INTRODUCTION Suppose thatf andg are real functions, such that0<R∞ 0 f2(t)dt &lt

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Volume 7, Issue 2, Article 76, 2006

FOUR INEQUALITIES SIMILAR TO HARDY-HILBERT’S INTEGRAL INEQUALITY

W.T. SULAIMAN

COLLEGE OFCOMPUTERSCIENCES ANDMATHEMATICS

UNIVERSITY OFMOSUL, IRAQ. [email protected]

Received 22 September, 2005; accepted 05 January, 2006 Communicated by B. Yang

ABSTRACT. Four new different types of inequalities similar to Hardy-Hilbert’s inequality are given.

Key words and phrases: Hardy-Hilbert integral inequality.

2000 Mathematics Subject Classification. 26D15.

1. INTRODUCTION

Suppose thatf andg are real functions, such that0<R∞

0 f²(t)dt < ∞and0<R∞

0 g²(t)dt

<∞, then (1.1)

Z ∞

0

Z ∞

0

f(x)g(y) x+y < π

Z ∞

0

f²(t)dt Z ∞

0

g²(t)dt ¹₂

,

where π is best possible. If (a_n) and (b_n) are sequences of real numbers such that 0 <

P∞

n=1a²_n<∞and0<P∞

n=1b²_n <∞, then (1.2)

∞

X

n=1

∞

X

m=1

ambn

m+n < π

∞

X

n=1

a²_n

∞

X

n=1

b²_n

!¹₂ .

The inequalities (1.1) and (1.2) are called Hilbert’s inequalities. These inequalities play an important role in analysis (cf. [1, Chap. 9]). In their recent papers Hu [5] and Gao [3] gave two distinct improvements of (1.1) and Gao [4] gave a strengthened version of (1.2).

The following definitions are given:

ϕ_λ(r) = r+λ−2

r (r=p, q), k_λ(p) =B(ϕ_λ(p), ϕ_λ(q)), andB is the beta function.

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

The author is grateful to the referee who read through the paper very carefully and correct many typing mistakes.

285-05

Recently, by introducing some parameters, Yang and Debnath [2] gave the following exten- sions:

Theorem A. Iff, g ≥0, p >1, ¹_p + ¹_q = 1, λ > 2−min{p, q},such that 0<

Z ∞

0

t^1−λf^p(t)dt <∞ and 0<

Z ∞

0

t^1−λg^q(t)dt <∞, then

(1.3) Z ∞

0

Z ∞

0

f(x)g(y)

(Ax+By)^λdxdy

< k_λ(p) A^ϕλ(p)B^ϕλ(q)

Z ∞

0

x^1−λf^p(x)dx

¹_pZ ∞

0

y^1−λg^q(y)dy ¹_q

, where the constant factor[k_λ(p)/A^ϕλ(p)B^ϕλ(q)]is the best possible.

Theorem B. If f ≥ 0, p > 1, ¹_p + ¹_q = 1, λ > 2−min{p, q}, A, B > 0 such that 0 <

R∞

0 t^1−λf^p(t)dt <∞,then (1.4)

Z ∞

0

y^{(λ−1)(p−1)} Z ∞

0

f(x) (Ax+By)^λdx

!p

dy <

k_λ(p) A^ϕλ(p)B^ϕλ(q)

pZ ∞

0

x^1−λf^p(x)dx, where the constant factor

k_λ(p)/A^ϕλ(p)B^ϕλ(q)p

is the best possible. The inequalities (1.3) and (1.4) are equivalent.

Theorem C. Ifa_n, b_n >0 (n∈N), p >1,¹_p + ¹_q = 1,2−min{p, q}< λ <2, A, B >0such that

0<

∞

X

n=1

n^1−λa^p_n <∞ and 0<

∞

X

n=1

n^1−λb^q_n<∞, then

(1.5)

∞

X

n=1

∞

X

m=1

a_mb_n

(Am+Bn)^λ < k_λ(p) A^ϕλ(p)B^ϕλ(q)

∞

X

n=1

n^1−λa^p_n

!¹_{p ∞} X

n=1

n^1−λb^q_n

!¹_q , where the constant factor

k_λ(p)/A^ϕλ(p)B^ϕλ(q)

is the best possible.

Theorem D. Ifa_n ≥ 0 (n ∈ N), p > 1,¹_p + ¹_q = 1,2−min{p, q} < λ ≤ 2, A, B > 0such that0<P∞

n=1n^1−λa^p_n <∞,then (1.6)

∞

X

n=1

n^{(λ−1)(p−1)}

∞

X

m=1

a_m (Am+Bn)^λ

!p

<

k_λ(p) A^ϕλ(p)B^ϕλ(q)

p ∞

X

n=1

n^1−λa^p_n, where the constant factor

k_λ(p)/A^ϕλ(p)B^ϕλ(q)p

is the best possible. The inequalities (1.5) and (1.6) are equivalent.

2. NEWINEQUALITIES

The aim of this paper is to give the following results:

Theorem 2.1. Let lnf, lng be convex for nonnegative functions f and g such that f(0) = g(0) = 0, f(∞) = g(∞) = ∞, f⁰(s) ≥ 0, g⁰(s) ≥ 0, s ∈ {x^p, y^q}. Let λ > max{p, q}, p >1, ¹_p + ¹_q = 1. Let

0<

Z ∞

0

t^−p2/q2[f(t^p)]^2−λ+p/q

[f⁰(t)]^pq dt <∞, 0<

Z ∞

0

t^−q2/p2[g(t^q)]^2−λ+q/p

[g⁰(t)]^pq dt <∞, then we have

(2.1) Z ∞

0

Z ∞

0

f(xy)g(xy)

(f(x^p) +g(y^q))^λdxdy

≤ 1

√p

p√^q

qB^1p(p, λ−p)B^1q (q, λ−q)

× Z ∞

0

t^−p2/q2[f(t^p)]^2−λ+p/q [f⁰(t)]^pq dt

!¹_p Z ∞

0

t^−q2/p2[g(t^q)]^2−λ+q/p [g⁰(t)]^pq dt

!¹_q . Proof. Sincelnf is convex andxy≤ ^x_p^p +^y_q^q,then

f(xy) = e^lnf(xy)≤e^lnf

xp p +^yq_q

≤e^{lnfp(xp)+lnf(yq)q} =f^1p(x^p)f^1q(y^q).

Therefore, we have Z ∞

0

Z ∞

0

f(xy)g(xy)

(f(x^p) +g(y^q))^λdxdy

≤ Z ∞

0

Z ∞

0

f^1p(x^p)g^1q(y^q)^[g0(yq)]

1p

[f⁰(x^p)]

1 q

y

q−1 p

x

p−1 q

(f(x^p) +g(y^q))^λp

f^1q(y^q)g^p1(x^p)^[f0(xp)]

1 q

[g⁰(y^q)]^1p x

p−1 q

y

q−1 p

(f(x^p) +g(y^q))^λq

dxdy

≤ Z ∞

0

Z ∞

0

f(x^p)g^p/q(y^q)g⁰(y^q)y^q−1

x^(p−1)p/q[f⁰(x^p)]^pq (f(x^p) +g(y^q))^λdxdy

!¹_p

× Z ∞

0

Z ∞

0

f(y^q)g^q/p(x^p)f⁰(x^p)x^p−1

y^(q−1)q/p[g⁰(y^q)]^qp(f(x^p) +g(y^q))^λdxdy

!¹_q

=M^1pN^1q, say.

Then

M = 1 q

Z ∞

0

x^(1−p)p/q[f(x^p)]^2−λ+p/q [f⁰(x)]^pq dx

Z ∞

0

g(y^q) f(x^p)

^p_q

g⁰(y^q)^qy_f(x^q−1p)

1 + _f^g(y_(x^qp⁾)

λ dy

= 1 q

Z ∞

0

x^−p2/q2[f(x^p)]^2−λ+p/q [f⁰(x)]^pq dx

Z ∞

0

u^p/q (1 +u)^λdu

= 1

qB(p, λ−p) Z ∞

0

x^−p2/q2[f(x^p)]^2−λ+p/q [f⁰(x)]^pq dx.

Similarly,

N = 1

pB(q, λ−q) Z ∞

0

y^−q2/p2[g(y^q)]^2−λ+q/p [g⁰(y)]^qp dy.

Therefore Z ∞

0

Z ∞

0

f(xy)g(xy)

(f(x^p) +g(y^q))^λdxdy

≤ 1

√q

p√^p

qB^1p (p, λ−p)B^1q (q, λ−q)

× Z ∞

0

tt^−p2/q2[f(t^p)]^2−λ+p/q [f⁰(t)]^pq dt

!¹_p Z ∞

0

t^−q2/p2[g(t^q)]^2−λ+q/p [g⁰(t)]^pq dt

!¹_q .

Theorem 2.2. Letf, g, hbe nonnegative functions,h(x, y)is homogeneous of ordernsuch that h(0,1) =h(1,0) = 0,h(∞,1) =h(1,∞) =∞. Letp >1, ¹_p +¹_q = 1,

0<1 +µ−r < λ, r∈n

p q,^q_po

, h_x(x, y)≥0, h_y(x, y)≥0, whereh_x =dh/dx,0<R∞

0 t^p/qf^p(t)dt <∞,0<R∞

0 t^q/pg^q(t)dt <∞,then (2.2)

Z ∞

0

Z ∞

0

f(x)g(y)h^µ(x, y) (1 +h(x, y))^λ dxdy

≤ 1 nB^1p

1 +µ−p

q, λ−1−µ+p q

B^1q

1 +µ−q

p, λ−1−µ−q p

× Z ∞

0

t^q−1p f^p(t)dt

¹_pZ ∞

0

t^p−1q g^q(t)dt ¹_q

. Proof. Observe that

Z ∞

0

Z ∞

0

f(x)g(y)h^µ(x, y) (1 +h(x, y))^λ dxdy

≤ Z ∞

0

Z ∞

0

f^p(x)h^µ(x, y)h_y(x, y) h^p/qx (x, y) (1 +h(x, y))^λdxdy

!¹_p

× Z ∞

0

Z ∞

0

g^q(y)h^µ(x, y)hx(x, y) h^q/py (x, y) (1 +h(x, y))^λdxdy

!¹_q

=M^1pN^1q, say.

Then

M =

Z ∞

0

f^p(x)dx Z ∞

0

h^µ(x, y)hy(x, y)

h^p/qx (x, y) (1 +h(x, y))^λdy.

Lety =xv,dy =xdvand hence h_y(x, y) = dh(x, xv)

dy =xⁿdh(1, v)

dy =xⁿ⁻¹dh(1, v)

dv =xⁿ⁻¹h_v(1, v), h_x(x, y) = dh(x, xv)

dx = d

dxxⁿh(1, v) =nxⁿ⁻¹h(1, v),

therefore

M = 1

n^p/q Z ∞

0

x^p/q−1f^p(x)dx Z ∞

0

[xⁿh(1, v)]^µ−p/qxⁿh_v(1, v) (1 +xⁿh(1, v))^λ dv

= 1

n^p/qB

1 +µ−p

q, λ−1−µ+p q

Z ∞

0

x^p/q−1f^p(x)dx.

Similarly,

N = 1

n^q/pB

1 +µ− q

p, λ−1−µ+ q p

Z ∞

0

y^q/p−1g^q(y)dy.

This implies Z ∞

0

Z ∞

0

f(x)g(y)h^µ(x, y) (1 +h(x, y))^λ dxdy

≤ 1 nB^p1

1 +µ− p

q, λ−1−µ+p q

B^1q

1 +µ− q

p, λ−1−µ+q p

× Z ∞

0

t^p/q−1f^p(t)dt

¹_pZ ∞

0

t^q/p−1g^q(t)dt ¹_q

. The following lemma is needed for the coming result.

Lemma 2.3. Lets≥1,0<1 +µ≤min{α, λ}and define f(s) =s^−α

Z s

0

t^µ (1 +t)^λdt, thenf(s)≤f(1).

Proof. We have

f⁰(s) =s^−α s^µ (1 +s)^λ +

Z s

0

t^µ

(1 +t)^λdt(−α)s^−α−1

≤ s^µ−α

(1 +s)^λ − αs^−α−1 (1 +s)^λ

Z s

0

t^µdt

= s^µ−α (1 +s)^λ

1− α 1 +µ

≤0.

This shows thatf is nonincreasing and hencef(s)≤f(1).

Theorem 2.4. Let f, g, F, G be nonnegative functions such that F(s) = Rs

0 f(t)dt, G(s) = Rs

0 g(t)dt,let,p > 1, ¹_p + ¹_q = 1,(1−λ/2)r≤λ/2 +α ≤2α, r∈n

p q,^q_po

, 0<

Z x

0

(x−t)t(1−λ/2)p/q−λ/2−α

F^p−1(t)f(t)dt <∞, 0<

Z x

0

(x−t)t(1−λ/2)q/p−λ/2−α

G^q−1(t)g(t)dt <∞,

then (2.3)

Z x

0

Z x

0

F(s)G(t) (s+t)^λ dsdt

≤

√p

p√^q q

2 x^αB ^λ₂,^λ₂ Z x

0

(x−t)t(1−λ/2)p/q−λ/2−α

F^p−1(t)f(t)dt ¹_p

× Z x

0

(x−t)t(1−λ/2)q/p−λ/2−α

G^q−1(t)g(t)dt ¹_q

. Proof. Observe that

Z x

0

Z x

0

F(s)G(t) (s+t)^λ dsdt

= Z x

0

Z x

0

F(s) t^1/p

s^1/q

λ/2−1

(s+t)^λ/p · G(t)

s^1/q t^1/p

λ/2−1

(s+t)^λ/q dsdt

≤ Z x

0

Z x

0

F^p(s)t^λ/2−1

(s+t)^λs^{(λ/2−1)p/q}dsdt

!¹_p Z x

0

Z x

0

G^q(t)s^λ/2−1

(s+t)^λt^{(λ/2−1)q/p}dsdt

!¹_q

=M^1pN^1q, say.

Then

M =

Z x

0

s(1−λ/2)p/q−λ/2

F^p(s)ds Z x

0 t s

λ/2−1 1 s

1 + _s^tλ dt

= Z x

0

s(1−λ/2)p/q−λ/2

F^p(s)dsx s

αx s

−αZ x/s

0

u^λ/2−1 (1 +u)^λdu

≤ Z x

0

s(1−λ/2)p/q−λ/2

F^p(s)dsx s

αZ 1

0

u^λ/2−1 (1 +u)^λdu

= x^α 2 B

λ 2,λ

2 Z x

0

s(1−λ/2)p/q−λ/2−αF^p(s)ds, by virtue of the lemma.

As

F^p(s) =p Z s

0

F^p−1(u)f(u)du, then

M = p 2x^αB

λ 2,λ

2 Z x

0

s(1−λ/2)p/q−λ/2−α

ds Z s

0

F^p−1(u)f(u)du

= p 2x^αB

λ 2,λ

2 Z x

0

Z s

0

u(1−λ/2)p/q−λ/2−α

F^p−1(u)f(u)dsdu

= p 2x^αB

λ 2,λ

2 Z x

0

(x−s)s(1−λ/2)p/q−λ/2−α

F^p−1(s)f(s)ds.

Similarly,

N = q 2x^αB

λ 2,λ

2 Z x

0

(x−t)t(1−λ/2)q/p−λ/2−α

G^q−1(t)g(t)dt.

Therefore, we have Z x

0

Z x

0

F(s)G(t) (s+t)^λ dsdt

≤

√p

p√^q q 2 x^αB

λ 2,λ

2

Z x

0

(x−t)t(1−λ/2)p/q−λ/2−α

F^p−1(t)f(t)dt ¹_p

× Z x

0

(x−t)t(1−λ/2)q/p−λ/2−αG^q−1(t)g(t)dt ¹_q

. Theorem 2.5. Letf, g be nonnegative functions, f is submultiplicative andg is concave non- increasing, f⁰(x), g⁰(y) ≥ 0, f(0) = g(0) = 0, f(∞) = g(∞) = ∞, p > 1, _p¹ + ¹_q = 1, 0< a+ 1< λ,0< b+ 1< λ,

0<

Z ∞

0

[f(x)]^µp−bp/q[g(x)]^1+a−λ

[f⁰(x)]^pq dx <∞, 0<

Z ∞

0

[g(y)]^µq−aq/p[f(y)]^1+b−λ

[g⁰(y)]^qp dy <∞, then

(2.4) Z ∞

0

Z ∞

0

f^µ(xy)

g^λ/2(xy)dxdy ≤2^λ/2B^p1 (a+ 1, λ−a−1)B^1q (b+ 1, λ−b−1)

× Z ∞

0

[f(t)]^µp[g(t)]^{1+a−λ−bp/q} [g⁰(t)]

p q

dt

!¹_p Z ∞

0

[f(t)]^µq[g(t)]^{1+b−λ−aq/p} [g⁰(t)]

q p

dt

!¹_q . Proof. Since√

xy≤ ^x+y₂ , then g(xy) = g

(√ xy)²

≥(g(√

xy))² ≥

g

x+y 2

2

≥

g(x) +g(y) 2

2

, and hence

Z ∞

0

Z ∞

0

f^µ(xy) g^λ/2(xy)dxdy

≤2^λ/2 Z ∞

0

Z ∞

0

f^µ(x)f^µ(y) (g(x) +g(y))^λdxdy

= 2^λ/2 Z ∞

0

Z ∞

0

f^µ(x)^[g(y)]a/p

[g(x)]^b/q

[g⁰(y)]^1/p [g⁰(x)]^1/q

(g(x) +g(y))^λp

·

f^µ(y)^[g(x)]b/q

[g(y)]^a/p ·^[g0(x)]1/q

[g⁰(y)]^1/p

(g(x) +g(y))^λq

≤2^λ/2 Z ∞

0

Z ∞

0

f^µp(x)g^a(y)g⁰(y)

[g(x)]^bp/q[g⁰(x)]^pq (g(x) +g(y))^λdxdy

!¹_p

× Z ∞

0

Z ∞

0

f^µq(y)g^b(x)g⁰(x)

[g(y)]^aq/p[g⁰(y)]^qp (g(x) +g(y))^λdxdy

!¹_q

= 2^λ/2M^1pN^1q, say.

Then

M =

Z ∞

0

[f(x)]^µp[g(x)]^{1+a−λ−bp/q} [g⁰(x)]^pq dx

Z ∞

0

g(y) g(x)

a g⁰(y)

g(x)

1 + ^g(y)_g(x)λdy

=B(a+ 1, λ−a−1) Z ∞

0

[f(x)]^ηp[g(x)]^{1+a−λ−bp/q} [g⁰(x)]^pq dx.

Similarly, we can show that

N =B(b+ 1, λ−b−1) Z ∞

0

[f(y)]^µq[g(y)]^{1+b−λ−aq/p} [g⁰(y)]^qp

dy.

The result follows.

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