R∞ 0 fp(t)dt &lt

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Volume 5, Issue 2, Article 25, 2004

ON HARDY-HILBERT’S INTEGRAL INEQUALITY

W.T. SULAIMAN MOSULUNIVERSITY

COLLEGE OFMATHEMATICS ANDCOMPUTERSCIENCE

[email protected]

Received 17 April, 2003; accepted 18 January, 2004 Communicated by L. Pick

ABSTRACT. In the present paper, by introducing some parameters, new forms of Hardy-Hilbert’s inequalities are given.

Key words and phrases: Hardy-Hilbert’s integral inequality.

2000 Mathematics Subject Classification. 26D15.

1. INTRODUCTION

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

0 f^p(t)dt <∞, and0<R∞

0 g^q(t)dt <∞, then, the Hardy-Hilbert integral inequality is given by:

(1.1)

Z ∞ 0

Z ∞ 0

f(x)g(y)

x+y dxdy≤ π sin^π_p

Z ∞ 0

f^p(t)dt

¹_pZ ∞ 0

g^q(t)dt ¹_q

, where the constant _sin^ππ

p

is the best possible (see [1]).

Yang [2] and [3] gave the following generalization of (1.1) (1.2)

Z ∞ 0

Z ∞ 0

f(x)g(y)

(x+y−2α)^λdxdy

≤K

1 p

λ(p)K

1 q

λ(q) Z ∞

0

(t−α)^1−λf^p(t)dt

¹_pZ ∞ 0

(t−α)^1−λg^q(t)dt ¹_q

, where

K_λ(r) = Z ∞

0

u^1r−1

(1 +u)^λdu=B 1

r, λ−1 r

0< λ≤1, λ > 1 r >0, B is the beta function defined by

B(p, q) = Z 1

0

x^p−1(1−x)^q−1dx, p, q >0

ISSN (electronic): 1443-5756

c 2004 Victoria University. All rights reserved.

053-03

and, for0< T <∞,

(1.3) Z T

0

Z T 0

f(x)g(y) (x+y)^λdxdy

≤β λ

2,λ 2

Z T α

"

1−1 2

t T

^λ₂#

t^1−λf²(t)dt

!¹₂

× Z T

α

"

1−1 2

t T

^λ₂#

t^1−λg²(t)dt

!¹₂ .

In the present paper, by introducing some parameters, new forms of Hardy-Hilbert’s inequalities are given.

2. NEWRESULTS

We state and prove the following:

Lemma 2.1. Letλ >0, p, q, r >1, ¹_p +¹_q + ¹_r = 1,f(t), g(t), h(t)≥0, λ_i(t)≥ 0,i=p, q, r, and assume that

0<

Z b a

λ^p_p(t)f^p(t)dt <∞, 0<

Z d c

λ^q_q(t)g^q(t)dt <∞

and

0<

Z d c

λ^r_r(t)h^r(t)dt <∞.

Then the two following inequalities are equivalent

(2.1) Z b

a

Z d c

Z k e

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

h^λ(x, y, z) dxdydz

≤K Z b

a

λ^p_p(t)f^p(t)dt

1

pZ d

c

λ^q_q(t)g^q(t)dt

1

q Z k

e

λ^r_r(t)h^r(t)dt

1 r

,

whereK =K(λ, p, q, r)is a constant, and

(2.2) Z b

a

λ

qr

pq+r(x) Z d

c

Z k e

g(y)h(z) h^λ(x, y, z)dydz

qr q+r

dx

≤K

qr q+r

Z d c

λ^q_q(t)g^q(t)dt

q+r^r Z k e

λ^r_r(t)h^r(t)dt

q q+r

.

Proof. Suppose (2.2) is satisfied, then Z b

a

Z d c

Z k e

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

h^λ(x, y, z) dxdydz

= Z b

a

λp(x)f(x)

λ⁻¹_p (x) Z d

c

Z k e

g(y)h(z) h^λ(x, y, z)dydz

dx

≤ Z b

a

λ^p_p(x)f^p(x)dx

¹p Z b a

λ

qr

pq+r

Z d c

Z k e

g(y)h(z) h^λ(x, y, z)dydz

qr q+r

dx

!

q+r qr

≤K Z b

a

λ^p_p(t)f^p(t)dt

¹pZ d c

λ^q_q(t)g^q(t)dt

¹q Z k e

λ^r_r(t)h^r(t)dt ¹r

. Now, suppose that (2.1) is satisfied, then

Z b a

λ

qr q+r

p

Z d c

Z k e

g(y)h(z) h^λ(x, y, z)dydz

qr q+r

dx

= Z b

a

Z d c

Z k e

g(y)h(z) h^λ(x, y, z).λ

qr

pq+r

Z d c

Z k e

g(y)h(z) h^λ(x, y, z)dydz

qr q+r−1

dxdydz

≤K Z d

c

λ^q_q(y)g^q(y)dy

1

q Z k

e

λ^r_r(z)h^r(z)dz

1 r

× Z b

a

λ^p_p(x)λ^−p

qr

p q+r(x) Z d

c

Z k e

g(y)h(z) h^λ(x, y, z)dydz

^p(

qr q+r−1)

dx

!¹_p

=K Z d

c

λ^q_q(y)g^q(y)dy

¹q Z k e

λ^r_r(z)h^r(z)dz ¹r

× Z b

a

λ

qr q+r

p (x) Z d

c

Z k e

g(y)h(z) h^λ(x, y, z)dydz

qr q+r

dx

!¹_p ,

therefore Z b

a

λ

qr

pq+r(x) Z d

c

Z k e

g(y)h(z) h^λ(x, y, z)dydz

qr q+r

dx

!¹_q+¹_r

≤k Z d

c

λ^q_q(t)g^q(t)dt

1

q Z k

e

λ^r_r(t)h^r(t)dt

1 r

,

and the desired equivalence is proved.

Lemma 2.2. (a) Let0≤y≤1, α >0,f ≥0. If we define the function g(y) =y^−α

Z y 0

f(x)dx, theng(y)≥g(1).

(b) Lety≥1, α >0, f ≥0.Defining the function, h(y) = y^−α

Z y 0

f(x)dx, we haveh(y)≥h(1).

Proof. (a) Letx= ¹_t,then

g(y) =y^−α Z ∞

y⁻¹

f ¹_t dt t² . We observe also that

g⁰(y) =y^−α

−y²f(y) +

Z ∞ y⁻¹

f ¹_t t² dt

!

(−α)y^−α−1 ≤0, thereforeg is non-increasing, which impliesg(y)≥g(1).

(b) We obviously have

h⁰(y) = y^αf(y) + Z y

0

f(x)dx

αy^α−1 ≥0, thereforehis non-decreasing, and henceh(y)≥h(1).

The following result may be stated as well.

Theorem 2.3. Let f(t), g(t), h(t) ≥ 0, p, q, r > 1, ¹_p + ¹_q + ¹_r = 1, 2 < λ < 3, γ >

µmax{p, q, r},and max

−1 p,−1

q,−1 r

< µ <min

λ−1

p ,λ−1

q ,λ−1 r

.

0<

Z T α

(t−α)^2−λf^p(t)dt < ∞, 0<

Z T α

(t−α)^2−λg^q(t)dt <∞ and

0<

Z T α

(t−α)^2−λh^r(t)dt <∞, then

(2.3) Z T

α

Z T α

Z T α

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

(x+y+z)^λ dxdydz

≤ Z T

α

φ(t, µ, λ, p)(t−α)^2−λf^p(t)dt ¹_p

× Z T

α

φ(t, µ, λ, q)(t−α)^2−λg^q(t)dt

¹q Z T α

φ(t, µ, λ, r)(t−α)^2−λh^r(t)dt ¹r

, where

φ(t, µ, λ, j)

=B(λ−µj−1, µj+1)

B(λ−2,1−µj)−(t−α T −α)^γ

Z 1 0

u^λ−3

(1 +u)^λ−µj−1du

−

t−α T −α

2γ

× Z 1

0

u^λ−µj−2 (1 +u)^λdu

Z 1 0

u^γ−µj−2

(1 +u)^λ−µj−1du, j =p, q, r.

Proof. The proof is as follows. We have Z T

α

Z T α

Z T α

f(x)g(x)h(z)

(x+y+z)^λ dxdydz

= Z T

α

Z T α

Z T α

f(x)

z−α y−α

µ

(x+y+z)^λ/p

g(y) ^x−α_z−αµ

(x+y+z)^λ/q

h(z) ^y−α_x−αµ

(x+y+z)^λ/rdxdydz

≤



 Z T

α

Z T α

Z T α

f^p(x)

z−α y−α

µp

(x+y+z)^λ dxdydz





1 p

Z T α

Z T α

Z T α

g^q(y) ^x−α_z−αµq

(x+y+z)^λ dxdydz

!¹_q

× Z T

α

Z T α

Z T α

h^r(z) ^y−α_x−αµr

(x+y+z)^λ dxdydz

!¹_r

=F^1pG^1qH^1r, say.

Then we have

F = Z T

α

(x−α)^2−λf^p(x)dx Z T

α

y−α x−α

−µp

(1 + ^y−α_x−α)^λ−µp−1 dy x−α

Z T α

z−α x+y−2α

−µp

(1 + _x+y−2α^z−α )^λ

dz x+y−2α. Now by Lemma 2.2, we can state that

Z T 0

z−α x+y−2α

µp

(1 + _x+y−2α^z−α )^λ

dz x+y−2α

=

Z _x+y−2α^T−α

0

u^µp (1 +u)^λdu

≤ Z ^T−α_y−α

0

u^µp (1 +u)^λdu

= Z ∞

y−α T−α

u^λ−µp−2 (1 +u)^λdu

= Z ∞

0

−

y−α T −α

γ y−α T −α

−γZ _T−α^y−α

0

! u^λ−µp−2 (1 +u)^λdu

≤

B(λ−µp−1, µp+ 1)−

y−α T −α

γZ 1 0

u^λ−µp−2 (1 +u)^λdu

.

Therefore F ≤

Z T α

(x−α)^2−λf^p(x)dx Z T

α

y−α x−α

−µp

(1 + _x−α^y−α)^λ−µp−1 dy x−α

×

B(λ−µp−1, µp+ 1)−

y−α T −α

γZ 1 0

u^λ−µp−2 (1 +u)^λdu

= Z T

α

(x−α)^2−λf^p(x)dx Z ^T−α_x−α

0

u^−µp

(1 +u)^λ−µp−1du

×

B(λ−µp−1, µp+ 1)−

x−α T −α

γ

u^γ Z 1

0

u^λ−µp−2 (1 +u)^λdu

= Z T

α

(x−α)^2−λf^p(x)dx×

"

B(λ−µp−1, µp+ 1)− Z ^T−α_x−α

0

u^−µp (1 +u)^λ−µp−1

−

x−α T −α

γZ 1 0

u^λ−µp−2 (1 +u)^λdu

Z ^T−α_x−α

0

u^γ−µp

(1 +u)^λ−µp−1du

#

= Z T

α

(x−α)^2−λf^p(x)dx×

"

B(λ−µp−1, µp+ 1)− Z ∞

x−α T−α

u^λ−3

(1 +u)^λ−µp−1du

−

x−α T −α

γZ 1 0

u^λ−µp−2 (1 +u)^λdu

Z ^T−α_x−α

0

u^γ−µp

(1 +u)^λ−µp−1du

#

= Z T

α

(x−α)^2−λf^p(x)dx

×

"

B(λ−µp−1, µp+ 1)− Z ∞

0

− Z _T−α^x−α

0

! u^λ−3

(1 +u)^λ−µp−1du

−

x−α T −α

γZ 1 0

u^λ−µp−2 (1 +u)^λdu

Z ^T−α_x−α

0

u^γ−µp

(1 +u)^λ−µp−1du

#

= Z T

α

(x−α)^2−λf^p(x)dx

×

"

B(λ−µp−1, µp+ 1) (

B(λ−2,1−µp)−

x−α T −α

γ x−α T −α

^−γ

× Z _T−α^x−α

0

u^λ−3

(1 +u)^λ−µp−1du

−

x−α T −α

2γZ 1 0

u^λ−µp−2 (1 +u)^λdu

×

T −α x−α

γZ ^T−α_x−α

0

u^γ−µp (1 +u)^λ−µp−1

≤ Z T

α

(x−α)^2−λf^p(x)dx×

B(λ−µp−1, µp+ 1)

B(λ−2,1−µp)−

x−α T −α

γ

× Z 1

0

u^λ−3

(1 +u)^λ−µp−1du

−

x−α T −α

2γZ 1 0

u^λ−µp−2 (1 +u)^λdu

Z 1 0

u^γ−µp (1 +u)^λ−µp−1

= Z T

α

φ(x, λ, µ, p)(x−α)^2−λf^p(x)dx, where

φ(x, λ, µ, p)

=

B(λ−µp−1, µp+ 1)

B(λ−2,1−µp)−

x−α T −α

γ Z 1 0

u^λ−3

(1 +u)^λ−µp−1du

−

x−α T −α

2γZ 1 0

u^λ−µp−2 (1 +u)^λdu

Z 1 0

u^γ−µp

(1 +u)^λ−µp−1du

# .

Similarly

G= Z T

α

φ(y, λ, µ, q)(y−α)^2−λg^q(y)dy, and

H = Z T

α

φ(z, λ, µ, r)(z−α)^2−λh^r(z)dz.

This completes the proof.

Corollary 2.4. Letf(t), g(t), h(z)≥0,p, q, r >1, ¹_p + ¹_q +¹_r = 1,2< λ <3,and max

−1 p,−1

q,−1 r

< µ <min

λ−1

p ,λ−1

q ,λ−1 r

.

If

0<

Z ∞ α

(t−α)^2−λf^p(t)dt <∞, 0<

Z ∞ α

(t−α)^2−λg^q(t)dt < ∞ and

0<

Z ∞ α

(t−α)^2−λh^r(t)dt <∞, then we have the inequality

(2.4) Z ∞

α

Z ∞ α

Z ∞ α

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

(x+y+z)^λ dxdydz

≤K Z ∞

α

(t−α)^2−λf^p(t)dt

¹_pZ ∞ α

(t−α)^2−λg^q(t)dt ¹_q

× Z ∞

α

(t−α)^2−λh^r(t)dt ¹_r

,

where

K = Y

j=p,q,r

B^1/j(λ−µj−1, µj+ 1)B^1/j(λ−2,1−µj) and

(2.5) Z ∞

α

(x−α)

qr(λ−2) p(q+r)

Z ∞ α

Z ∞ α

g(y)h(z)

(x+y+z)^λdxdydz _q+r^qr

dx

≤K^q+rqr Z ∞

α

(t−α)^2−λg^q(t)dt

_q+r^r Z ∞ α

(t−α)^2−λh^r(t)dt _q+r^q

.

The inequalities (2.4)and (2.5) are equivalent.

Proof. Follows from Theorem 2.3 and Lemma 2.1, on choosingγ = 1, T = ∞, andλ_j(t) =

(t−α)^2−λj . We omit the details.

Corollary 2.5. Letf(t), g(t), h(t)≥0,2< λ <3,γ >3µ, and−¹₃ < µ < ^λ−1₃ . If 0<

Z T α

(t−α)^2−λf³(t)dt < ∞, 0<

Z T α

(t−α)^2−λg³(t)dt <∞ and

0<

Z T α

(t−α)^2−λh³(t)dt <∞, then

(2.6) Z T

α

Z T α

Z T α

f(x)g(x)h(z)

(x+y+z)^λ dxdydz

≤ Z T

α

φ(t, λ, µ,3)(t−α)^2−λf³(t)dt

¹₃ Z T α

φ(t, λ, µ,3)(t−α)^2−λg³(t)dt ¹₃

× Z T

α

φ(t, λ, µ,3)(t−α)^2−λh³(t)dt ¹3

, and

(2.7) Z T

α

φ⁻¹²(t, λ, µ,3)(x−α)^λ2−1 Z T

α

Z T α

g(y)h(z)

(x+y+z)^λdydz ¹₂

dx

≤ Z T

α

φ(t, λ, µ,3)(t−α)^2−λg³(t)dt

¹3 Z T α

φ(t, λ, µ,3)(t−α)^2−λh³(t)dt ¹3

.

The inequalities (2.6) and (2.7) are equivalent.

Proof. Follows from Theorem 2.3 and Lemma 2.1, by putting p = q = r = 3, andλ_j(t) =

(t−α)^2−λ3 φ¹³(t, λ, µ,3), j=p, q, r.

Note. In Corollary 2.5, we may take as a special caseµ= 1− ^λ₃ to obtain φ(t, λ,1−λ/3,3)

=B(2λ−4,4−λ)B(λ−2, λ−2)

1− 1 2

t−α T −α

γ

−

t−α T −α

2γZ 1 0

u^2λ−5 (1 +u)^λdu

Z 1 0

u^γ+λ−3 (1 +u)^2λ−4du.

Corollary 2.6. Letf(t), g(t), h(t)≥0,2< λ <3, and−¹₃ < µ < ^λ−1₃ . If 0<

Z ∞ α

(t−α)^2−λf³(t)dt <∞, 0<

Z ∞ α

(t−α)^2−λg³(t)dt <∞ and

0<

Z ∞ α

(t−α)^2−λh³(t)dt <∞,

then (2.8)

Z ∞ α

Z ∞ α

Z ∞ α

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

(x+y+z)^λ dxdydz

≤K Z ∞

α

(t−α)^2−λf³(t)dt

¹₃ Z ∞ α

(t−α)^2−λg³(t)dt ¹₃

× Z ∞

α

(t−α)^2−λh³(t)dt ¹₃

, where

K = (B(λ−3µ−1,3µ+ 1)B(λ−2,1−3µ) and

(2.9) Z ∞

α

(x−α)⁻¹² Z ∞

α

Z ∞ α

g(y)h(z)

(x+y+z)^λdydz ¹₂

dx

≤K^3/2 Z ∞

α

(t−α)^2−λg³(t)dt ¹₂

× Z ∞

α

(t−α)^2−λh³(t)dt ¹₂

. The inequalities (2.8) and (2.9) are equivalent.

Proof. Follows from Corollary 2.4 and Lemma 2.1, by puttingp=q=r= 3, T =∞.

REFERENCES

[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge Univ., Cambridge, UK, (1952).

[2] B. YANG, On generalization of Hardy-Hilbert’s integral inequality, Acta. Math. Sinica, 41 (1998), 839–844.

[3] B. YANG, On Hilbert’s integral inequality, J. Math. Anal. Appl., 220 (2000), 778–785.