σ)

α

1α#

×

"

Ψ(Q(t)) Q(t)

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

γ1#

fors∈(0, x), t∈(0, y).Taking the integral on both sides of (3.12) first overtfrom 0 to y and then over s from 0 to x of the resulting inequality and using Hölder’s inequality with indicesα,α/(α−1)andγ,γ/(γ−1)and interchanging the order of integrals, we observe that

Z x 0

Z y 0

Φ(F(s))Ψ(G(t)) γs

(α−1)(α+γ)

αγ +αt

(γ−1)(α+γ) αγ

dsdt

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≤ 1 α+γ

"

Z x 0

Φ(P(s)) P(s)

Z s 0

p(σ)Φ

f(σ) p(σ)

α

1α

ds

#

×

"

Z y 0

Ψ(Q(t)) Q(t)

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

1γ

dt

#

≤ 1 α+γ

(Z x 0

Φ(P(s)) P(s)

α−1α ds

)α−1α Z x

0

Z s 0

p(σ)Φ

f(σ) p(σ)

α

dσds α1

× (Z y

0

Ψ(Q(t)) Q(t)

γ−1γ dt

)γ−1γ Z y

0

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

dτ dt 1γ

= 1

α+γ (Z x

0

Φ(P(s)) P(s)

α−1α ds

)α−1α ( Z y

0

Ψ(Q(t)) Q(t)

γ−1γ dt

)γ−1γ

× Z x

0

(x−s)

p(s)Φ f(s)

p(s) α

ds

α1 Z y 0

(y−t)

q(t)Ψ g(t)

q(t) γ

dt γ1

.

Remark 6. From the inequality (2.7), we obtain (3.13) sω11sω22 ≤ 1

ω12 ω1sω1122sω212

forω1 >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

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inequalities (3.14)

Z x 0

Z y 0

Fp(s)Gq(t) γs

(α−1)(α+γ)

αγ +αt

(γ−1)(α+γ) αγ

dsdt

≤ αγD(p, q, x, y;α, γ) α+γ

"

1 α

Z x 0

(x−s)(Fp−1(s)f(s))αds α+γαγ

+ 1 γ

Z y 0

(y−t)(Gq−1(t)g(t))γdt α+γαγ #

,

whereD(p, q, x, y;α, γ) = α+γpq xα−1α yγ−1γ . Also, Z x

0

Z y 0

Φ(F(s))Ψ(G(t)) γs(α−1)(α+γ)αγ +αt(γ−1)(α+γ)αγ

dsdt

≤ αγL(x, y;α, γ) α+γ

"

1 α

Z x 0

(x−s)

p(s)Φ f(s)

p(s) α

ds α+γαγ

+ 1 γ

Z y 0

(y−t)

q(t)Ψ g(t)

q(t) γ

dt α+γαγ #

,

where

L(x, y;α, γ) = 1 α+γ

x

Z

0

Φ(P(s)) P(s)

α−1α ds

α−1

α (

Z y 0

Ψ(Q(t)) Q(t)

γ−1γ dt

)γ−1γ . The following theorems deal with slight variants of (3.8) given in Theorem3.2.

Before we state our next theorem, we point out that “F(s) = Rs

0 f(σ)dσandG(t) =

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