A NEW EXTENSION OF HILBERT’S INEQUALITY FOR MULTIFUNCTIONS WITH BEST CONSTANT FACTORS

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A NEW EXTENSION OF HILBERT’S INEQUALITY FOR MULTIFUNCTIONS WITH BEST CONSTANT FACTORS

H. A. AGWO

Abstract. The aim of this paper is to establish a new extension of Hilbert’s inequality and Hardy- Hilbert’s inequality for multifunctions with best constant factors. Also, we present some applications for Hilbert’s inequality which give new integral inequalities.

1. Introduction

Hilbert’s inequality has a great interest in analysis and its applications (see [10], [11]). The original Hilbert’s inequality can be stated as follows

Iff(x), g(x)≥0, such that 0<R∞

0 f²(x)dx <∞and 0<R∞

0 g²(x)dx <∞, then (see [6])

∞

Z

0

∞

Z

0

f(x)g(y)

x+y dxdy < π







∞

Z

0

f²(x)dx

∞

Z

0

g²(x)dx







1 2

, (1)

where the constant factorπis the best possible. This inequality was extended by Hardy-Riesz as (see [5]):

Received May 25, 2008.

2000Mathematics Subject Classification. Primary 26D15, 49C99.

Key words and phrases. Hilbert’s inequality; integral inequalities.

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Ifp >1, ¹_p +¹_q = 1, f(x),g(x)≥0, such that 0<R∞

0 f^p(x)dx <∞and 0<R∞

0 g^q(x)dx <∞, then

Z ∞ 0

f(x)g(y)

x+y dxdy < π sin(^π_p)

Z ∞ 0

f^p(x)dx

¹_pZ ∞ 0

g^q(x)dx ¹_q

, (2)

where the constant factor _sin(^ππ

p) is the best possible.

Hardy-Hilbert’s integral inequality is important in analysis and its applications (see[10], [11]).

In recent years, the various improvements and extensions on the inequality (1) and (2) appeared in some papers (such as [1]–[4], [7], [9], [12]–[14]) and bibliography therein. They focalize on changing the denominator of the function of the left-hand side of (2). Such as the denominator (x+y) is replaced by (Ax+By)^λ in paper [13], the denominator (x+y) is replaced by (x^t+y^t) (tis a parameter which is independent ofxandy) in paper [7]. Generally, the denominator (x+y) is replaced by (xu(x) +yv(y))^λin paper [9].

The main objective of this paper is to build some new Hilbert-type integral inequalities with best constant factors which are extensions of above results for multi-functionsf, gandh. Moreover the denominator is (m(x) +n(y) +r(z)), wherem, nandrare arbitrary functions.

2. Main results

We need the formula of theβ function as (see [8]):

β(u, v) =

∞

Z

0

t^u−1

(1 +t)^u+vdt=β(v, u) u, v >0.

(3)

Before stating our results we need the following lemmas.

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Lemma 2.1. Let f(x, y, z)∈L^p[0,∞]×[0,∞]×[0,∞] andg(x, y, z)∈L^q_[0,∞]_×_[0,∞]×_[0,∞], where ¹_p+¹_q = 1. Then

∞

Z

0

∞

Z

0

∞

Z

0

|f(x, y, z)g(x, y, z)|dxdydz

≤





∞

Z

0

∞

Z

0

∞

Z

0

|f(x, y, z)|^pdxdydz





1 p



∞

Z

0

∞

Z

0

∞

Z

0

|g(x, y, z)|^qdxdydz





1 q

. (4)

Lemma 2.2. Letf(x, y, z)∈L^p[0,∞]×[0,∞]×[0,∞],g(x, y, z)∈L^q[0,∞]×[0,∞]×[0,∞] and h(x, y, z)∈ L^k[0,∞]×[0,∞]×[0,∞] where 1

p+1 q +1

k = 1. Then

∞

Z

0

∞

Z

0

∞

Z

0

|f(x, y, z)g(x, y, z)h(x, y, z)|dxdydz

≤





∞

Z

0

∞

Z

0

∞

Z

0

|f(x, y, z)|^pdxdydz





1 p

×





∞

Z

0

∞

Z

0

∞

Z

0

|g(x, y, z)|^qdxdydz





1 q

×





∞

Z

0

∞

Z

0

∞

Z

0

|h(x, y, z)|^rdxdydz





1 k

. (5)

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Lemma 2.3. Ifp1>1, 1 p₁ + 1

q₁ = 1,p >1, 1 p+1

q +1

k = 1then for0< ε < 1

qk, we have

∞

Z

0

v^p¹¹^qk⁻^p^ε¹⁻¹ (1 +v)^qk¹

dv=β 1

p1qk, 1 q1qk

+ 0(1) ε→0⁺. (6)

Proof. Since

∞

Z

0

v^p¹¹^qk⁻^p^ε¹⁻¹ (1 +v)^qk¹

dv−β 1

p₁qk, 1 q₁qk

=

∞

Z

0

v^p¹¹^qk⁻^p^ε¹⁻¹−v^p¹¹^qk⁻¹ (1 +v)^qk¹

dv

≤

1

Z

0

v^p¹¹^qk⁻^p^ε¹⁻¹−v^p¹¹^qk⁻¹ (1 +v)^qk¹

dv+

∞

Z

1

v^p¹¹^qk⁻^p^ε¹⁻¹−v^p¹¹^qk⁻¹ (1 +v)^qk¹

dv

≤

1

Z

0

v^p¹¹^qk⁻^p^ε¹⁻¹−v^p¹¹^qk⁻¹ dv+

∞

Z

1

v^p¹¹^qk⁻¹−v^p¹¹^qk⁻^p^ε¹⁻¹ v^qk¹ dv

= 1

1 p1qk− ε

p1

− 1

1 p1qk

+ −1 1 q1qk

+ 1

1 q1qk− ε

q1

→0 for ε→0.

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Lemma 2.4. Ifp1>1, 1 p₁ + 1

q₁ = 1,p >1, 1 p+1

q +1

k = 1and0< ε < 1

qk, setting

J₁:=

∞

Z

1

∞

Z

1

n(y) m(x) +n(y)

_qk¹

(m(x))^p¹¹^qk⁻^p^ε¹⁻¹

×(n(y))⁻^p¹¹^qk⁻^q^ε¹⁻¹dm(x) dx

dn(y) dy dxdy, then we have

1 ε

β

1 p1qk, 1

q1qk

+o(1)

−O(1)

≤J₁≤ 1 ε

β

1 p₁qk, 1

q₁qk

+o(1)

, ε→0⁺. (7)

Proof. For fixedy,settingm(x) =n(y)v,then by (6), we obtain J₁ =

∞

Z

1

(n(y))⁻^p¹¹^qk⁻^q^ε¹⁻¹dn(y) dy

×





∞

Z

1

n(y) m(x) +n(y)

_qk¹

(m(x))^p¹¹^qk⁻^p^ε¹⁻¹dm(x) dx dx



dy

=

∞

Z

1

(n(y))^−ε−1dn(y) dy







∞

Z

1 n(y)

v^p¹¹^qk⁻^p^ε¹⁻¹ (1 +v)^qk¹ dv





 dy

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=

∞

Z

1





∞

Z

0

v^p¹¹^qk⁻^p^ε¹⁻¹ (1 +v)^qk¹

dv



dy

−

∞

Z

1







1 n(y)

Z

0

v^p¹¹^qk⁻^p^ε¹⁻¹ (1 +v)^qk¹

dv





dy

≥ 1 ε

β

1 p1qk, 1

q1qk

+ 0(1)

−

∞

Z

1





1

Z

0

v^p¹¹^qk⁻^p^ε¹⁻¹dv



dy

= 1

ε

β 1

p₁qk, 1 q₁qk

+ 0(1)

−1 ε

1 1

p₁qk− ε p₁

= 1

ε

β 1

p₁qk, 1 q₁qk

+o(1)

−O(1).

By the same way, we have J1 ≤

∞

Z

1

∞

Z

0

n(y) m(x) +n(y)

_qk¹

(m(x))^p¹¹^qk⁻^p^ε¹⁻¹

×(n(y))⁻^p¹¹^qk⁻^q^ε¹⁻¹dm(x) dx

dn(y) dy dxdy,

= 1

ε

β 1

p1qk, 1 q1qk

+o(1)

.

The lemma is proved.

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Theorem 2.5. Assume thatm,n,rare increasing functions defined on[0,∞[such thatm(0) = n(0) =r(0) = 0, lim

x→∞m(x) = lim

x→∞n(x) = lim

x→∞r(x) =∞, andf,g,hsatisfy

∞

Z

0

(m(x))

q1

p1(1+_pq¹) dm(x)

dx

−q1(^p_k+_p¹

1)

|f(x)|^kq²¹ dx <∞ (8)

∞

Z

0

(m(x))

p1

q1−_qk¹ dm(x) dx

^−p_q¹

1 |f(x)|^pp²¹ dx <∞, (9)

∞

Z

0

(n(y))^p^q¹¹⁽¹⁺^qk¹⁾

dn(y) dy

−q1(^p_k+_p¹

1)

|g(y)|

pq1

2 dy <∞, (10)

∞

Z

0

(n(y))

p1

q1−_pk¹ dn(y) dy

^−p_q₁¹

|g(y)|^qp²¹dy <∞, (11)

∞

Z

0

(r(z))

q1

p1(1+_pk¹)dr(z) dz

−q1(_k^p+_p¹

1)

|h(z)|^qq²¹ dz <∞, (12)

∞

Z

0

(r(z))

p1 q1−_pk¹

dr(z) dz

^−p_q¹

1 |h(z)|^kp²¹ dz <∞, (13)

where 1 p+1

q +1

k = 1, 1 p1

+ 1 q1

= 1, then

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∞

Z

0

∞

Z

0

∞

Z

0

|f(x)g(y)h(z)|

m(x) +n(y) +r(z)dxdydz≤ π sin_qk^π β

1 q₁qk, 1

p₁qk !¹_p

×





∞

Z

0

(m(x))

p1

q1−_qk¹ dm(x) dx

−p1 q1

|f(x)|^pp²¹ dx





1 pp1

×





∞

Z

0

(n(y))^q^p¹¹⁽¹⁺^qk¹⁾

dn(y) dy

−q1(^p_k+_p¹

1)

|g(y)|

pq1 2 dy





1 pq1

× π

sin_pk^π β 1

q1pk, 1 p1pk

!¹_q

×





∞

Z

0

(n(y))^p^q¹¹⁻^pk¹

dn(y) dy

^−p_q¹

1 |g(y)|^qp²¹ dy





1 qp1

×





∞

Z

0

(r(z))

q1

p1(1+_pk¹)dr(z) dz

−q1(^p_k+_p¹

1)

|h(z)|^qq²¹ dz





1 qq1

× π

sin_pq^π β 1

q1pq, 1 p1pq

!¹_k



∞

Z

0

(r(z))

p1 q1−_pk¹

dr(z) dz

^−p_q¹

1 |h(z)|^kp²¹ dz





1 kp1

×





∞

Z

0

(m(x))

q1

p1(1+_pq¹) dm(x)

dx

−q1(^p_k+_p¹

1)

|f(x)|^kq²¹ dx





1 kq1

, (14)

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where the constant factors π

sin_qk^π β 1

q1qk, 1 p1qk

, π

sin_pk^π β 1

q1pk, 1 p1pk

, π

sin_pq^π β 1

q1pq, 1 p1pq

are best possible.

Proof. Since I=

∞

Z

0

∞

Z

0

∞

Z

0

|f(x)g(y)h(z)|

m(x) +n(y) +r(z)dxdydz

=

∞

Z

0

∞

Z

0

∞

Z

0

|f(x)|¹² |g(y)|¹² (m(x) +n(y) +r(z))¹^p

m(x) +n(y) r(z)

_pqk¹ n(y) m(x) +n(y)

_pqk¹ dr(z) dz

¹_pdn(y) dy

⁻¹_k

× |g(y)|¹² |h(z)|¹² (m(x) +n(y) +r(z))¹^q

n(y) +r(z) m(x)

_pqk¹ r(z) n(y) +r(z)

_pqk¹ dm(x) dx

¹_qdr(z) dz

⁻¹_p

× |f(x)|¹² |h(z)|¹² (m(x) +n(y) +r(z))¹^k

r(z) +m(x) n(y)

_pqk¹ m(x) r(z) +m(x)

_pqk¹

×

dn(y) dy

¹_kdm(x) dx

⁻¹_q

dxdydz.

(15)

Applying H¨older’s inequality on (15) we get I≤I

1 p

1I

1 q

2I

1 k

(16) 3

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where

I1=

∞

Z

0

∞

Z

0

∞

Z

0

|f(x)|^p² |g(y)|^p² m(x) +n(y) +r(z)

m(x) +n(y) r(z)

_qk¹ n(y) m(x) +n(y)

_qk¹

×dr(z) dz

dn(y) dy

^−p_k

dxdydz, (17)

I2=

∞

Z

0

∞

Z

0

∞

Z

0

|g(y)|^q² |h(z)|^q² m(x) +n(y) +r(z)

n(y) +r(z) m(x)

_pk¹ r(z) n(y) +r(z)

_pk¹

×dm(x) dx

dr(z) dz

^−q_p

dxdydz, (18)

I3=

∞

Z

0

∞

Z

0

∞

Z

0

|f(x)|^k² |h(z)|^k² m(x) +n(y) +r(z)

r(z) +m(x) n(y)

_pq¹

m(x) r(z) +m(x)

_pq¹

×dn(y) dy

dm(x) dx

^−k_q

dxdydz.

(19)

Consider the weight coefficient w1(x, y) =

Z ∞ 0

1

m(x) +n(y) +r(z)

m(x) +n(y) r(z)

_qk¹ dr(z) dz dz.

(20)

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Letv= r(z)

m(x) +n(y) in (20) then we obtain

w₁(x, y) = π sin π

qk . (21)

Similarly,

w₂(y, z) =

∞

Z

0

1

m(x) +n(y) +r(z)

n(y) +r(z) m(x)

_pk¹

dx= π sin π

pk (22)

and

w3(z, x) =

∞

Z

0

1

m(x) +n(y) +r(z)

r(z) +m(x) n(y)

_pq¹

dx= π sin π

pq . (23)

Combining (21), (22), (23) and (16) we get

I≤



 π sin_qk^π

∞

Z

0

∞

Z

0

n(y) m(x) +n(y)

_qk¹

|f(x)|^p² |g(y)|^p²

dn(y) dy

^−p_k dxdy





1 p

×



 π sin_pk^π

∞

Z

0

∞

Z

0

r(z) n(y) +r(z)

_pk¹

|g(y)|^q² |h(z)|^q²

dr(z) dz

^−q_p dxdy





1 q

×



 π sin_pq^π

∞

Z

0

∞

Z

0

m(x) r(z) +m(x)

_pq¹

|f(x)|^k² |h(z)|^k²

dm(x) dx

^−k_q dxdy





1 k

. (24)

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Applying H¨older’s inequality with p1 > 1,_p¹

1 +_q¹

1 = 1 on the first integral on the right side in (24),we have

∞

Z

0

∞

Z

0

n(y) m(x) +n(y)

_qk¹

|f(x)|^p²|g(y)|^p²

dn(y) dy

^−p_k dxdy

=

∞

Z

0

∞

Z

0

|f(x)|^p² _n(y)

m(x)

_p ¹

1q1qk−_p¹

1 m(x)^q¹¹⁻^p¹¹ (m(x) +n(y))^p¹¹^qk

dn(y) dy

_p¹

1

dm(x) dx

⁻¹_q

1

×|g(y)|^p² (n(y))^qk¹⁺^p¹¹⁻^q¹¹ (m(x) +n(y))^q¹¹^qk

m(x) n(y)

_p₁_q¹₁_qk−_q¹

1 dn(y) dy

^−p_k−_p¹

1 dm(x) dx

_q¹₁ dydx

≤





∞

Z

0

∞

Z

0

|f(x)|^pp²¹ (m(x) +n(y))^qk¹

n(y) m(x)

_q₁¹_qk−1

(m(x))

p1

q1−1dm(x) dx

^−p_q₁¹ dn(y) dy dydx





1 p1

×





∞

Z

0

∞

Z

0

|g(y)|^pq²¹ (n(y))

q1 qk+^q_p¹

1−1

(m(x) +n(y))^qk¹

m(x) n(y)

_p¹

1qk−1 dn(y)

dy

−q1(^p_k+_p¹

1)

dm(x) dx dydx





1 q1

. (25)

Let

w₄(x) =

∞

Z

0

1

1 + n(y) m(x)

_qk¹

n(y) m(x)

_q₁¹_qk−1

dn(y) dy dy (26)

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Puttingv1= n(y)

m(x) in (26) we have w4(x) =m(x)β

1 q₁qk, 1

qk − 1 q₁qk

=m(x)β 1

q₁qk, 1 p₁qk

. (27)

Similarly

w₅(y) =

∞

Z

0

1

1 +m(x) n(y)

_qk¹

m(x) n(y)

_p₁¹_qk−1

dm(x) dx dx

=n(y)β 1

q1qk, 1 p1qk

. (28)

From (27), (28) and (25) we get

∞

Z

0

∞

Z

0

n(y) m(x) +y

_qk¹

|f(x)|^p² |g(y)|^p²

dn(y) dy

^−p_k dxdy

≤β 1

q1qk, 1 p1qk





∞

Z

0

(m(x)

p1 q1−_qk¹

dm(x) dx

^−p_q¹

1 |f(x)|^pp²¹ dx





1 p1

×





∞

Z

0

(n(y))

q1

p1(1+_qk¹)dn(y) dy

−q1(^p_k+_p¹

1)

|g(y)|^pq²¹ dy





1 q1

. (29)

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For the second and third integrals on the right side of (24), following the same steps used for obtaining (29), we obtain

∞

Z

0

∞

Z

0

r(z) n(y) +r(z)

_pk¹

|g(y)|^q² |h(z)|^q²

dr(z) dz

^−q_p dydz

≤β 1

q₁qk, 1 p₁qk





∞

Z

0

(n(y))

p1

q1−_pk¹ dn(y) dy

−p1 q1

|g(y)|^qp²¹ dy





1 p1

×





∞

Z

0

(r(z))

q1

p1(1+_pk¹)dr(z) dz

−q1(_k^p+_p¹

1)

|h(z)|^qq²¹ dz





1 q1

, (30)

and

∞

Z

0

∞

Z

0

m(x) r(z) +m(x)

_pq¹

|h(z)|^k² |f(x)|^k²

dm(x) dx

^−k_q dzdx

≤β 1

q1pq, 1 p1pq





∞

Z

0

(r(z))

p1 q1−_pk¹

dr(z) dz

^−p_q¹

1 |h(z)|^kp²¹ dz





1 p1

×





∞

Z

0

(m(x))

q1

p1(1+_pq¹)dm(x) dx

−q1(^p_k+_p¹

1)

|f(x)|^kq²¹ dx





1 q1

. (31)

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Let

γ1=β 1

q₁qk, 1 p₁qk

, γ2=β 1

q₁qk, 1 p₁qk

, γ3=β 1

q₁pq, 1 p₁pq

.

To prove the constant factorsγ1, γ2 andγ3 are best possible. Assume that the constant factor γ1 is not the best possible, then there exists a positive constantλ1 with λ1 < γ1, such that (29) is still valid if we replaceγ1 byλ1. Without loss of generality, we assume thatm(1) =n(1) = 1.

For 0< ε <1,settingfεandgε asfε(x) =gε(x) = 0,forx∈(0,1), and forx∈[1,∞)

|fε(x)|= (m(x))

2 pp1

“−^p_q¹

1+_qk¹−ε−1” dm(x)

dx _pp²

1

“_p

q11+1”

|gε(x)|= (n(x))^pq²¹

“−^q_p¹

1(¹⁺qk¹)^−ε−1^”dn(x) dx

_pq²₁

“ q₁“_p

k+_p¹

1

” +1”

,

then we obtain λ1





∞

Z

0

(m(x))

p1

q1−_qk¹ dm(x) dx

^−p_q¹

1 |f(x)|^pp²¹dx_p¹₁

×





∞

Z

0

(n(y))

q1

p1(1+_qk¹) dn(y)

dy

−q1(^p_k+_p¹

1)

|g(y)|^pq²¹ dy_q¹₁

=λ1





∞

Z

1

(m(x))^−ε−1dm(x) dx dx





1 p1 



∞

Z

1

(n(y))^−ε−1dn(y) dy dy





1 q1

= λ1

ε .

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But we have

∞

Z

0

∞

Z

0

n(y) m(x) +y

_qk¹

|f(x)|^p² |g(y)|^p²

dn(y) dy

^−p_k dxdy

=

∞

Z

1

∞

Z

1

n(y) m(x) +y

_qk¹

(m(x))^p¹¹^qk⁻^p^ε¹⁻¹(n(y))⁻^p¹¹^qk⁻^q^ε¹⁻¹ dm(x)

dx

dn(y) dy

dxdy=J1

≥1

ε(γ1+o(1))−O(1).

Hence we find

1

ε(γ1+o(1))−O(1)< λ1

(32) ε or

γ1+o(1)−εO(1)< λ1. (33)

Forε→0⁺, it follows that γ1≤λ1. This contradicts the fact thatλ1< γ1. Hence the constant factorγ1 in (29) is the best possible. Similarlyγ2 andγ3are the best possible. Substituting from

(29), (30), (31) in (24), the result of the theorem follows.

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Remark 1. Puttingp=q=k=¹₃ in (14), we get a new inequality in the form

∞

Z

0

∞

Z

0

∞

Z

0

|f(x)g(y)h(z)|

≤ π

sin^π₉β 1

9q₁, 1 9p₁





∞

Z

0

(m(x))

p1

q1−¹₉dm(x) dx

−p1 q1

|f(x)|^3p²¹dx





1 3p1

×





∞

Z

0

(n(y))

10q1 9p1

dn(y) dy

−q1(1+_p¹

1)

|g(y)|^3q²¹ dy





1 3q1

×





∞

Z

0

(n(y))

p1

q1−¹₉dn(y) dy

−p1 q1

|g(y)|^3p²¹ dy





1 3p1

×





∞

Z

0

(r(z))^10q^9p¹¹

dr(z) dz

−q1(1+_p¹

1)

|h(z)|

3q1 2 dz





1 3q1

×





∞

Z

0

(r(z))

p1 q1−¹₉

dr(z) dz

^−p_q¹

1|h(z)|^kp²¹ dz





1 3p1

×





∞

Z

0

(m(x))

10q1 9p1

dm(x) dx

−q1(1+_p¹

1)

|f(x)|^3q²¹ dx





1 3q1

. (34)

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Remark 2. Puttingp1=q1= 2 in (14) we obtain

∞

Z

0

∞

Z

0

∞

Z

0

|f(x)g(y)h(z)|

≤ π

sin_qk^π β 1

2qk, 1 2qk

!_p¹



∞

Z

0

(m(x))¹⁻^qk¹

dm(x) dx

−1

|f(x)|^qdx





1 2p

×





∞

Z

0

(n(y))⁽¹⁺^qk¹⁾

dn(y) dy

−2(^p_k+¹₂)

|g(y)|^pdy





1 2p

× π

sin_pk^π β 1

2pk, 1 2pk

!¹_q



∞

Z

0

(n(y))¹⁻^pk¹

dn(y) dy

^−p_q¹

1 |g(y)|^pdy





1 2q

×





∞

Z

0

(r(z))⁽¹⁺^pk¹⁾

dr(z) dz

−2(^p_k+¹₂)

|h(z)|^qdz





1 2q

(35)

× π

sin_pq^π β 1

2pq, 1 2pq

!_k¹



∞

Z

0

(r(z))¹⁻^pk¹

dr(z) dz

−1

|h(z)|^kdz





1 2k

×





∞

Z

0

(m(x))⁽¹⁺^pq¹⁾

dm(x) dx

−2(^p_k⁺¹₂)

|f(x)|^kdx





1 2k

,

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which is a new inequality.

Remark 3. Letm(x) =x,n(y) =y andr(z) =z in (34) and (35) we get respectively

∞

Z

0

∞

Z

0

∞

Z

0

|f(x)g(y)h(z)|

x+y+z dxdydz

≤ π

sin^π₉β 1

9q₁, 1 9p₁









∞

Z

0

x

p1 q1−¹₉

|f(x)|^3p²¹ dx









∞

Z

0

y

p1 q1−¹₉

|g(y)|^3p²¹ dy





×





∞

Z

0

z^p^q¹¹⁻¹⁹|h(z)|

3p1 2 dz









1 3p1

×









∞

Z

0

x

10q1

9p1 |f(x)|^3q²¹ dx









∞

Z

0

y

10q1

9p1 |g(y)|^3q²¹ dy





×





∞

Z

0

z

10q1

9p1 |h(z)|^3q²¹ dz









1 3q1

.

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and

∞

Z

0

∞

Z

0

∞

Z

0

|f(x)g(y)h(z)|

x+y+z dxdydz

≤ π

sin_qk^π β 1

2qk, 1 2qk

!¹_p







∞

Z

0

x¹⁻^qk¹ |f(x)|^pdx









∞

Z

0

y(¹⁺_qk¹)|g(y)|^pdy









1 2p

× π

sin_pk^π β 1

2pk, 1 2pk

!¹_q







∞

Z

0

y¹⁻^pk¹ |g(y)|^qdy









∞

Z

0

z¹⁺^pk¹ |h(z)|^qdz









1 2q

× π

sin_pq^π β 1

2pq, 1 2pq

!¹_k







∞

Z

0

z¹⁻^pk¹ |h(z)|^kdz









∞

Z

0

x¹⁺^pq¹ |f(x)|^kdx









1 2k

.

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7. Kuang Jichang,On new extensions of Hilbert’s integral inequality, J. Math. Anal. Appl.,235(2) (1999), 608–

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12. Peˇcari´c J. E.,Generalization of inequalities of Hardy-Hilbert type, Math. Inequal. Appl.7(2) (2004), 217–225.

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H. A. Agwo, Department of Mathematics, Faculty of Education, Ain Shams University, Roxy, Cairo, Egypt,e-mail:

Hassanagwa@yahoo.com