Hölder’s inequality

SOME NEW HILBERT-PACHPATTE’S INEQUALITIES

WENGUI YANG

SCHOOL OFMATHEMATICS ANDCOMPUTATIONALSCIENCE

XIANGTANUNIVERSITY, XIANGTAN, 411105 HUNAN, P.R. CHINA

yangwg8088@163.com

Received 05 September, 2008; accepted 2 January, 2009 Communicated by W.S. Cheung

ABSTRACT. 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. INTRODUCTION

Letp≥1, q ≥1and{a_m}and{b_n}be two nonnegative sequences of real numbers defined form = 1,2, . . . , k andn = 1,2, . . . , r, wherek andrare 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−1b_n)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

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

,

031-09

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].

Young-Ho Kim [9] gave new inequalities similar to the Hilbert-Pachpatte inequalities as fol- lows.

Letp ≥1, q ≥1, α >0, and{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 define Am =Pm

s=1asandBn=Pn

t=1bt. 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−1b_n)² )¹₂

,

unless{a_m}or{b_n}is null, whereC(p, q, k, r;α) = ¹_2α¹ pq√

kr.

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

Let p ≥ 1, q ≥ 1, α > 0and f(σ) ≥ 0, g(τ) ≥ 0for σ ∈ (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,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) = ¹_2α¹

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].

2. MAINRESULTS

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 nonnegative sequences of real numbers defined form = 1,2, . . . , k andn = 1,2, . . . , r, wherek andrare natural numbers and defineA_m =Pm

s=1a_sandB_n =Pn

t=1b_t. 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^q−1_n b_n)^γ )¹_γ

,

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

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]), (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 as, m= 1,2, . . . , k, (2.3)

B_n^q ≤q

n

X

t=1

B_t^q−1b_t, n = 1,2, . . . , r.

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

m

X

s=1

A^p−1_s a_s ≤m^α−1α ( _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−1bt≤n

γ−1 γ

( _n X

t=1

(B_t^q−1bt)^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 )¹_r

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 = _γ¹ andr =ω₁ +ω₂,from (2.3) – (2.5) and (2.7), we have

A^p_mB_n^q ≤pqm^α−1α n^γ−1γ ( _m

X

s=1

(A^p−1_s a_s)^α

)¹_α( _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 )

× ( _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 as)^α

)_α¹ ( _n X

t=1

(B_t^q−1bt)^γ )¹_γ

, form = 1,2, . . . , k, n = 1,2, . . . , r.Taking the sum on both sides of (2.9) first overn from1 torand then overmfrom1tok of 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^q−1_t b_t)^γ )_γ¹



≤ pq α+γk^α−1α

( _k X

m=1 m

X

s=1

(A^p−1_s a_s)^α )_α¹

r^γ−1γ ( _r

X

n=1 n

X

t=1

(B_t^q−1b_t)^γ )_γ¹

= pq

α+γk^α−1α r^γ−1γ ( _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 )_γ¹

.

Remark 1. In Theorem 2.1, setting α = γ = 2, we have (1.1). In Theorem 2.1, setting

1

α +_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^q−1_n b_n)^γ )¹_γ

,

unless{a_m}or{b_n}is null, whereC(p, q, k, r;α, γ) = _α+γ^pq k^α−1α r^γ−1γ . Remark 2. In Theorem 2.1, 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{a_m}or{b_n}is null, whereC(1,1, k, r;α, γ) = _α+γ¹ k^α−1α r^γ−1γ .

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{p_m}and{q_n}should be two positive sequences form= 1,2, . . . , k andn= 1,2, . . . , r in Theorem 2.3 of [9].

Theorem 2.2. Letα >1, γ >1and{a_m}and{b_n}be two nonnegative sequences of real num- bers and{p_m}and{q_n}be positive sequences defined form = 1,2, . . . , kandn= 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_s

andQ_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(A_m)Ψ(B_n) 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 a_m

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(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 .

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

Φ(Am) = Φ

P_mPm

s=1p_sa_s/p_s Pm

s=1p_s (2.12)

≤Φ(P_m)Φ Pm

s=1p_sa_s/p_s Pm

s=1p_s

≤ Φ(P_m) P_m

m

X

s=1

p_sΦ a_s

p_s

≤ Φ(P_m) P_m m^α−1α

( _m X

s=1

psΦ

a_s p_s

α)_α¹ ,

and similarly,

(2.13) Ψ(B_n)≤ Ψ(Q_n)

Q_n n^γ−1γ ( _n

X

t=1

q_tΨ

b_t q_t

γ)¹_γ .

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

Φ(A_m)Ψ(B_n) (2.14)

≤m^α−1α n^γ−1γ





Φ(P_m) P_m

( _m X

s=1

p_sΦ

a_s p_s

α)_α¹



×





Ψ(Q_n) Qn

( _n X

t=1

q_tΨ

b_t qt

γ)¹_γ



≤ αγ α+γ

(m

(α−1)(α+γ) αγ

α +n

(γ−1)(α+γ) αγ

γ

)



Φ(P_m) P_m

( _m X

s=1

p_sΦ

a_s p_s

α)_α¹



×





Ψ(Q_n) Q_n

( _n X

t=1

q_tΨ

b_t 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(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)_γ¹



form= 1,2, . . . , k, n= 1,2, . . . , r.Taking the sum on both sides of (2.15) first overnfrom1 torand then overmfrom1tok of 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)¹_γ



≤ 1 α+γ

( _k X

m=1

Φ(P_m) P_m

_α−1^α )^α−1_α ( _k X

m=1 m

X

s=1

p_sΦ

a_s p_s

α)_α¹

× ( _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) Pm

_α−1^α )^α−1_α ( _r X

n=1

Ψ(Q_n) Qn

_γ−1^γ )^γ−1_γ

× ( _k

X

m=1

(k−m+ 1)

p_sΦ a_s

p_s

α)¹_α( _r X

n=1

(r−n+ 1)

q_tΨ b_t

q_t

γ)_γ¹ .

Remark 3. From the inequality (2.7), we obtain

(2.16) s^ω₁¹s^ω₂² ≤ 1 ω1+ω2

ω₁s^ω₁^1+ω2 +ω₂s^ω₂^1+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

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 am)^α )

α+γ αγ

+1 γ

( _r X

n=1

(r−n+ 1)(B_n^q−1b_n)^γ

)^α+γ_αγ 

,

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

pmΦ a_m

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 Theorem 2.2.

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

s=1a_s andB_n = ¹_nPn

t=1b_t. Let ΦandΨbe 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)

× ( _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−1α r^γ−1γ .

Proof. From the hypotheses and by using Jensen’s inequality and Hölder’s inequality, 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−1α

( _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(bt)≤ 1 nn

γ−1 γ

( _n X

t=1

Ψ^γ(bt) )^γ−1_γ

.

The rest of the proof can be completed by following the same steps as in the proofs of Theo- rems 2.1 and 2.2 with suitable changes and hence we omit the details.

Theorem 2.4. Let α > 1, γ > 1 and {a_m} and {b_n} be two nonnegative sequences of real numbers and {p_m} and {q_n} be positive sequences defined for m = 1,2, . . . , k and n = 1,2, . . . , r, wherekandrare natural numbers and defineP_m =Pm

s=1p_s, Q_n =Pn t=1q_t, Am = _P¹

m

Pm

s=1pmas andBn = _Q¹

n

Pn

t=1qnbt. 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}

≤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) [q_nΨ (b_n)]^γ )¹_γ

,

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

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

Φ(A_m) = Φ 1 P_m

m

X

s=1

p_sa_s

!

≤ 1 P_m

m

X

s=1

p_sΦ(a_s)

≤ 1 P_mm^α−1α

( _m X

s=1

[p_sΦ(a_s)]^α )^α−1_α

,

Ψ(B_n) = Ψ 1 Qn

n

X

t=1

q_tb_t

!

≤ 1 Q_n

n

X

t=1

q_tΨ(b_t)

≤ 1 Q_nn^γ−1γ

( _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 Theo- rems 2.1 and 2.2 with suitable changes and hence we omit the details.

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) ≥ 0 for σ ∈ (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 ¹_α

× 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−1α y^γ−1γ .

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−1α Z s

0

(F^p−1(σ)f(σ))^αdσ _α¹

, s ∈(0, s),

and (3.4)

Z y 0

G^q−1(t)g(t)dt ≤t^γ−1γ Z t

0

(G^q−1(τ)g(τ))^γdτ ¹_γ

, t∈(0, t).

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

F^p(s)G^q(t)≤pqs^α−1α t^γ−1γ Z s

0

(F^p−1(σ)f(σ))^αdσ

_α¹ Z t 0

(G^q−1(τ)g(τ))^γdτ _γ¹ (3.5)

≤ pqαγ α+γ

(m^{(α−1)(α+γ)αγ}

α +n^{(γ−1)(α+γ)αγ} γ

)

× Z s

0

(F^p−1(σ)f(σ))^αdσ

_α¹ Z t 0

(G^q−1(τ)g(τ))^γdτ _γ¹

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

(F^p−1(σ)f(σ))^αdσ

_α¹ Z t 0

(G^q−1(τ)g(τ))^γdτ _γ¹

fors ∈ (0, x), t ∈ (0, y).Taking the integral on both sides of (3.6) first over tfrom0toyand then oversfrom0toxof 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

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 ¹_α

ds

# "

Z y 0

Z t 0

(G^q−1(τ)g(τ))^γdτ ¹_γ

dt

#

≤ pq α+γx^α−1α

Z x 0

Z s 0

(F^p−1(σ)f(σ))^αdσds _α¹

y^γ−1γ Z y

0

Z t 0

(G^q−1(τ)g(τ))^γdτ dt ¹_γ

= pq

α+γx^α−1α y

γ−1 γ

Z x 0

(x−s)(F^p−1(s)f(s))^αds

_α¹ Z t 0

(y−t)(G^q−1(t)g(t))^γdt ¹_γ

.

Remark 4. In Theorem 3.1, setting α = γ = 2, we have (1.2). In Theorem 3.1, setting

1

α +_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

¹_αZ 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−1α y^γ−1γ . Remark 5. In Theorem 3.1, 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−1α y

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 and q(τ) > 0 for σ ∈ (0, x), τ ∈ (0, y), wherex, y are positive real numbers. DefineF(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). Let Φand Ψbe two real-valued, nonnegative, convex, and submultiplicative functions defined on R⁺ = [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 . Proof. From the hypotheses ofΦandΨand by using Jensen’s inequality and Hölder’s inequal- ity, 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−1α

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−1γ 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 = ¹_γ, r = ω1 +ω2,from (3.9), (3.10) and (2.7), we observe that

Φ(F(s))Ψ(G(t))≤s^α−1α t^γ−1γ

"

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#

≤ 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 ¹_γ#

fors∈(0, x), t∈(0, y).Taking the integral on both sides of (3.12) first overtfrom0toyand then oversfrom0toxof 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

≤ 1 α+γ

"

Z x 0

Φ(P(s)) P(s)

Z s 0

p(σ)Φ

f(σ) p(σ)

α

dσ _α¹

ds

#

×

"

Z y 0

Ψ(Q(t)) Q(t)

Z t 0

q(τ)Ψ

g(τ) q(τ)

γ

dτ _γ¹

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^ω₁^1+ω2 +ω2s^ω₂^1+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

(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−1α y^γ−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+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 Theorem 3.2. Before we state our next theorem, we point out that “F(s) = Rs

0 f(σ)dσ and G(t) = Rt

0g(τ)dτ” are replaced by “F(s) = ¹_sRs

0 f(σ)dσ andG(t) = ¹_tRt

0 g(τ)dτ” in Theorem 3.4 in [9].

Theorem 3.3. Let α > 1, γ > 1 and f(σ) ≥ 0, g(τ) ≥ 0 for σ ∈ (0, x), τ ∈ (0, y), where x, y are positive real numbers. Define F(s) = ¹_sRs

0 f(σ)dσ, G(t) = ¹_tRt

0 g(τ)dτ for s ∈ (0, x), t ∈ (0, y). LetΦandΨbe two real-valued, nonnegative, convex functions defined onR⁺ = [0,∞).Then

Z x 0

Z y 0

stΦ(F(s))Ψ(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)Φ^α(f(s))ds

_α¹ Z y 0

(y−t)Ψ^γ(g(t))dt ¹_γ

, whereD(1,1, x, y;α, γ) = _α+γ¹ x^α−1α y

γ−1 γ .

Theorem 3.4. Letα >1, γ >1andf(σ)≥0, g(τ)≥0, p(σ)>0andq(τ)>0forσ∈(0, x), τ ∈(0, y), wherex,yare positive real numbers. DefineP(s) =Rs

0 p(σ)dσ,Q(t) =Rt

0q(τ)dτ, F(s) = _P¹_(s)Rs

0 p(σ)f(σ)dσ andG(t) = _Q(t)¹ Rt

0 q(τ)g(τ)dτ for s ∈ (0, x), t ∈ (0, y). Let Φ andΨbe two real-valued, nonnegative, convex functions defined onR+ = [0,∞).Then

Z x 0

Z y 0

P(s)Q(t)Φ(F(s))Ψ(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) [p(s)Φ (f(s))]^αds

¹_αZ y 0

(y−t) [q(t)Ψ (g(t))]^γdt ¹_γ

, whereD(1,1, k, r;α, γ) = _α+γ¹ x^α−1α y

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.

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