SOME NEW HILBERT-PACHPATTE’S INEQUALITIES

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Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009

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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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1. Introduction

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

s=1a_sandB_n=Pn

t=1b_t. Then (1.1)

k

X

m=1 r

X

n=1

A^p_mB_n^q

m+n ≤C(p, q, k, r) ( _k

X

m=1

(k−m+ 1)(A^p−1_m a_m)² )¹₂

× ( _r

X

n=1

(r−n+ 1)(B_n^q−1bn)² )¹₂

, unless{a_m}or{b_n}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

F^p(s)G^q(t)dsdt

s+t ≤D(p, q, x, y) Z x

0

(x−s)(F^p−1(s)f(s))²ds ¹₂

× Z y

0

(y−t)(G^q−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 inequalities as follows.

Letp ≥ 1, q ≥ 1, α > 0, and{a_m} and{b_n}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 defineA_m =Pm

s=1a_sandB_n=Pn

t=1b_t. Then (1.3)

k

X

m=1 r

X

n=1

A^p_mB_n^q

(m^α+n^α)^α¹ ≤C(p, q, k, r;α) ( _k

X

m=1

(k−m+ 1)(A^p−1_m a_m)² )¹₂

× ( _r

X

n=1

(r−n+ 1)(B_n^q−1bn)² )¹₂

,

unless{a_m}or{b_n}is null, whereC(p, q, k, r;α) = ¹₂¹_α 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

F^p(s)G^q(t)dsdt (s^α+t^α)^α¹

≤D(p, q, x, y;α) Z x

0

(x−s)(F^p−1(s)f(s))²ds ¹₂

× Z y

0

(y−t)(G^q−1(t)g(t))²dt ¹₂

,

unlessf(σ)≡0org(τ)≡0, whereD(p, q, x, y;α) = ¹₂¹_α 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 inequality, 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{a_m}and {b_n}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

A^p_mB_n^q 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)(A^p−1_m a_m)^α

)_α¹ ( _r X

n=1

(r−n+ 1)(B_n^q−1b_n)^γ )¹_γ

,

unless{a_m}or{b_n}is null, whereC(p, q, k, r;α, γ) = _α+γ^pq k^α−1^α r^γ−1^γ .

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 inequality (see [10,11]),

(2.2)

( _n X

m=1

z_m )β

≤β

n

X

m=1

z_m ( _m

X

k=1

z_k )β−1

,

whereβ ≥1is a constant andz_m ≥0,(m= 1,2, . . . , n), it is easy to observe that A^p_m ≤p

m

X

s=1

A^p−1_s a_s, m= 1,2, . . . , k, (2.3)

B_n^q ≤q

n

X

t=1

B_t^q−1b_t, n = 1,2, . . . , r.

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

m

X

s=1

A^p−1_s a_s ≤m^α−1^α ( _m

X

s=1

(A^p−1_s a_s)^α )_α¹

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

and (2.5)

n

X

t=1

B_t^q−1b_t≤n^γ−1^γ ( _n

X

t=1

(B_t^q−1b_t)^γ )¹_γ

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

Using the inequality of means [12]

(2.6)

( _n Y

i=1

s^ω_iⁱ )_Ωn¹

≤ ( 1

Ω_n

n

X

i=1

ωis^r_i )_r¹

forr >0, ω_i >0,Pn

i=1ω_i = Ω_n,we observe that (2.7) (s^ω₁¹s^ω₂²)^r/(ω¹^+ω²⁾ ≤ 1

ω₁+ω₂ (ω₁s^r₁+ω₂s^r₂).

Lets₁ =m^α−1, s₂ =n^γ−1, ω₁ = _α¹, ω₂ = ¹_γ andr=ω₁+ω₂,from (2.3) – (2.5) and (2.7), we have

A^p_mB_n^q ≤pqm^α−1^α n^γ−1^γ ( _m

X

s=1

(A^p−1_s a_s)^α

)_α¹ ( _n X

t=1

(B_t^q−1b_t)^γ )¹_γ (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

(A^p−1_s a_s)^α

)_α¹ ( _n X

t=1

(B_t^q−1b_t)^γ )¹_γ

,

form= 1,2, . . . , k, n= 1,2, . . . , r.From (2.8), we observe that (2.9) A^p_mB_n^q

γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

≤ pq α+γ

( _m X

s=1

(A^p−1_s a_s)^α

)_α¹ ( _n X

t=1

(B_t^q−1b_t)^γ )¹_γ

, 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

A^p_mB_n^q γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

≤ pq α+γ





k

X

m=1

( _m X

s=1

(A^p−1_s a_s)^α )¹_α







r

X

n=1

( _n X

t=1

(B_t^q−1b_t)^γ )¹_γ



≤ pq α+γk^α−1^α

( _k X

m=1 m

X

s=1

(A^p−1_s as)^α )_α¹

r

γ−1 γ

( _r X

n=1 n

X

t=1

(B_t^q−1bt)^γ )_γ¹

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= pq

α+γk^α−1^α r^γ−1^γ ( _k

X

m=1

(k−m+ 1)(A^p−1_m a_m)^α )¹_α

× ( _r

X

n=1

(r−n+ 1)(B_n^q−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

A^p_mB_n^q

γm^α−1+αn^γ−1 ≤C(p, q, k, r;α, γ)

× ( _k

X

m=1

(k−m+ 1)(A^p−1_m a_m)^α

)_α¹ ( _r X

n=1

(r−n+ 1)(B_n^q−1b_n)^γ )¹_γ

,

unless{a_m}or{b_n}is null, whereC(p, q, k, r;α, γ) = _α+γ^pq k^α−1^α r^γ−1^γ . Remark 2. In Theorem2.1, settingp=q= 1, we have

(2.10)

k

X

m=1 r

X

n=1

A_mB_n

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)a^α_m

)_α¹ ( _r X

n=1

(r−n+ 1)b^γ_n )¹_γ

,

unless{am}or{bn}is null, whereC(1,1, k, r;α, γ) = _α+γ¹ k^α−1^α 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{p_m}and {q_n} should be two positive sequences form= 1,2, . . . , kandn= 1,2, . . . , rin Theorem 2.3 of [9].

Theorem 2.2. Letα >1, γ >1and{a_m}and{b_n}be two nonnegative sequences of real numbers and{p_m}and{q_n}be positive sequences defined form = 1,2, . . . , k andn = 1,2, . . . , r, wherek andr are natural numbers and defineA_m =Pm

s=1a_s, B_n =Pn

t=1b_t,P_m =Pm

s=1p_sandQ_n =Pn

t=1q_t. 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)

p_mΦ am

p_m

α)_α¹

× ( _r

X

n=1

(r−n+ 1)

q_nΨ b_n

q_n

γ)¹_γ , where

M(k, r;α, γ) = 1 α+γ

( _k X

m=1

Φ(Pm) P_m

_α−1^α )^α−1_α ( _r X

n=1

Ψ(Qn) Q_n

_γ−1^γ )^γ−1_γ . 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(A_m) = Φ

PmPm

s=1psas/ps

Pm s=1p_s

(2.12)

≤Φ(P_m)Φ Pm

s=1p_sa_s/p_s Pm

s=1p_s

≤ Φ(Pm) P_m

m

X

s=1

p_sΦ as

p_s

≤ Φ(P_m) P_m m^α−1^α

( _m X

s=1

p_sΦ

a_s p_s

α)_α¹ ,

and similarly,

(2.13) Ψ(B_n)≤ Ψ(Q_n) Q_n n^γ−1^γ

( _n X

t=1

q_tΨ

b_t q_t

γ)¹_γ .

Lets1 =m^α−1, s2 =n^γ−1, ω1 = _α¹, ω2 = ¹_γ andr=ω1+ω2,from (2.7), (2.12) and (2.13), we have

Φ(Am)Ψ(Bn)≤m^α−1^α n

γ−1 γ





Φ(P_m) P_m

( _m X

s=1

psΦ

a_s p_s

α)_α¹ (2.14) 

×





Ψ(Q_n) Q_n

( _n X

t=1

qtΨ

b_t q_t

γ)¹_γ



≤ αγ α+γ

(m

(α−1)(α+γ) αγ

α + n

(γ−1)(α+γ) αγ

γ )

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×





Φ(Pm) P_m

( _m X

s=1

p_sΦ

as

p_s

α)_α¹







Ψ(Qn) Q_n

( _n X

t=1

q_tΨ

bt

q_t

γ)¹_γ



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

γm^{(α−1)(α+γ)}^αγ +αn^{(γ−1)(α+γ)}^αγ

≤ 1 α+γ





Φ(P_m) Pm

( _m X

s=1

p_sΦ

a_s ps

α)_α¹



×





Ψ(Q_n) Qn

( _n X

t=1

q_tΨ

b_t qt

γ)¹_γ



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(A_m)Ψ(B_n) γm

(α−1)(α+γ)

αγ +αn

(γ−1)(α+γ) αγ

≤ 1 α+γ





k

X

m=1

Φ(P_m) Pm

( _m X

s=1

p_sΦ

a_s ps

α)_α¹







r

X

n=1

Ψ(Q_n) Qn

( _n X

t=1

q_tΨ

b_t qt

γ)_γ¹



≤ 1 α+γ

( _k X

m=1

Φ(Pm) P_m

_α−1^α )^α−1_α ( _k X

m=1 m

X

s=1

p_sΦ

as

p_s

α)_α¹

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× ( _r

X

n=1

Ψ(Q_n) Q_n

_γ−1^γ )^γ−1_γ ( _r X

n=1 n

X

t=1

q_tΨ

b_t q_t

γ)¹_γ

= 1

α+γ ( _k

X

m=1

Φ(P_m) P_m

_α−1^α )^α−1_α ( _r X

n=1

Ψ(Q_n) Q_n

_γ−1^γ )^γ−1_γ

× ( _k

X

m=1

(k−m+ 1)

p_sΦ a_s

ps

α)¹_α( _r X

n=1

(r−n+ 1)

q_tΨ b_t

qt

γ)¹_γ .

Remark 3. From the inequality (2.7), we obtain (2.16) s^ω₁¹s^ω₂² ≤ 1

ω1+ω2

ω₁s^ω₁¹^+ω² +ω₂s^ω₂¹^+ω²

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

A^p_mB_n^q 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)(A^p−1_m a_m)^α )^α+γ_αγ

+1 γ

( _r X

n=1

(r−n+ 1)(B_n^q−1b_n)^γ

)^α+γ_αγ 

,

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whereC(p, q, k, r;α, γ) = _α+γ^pq k^α−1^α r^γ−1^γ .Also,

k

X

m=1 r

X

n=1

Φ(A_m)Ψ(B_n) 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)

p_mΦ am

p_m

α)^α+γ_αγ

+ 1 γ

( _r X

n=1

(r−n+ 1)

q_nΨ b_n

q_n

γ)^α+γ_αγ 

,

where

M(k, r;α, γ) = 1 α+γ

( _k X

m=1

Φ(P_m) P_m

_α−1^α )^α−1_α ( _r X

n=1

Ψ(Q_n) Q_n

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

Theorem 2.3. Letα > 1, γ >1and{a_m}and{b_n}be two nonnegative sequences of real numbers defined form = 1,2, . . . , k andn = 1,2, . . . , r, wherekandrare natural numbers and defineA_m = _m¹ Pm

s=1a_sandB_n = _n¹ Pn

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

k

X

m=1 r

X

n=1

mnΦ(A_m)Ψ(B_n) 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)Φ^α(a_m)

)_α¹ ( _r X

n=1

(r−n+ 1)Ψ^γ(b_n) )¹_γ

,

whereC(1,1, k, r;α, γ) = _α+γ¹ k^α−1^α r^γ−1^γ .

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

Φ(A_m) = Φ 1 m

m

X

s=1

a_s

!

≤ 1 m

m

X

s=1

Φ(a_s)≤ 1 mm^α−1^α

( _m X

s=1

Φ^α(a_s) )^α−1_α

,

Ψ(B_n) = Ψ 1 n

n

X

t=1

b_t

!

≤ 1 n

n

X

t=1

Ψ(b_t)≤ 1 nn^γ−1^γ

( _n X

t=1

Ψ^γ(b_t) )^γ−1_γ

. 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, Q_n=Pn

t=1q_t, A_m = _P¹

m

Pm

s=1p_ma_sandB_n = _Q¹

n

Pn

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

k

X

m=1 r

X

n=1

P_mQ_nΦ(A_m)Ψ(B_n) 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) [p_mΦ (a_m)]^α )_α¹

× ( _r

X

n=1

(r−n+ 1) [qnΨ (bn)]^γ )¹_γ

,

whereC(1,1, k, r;α, γ) = _α+γ¹ k^α−1^α r^γ−1^γ .

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

Φ(A_m) = Φ 1 Pm

m

X

s=1

p_sa_s

!

≤ 1 P_m

m

X

s=1

p_sΦ(a_s)≤ 1 P_mm^α−1^α

( _m X

s=1

[p_sΦ(a_s)]^α )^α−1_α

,

Ψ(B_n) = Ψ 1 Q_n

n

X

t=1

q_tb_t

!

≤ 1 Qn

n

X

t=1

q_tΨ(b_t)≤ 1 Qn

n^γ−1^γ ( _n

X

t=1

[q_tΨ(b_t)]^γ )^γ−1_γ

.

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

F^p(s)G^q(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)(F^p−1(s)f(s))^αds _α¹

× Z y

0

(y−t)(G^q−1(t)g(t))^γdt ¹_γ

,

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

F^p(s) = p Z s

0

F^p−1(σ)f(σ)dσ, s∈(0, x), (3.2)

G^q(t) = q Z t

0

G^q−1(τ)g(τ)dτ, t ∈(0, y).

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

Z x 0

F^p−1(σ)f(σ)dσ≤s^α−1^α Z s

0

(F^p−1(σ)f(σ))^αdσ _α¹

, s∈(0, s),

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and (3.4)

Z y 0

G^q−1(t)g(t)dt≤t^γ−1^γ Z t

0

(G^q−1(τ)g(τ))^γdτ _γ¹

, t∈(0, t).

Lets₁ =s^α−1, s₂ =t^γ−1, ω₁ = _α¹, ω₁ = _γ¹, r =ω₁+ω₂,from (3.2) – (3.4) and (2.7), we observe that

F^p(s)G^q(t) (3.5)

≤pqs^α−1^α t^γ−1^γ Z s

0

(F^p−1(σ)f(σ))^αdσ

_α¹ Z t 0

(G^q−1(τ)g(τ))^γdτ _γ¹

≤ pqαγ α+γ

(m^{(α−1)(α+γ)}^αγ

α + n^{(γ−1)(α+γ)}^αγ γ

)

× Z s

0

_α¹ Z t 0

(G^q−1(τ)g(τ))^γdτ ¹_γ

fors∈(0, x), t∈(0, y).From (3.5), we observe that (3.6) F^p(s)G^q(t)

γs

(α−1)(α+γ)

αγ +αt

(γ−1)(α+γ) αγ

≤ pq α+γ

Z s 0

_α¹ Z t 0

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

F^p(s)G^q(t) γs^{(α−1)(α+γ)}^αγ +αt^{(γ−1)(α+γ)}^αγ

dsdt

≤ pq α+γ

"

Z x 0

Z s 0

(F^p−1(σ)f(σ))^αdσ _α¹

ds

# "

Z y 0

Z t 0

dt

#

≤ pq α+γx^α−1^α

Z x 0

Z s 0

(F^p−1(σ)f(σ))^αdσds _α¹

y^γ−1^γ Z y

0

Z t 0

(G^q−1(τ)g(τ))^γdτ dt ¹_γ

= pq

α+γx^α−1^α y^γ−1^γ Z x

0

(x−s)(F^p−1(s)f(s))^αds

¹_αZ t 0

.

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

F^p(s)G^q(t) γs^α−1+αt^γ−1dsdt

≤D(p, q, x, y;α, γ) Z x

0

(x−s)(F^p−1(s)f(s))^αds _α¹

× Z y

0

,

unlessf(σ)≡0org(τ)≡0, whereD(p, q, x, y;α, γ) = _α+γ^pq x^α−1^α y^γ−1^γ .

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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)f^α(s)ds

_α¹ Z y 0

(y−t)g^γ(t)dt ¹_γ

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

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 ¹_γ

, where

L(x, y;α, γ) = 1 α+γ

(Z x 0

Φ(P(s)) P(s)

_α−1^α ds

)^α−1_α ( Z y

0

Ψ(Q(t)) Q(t)

_γ−1^γ dt

)^γ−1_γ .

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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)

dσ

≤ Φ(P(s)) P(s) s^α−1^α

Z s 0

p(σ)Φ

f(σ) p(σ)

α

dσ _α¹

,

and similarly,

(3.10) Ψ(G(t))≤ Ψ(Q(t)) Q(t) t^γ−1^γ

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

dτ ¹_γ

.

Lets₁ =s^α−1, s₂ =t^γ−1, ω₁ = _α¹, ω₁ = _γ¹, r=ω₁ +ω₂,from (3.9), (3.10) and (2.7), we observe that

Φ(F(s))Ψ(G(t))≤s^α−1^α t^γ−1^γ

"

Φ(P(s)) P(s)

Z s 0

p(σ)Φ

f(σ) p(σ)

α

dσ _α¹# (3.11)

×

"

Ψ(Q(t)) Q(t)

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

dτ ¹_γ#

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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

dσ _α¹#

×

"

Ψ(Q(t)) Q(t)

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

dτ ¹_γ#

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

dσ ¹_α#

×

"

Ψ(Q(t)) Q(t)

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

dτ _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

dσ ¹_α

ds

#

×

"

Z y 0

Ψ(Q(t)) Q(t)

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

dτ ¹_γ

dt

#

≤ 1 α+γ

(Z x 0

Φ(P(s)) P(s)

_α−1^α ds

)^α−1_α 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−1^γ dt

)^γ−1_γ Z y

0

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

dτ dt ¹_γ

= 1

α+γ (Z x

0

Φ(P(s)) P(s)

_α−1^α ds

)^α−1_α ( Z y

0

Ψ(Q(t)) Q(t)

_γ−1^γ dt

)^γ−1_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 _γ¹

.

Remark 6. From the inequality (2.7), we obtain (3.13) s^ω₁¹s^ω₂² ≤ 1

ω₁+ω₂ ω₁s^ω₁¹^+ω² +ω₂s^ω₂¹^+ω²

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

F^p(s)G^q(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)(F^p−1(s)f(s))^αds ^α+γ_αγ

+ 1 γ

Z y 0

(y−t)(G^q−1(t)g(t))^γdt ^α+γ_αγ #

,

whereD(p, q, x, y;α, γ) = _α+γ^pq x^α−1^α y^γ−1^γ . 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+γ_αγ

+ 1 γ

Z y 0

(y−t)

q(t)Ψ g(t)

q(t) γ

dt ^α+γ_αγ #

,

where

L(x, y;α, γ) = 1 α+γ







x

Z

0

Φ(P(s)) P(s)

_α−1^α ds







α−1

α (

Z y 0

Ψ(Q(t)) Q(t)

_γ−1^γ dt

)^γ−1_γ . 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) =