On the norm of an integral operator and applications

J. Math. Anal. Appl. 321 (2006) 182–192 www.elsevier.com/locate/jmaa

On the norm of an integral operator and applications Bicheng Yang Department of Mathematics, Guangdong Institute of Education, Guangzhou, Guangdong 510303, PR China Received 5 June 2005 Available online 8 September 2005 Submitted by L. Debnath

Abstract In this paper, the norm of an integral operator T : Lr (0, ∞) → Lr (0, ∞) (r > 1) is obtained. As applications, a new bilinear integral operator inequality with the norm and the equivalent forms are given, and some new Hilbert’s type inequalities with the best constant factors as the particular cases are established. © 2005 Elsevier Inc. All rights reserved. Keywords: Norm; Integral operator; Bilinear integral inequality; Beta function; Hilbert’s type inequality

Let H be a real separable Hilbert space and T : H → H be a bounded self-adjoint semipositive definite operator. Then one has the following new inequality (see [1, (17)]):  T 2  a2 b2 + (a, b)2 (a, b ∈ H ), (1) 2 √ where (a, b) is the inner product of a and b, and a = (a, a) is the norm of a. In particular, set H = L2 (0, ∞) and define T : L2 (0, ∞) → L2 (0, ∞) as: for f ∈ L2 (0, ∞), ∞ 1 f (x) dx, y ∈ (0, ∞), (2) (Tf )(y) := x +y (a, T b)2 

0

then by (1), one has the sharper form of Hilbert’s integral inequality as (see [1, p. 292])  ∞ 2  1  ∞ ∞∞ ∞ 2 π f (x)g(y) dx dy  √ f 2 (x) dx g 2 (x) dx + f (x)g(x) dx . x+y 2 0 0

0

0

E-mail address: [email protected]. 0022-247X/$ – see front matter © 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2005.07.071

0

(3)

B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

183

∞ And by using Cauchy’s inequality in the term ( 0 f (x)g(x) dx)2 of (3), the Hilbert’s integral inequality is obtained as follows:  ∞ 1 ∞∞ ∞ 2 f (x)g(y) dx dy  π f 2 (x) dx g 2 (x) dx . (4) x +y 0 0

0

0

In 1925, Hardy–Riesz [2] gave an extension of (4) as: if p > 1, g ∈ Lq (0, ∞) and f p , gq > 0, then ∞∞ 0 0

f (x)g(y) π dx dy < f p gq , x +y sin( πp )

1 p

+

1 q

= 1, f ∈ Lp (0, ∞),

(5)

∞ π where the constant factor sin(π/p) is the best possible, and f p = { 0 f p (x) dx}1/p . Inequalities (4) and (5) are important in analysis and its applications (see [3,4]). Recently, (4) and (5) have been extended by [5,6] by using the way of weight function and introducing a parameter λ. In 2003, Yang and Rassias [7] summarized how to use the way of weight coefficient in research for the Hilbert’s type inequalities. In view of (2), one may rewrite (5) as π f p gq , (Tf, g) < sin(π/p) where (Tf, g) is the formal inner product of Tf and g, which is defined by  ∞ ∞ ∞∞ f (x) f (x)g(y) dx g(y) dy = dx dy. (Tf, g) := x+y x +y 0

0

0 0

In this paper, the characterization of the norm of an integral operator T : Lr (0, ∞) → Lr (0, ∞) (r > 1; r = p, q) is considered. As applications, a new bilinear integral operator inequality with the norm and the equivalent forms are established. Some new Hilbert’s type integral inequalities with the best constant factors as the particular cases are given. Let p > 1, p1 + q1 = 1, Lr (0, ∞) (r = p, q) be two real normal spaces, and k(x, y) be continuous in (0, ∞) × (0, ∞), satisfying k(x, y) = k(y, x) > 0, for x, y ∈ (0, ∞). Define the integral operator T as: for f ∈ Lp (0, ∞), ∞ (6) (Tf )(y) := k(x, y)f (x) dx, y ∈ (0, ∞), or for g

0 q ∈ L (0, ∞),

∞ (T g)(x) :=

k(x, y)g(y) dy,

x ∈ (0, ∞).

(7)

0

For ε ( 0) small enough and x > 0, setting k ε (r, x) as  1+ε ∞ x r

dt (r = p, q), kε (r, x) := k(x, t) t 0

one has the following theorem:

(8)

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B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

Theorem 1. ∞ 1 (i) If k 0 (r, x) = 0 k(x, t)( xt ) r dt = k0 (p) (r = p, q; x > 0), and k0 (p) is a constant indepenr dent of x, then T ∈ B(L (0, ∞) → Lr (0, ∞)) and T r  k0 (p) (r = p, q); (ii) if k ε (r, x) = kε (p) (r = p, q; x > 0) is independent of x, and kε (p) = k0 (p) + o(1) (ε → 0+ ), then T r = k0 (p) (r = p, q). Proof. (i) For f ( 0) ∈ Lp (0, ∞), by Hölder’s inequality with weight (see [8]), one has from condition (i) that  ∞ p  ∞

 1   1  p x pq y pq dx k(x, y)f (x) dx = k(x, y) f (x) y x 0

 ∞



0

 ∞  1  1 p−1 x q p y p k(x, y) f (x) dx k(y, x) dx y x

0

∞ =

0

 1  p−1 x q p k(x, y) f (x) dx k0 (p) , y

0

and then one obtains that  ∞ ∞ p  1 p k(x, y)f (x) dx dy Tf p = 0

0

 1/q  k0 (p)

 ∞∞

1  1 p x q p k(x, y) f (x) dx dy y

0 0

 1/q = k0 (p)

 ∞ ∞ 0

1  1  p x q k(x, y) dy f p (x) dx = k0 (p)f p . y

0

It follows that Tf ∈ Lp (0, ∞) and T p  k0 (p) (cf. [9]). By the same way, one has T g ∈ Lq (0, ∞) and T q  k0 (p). (ii) It is obvious that condition (ii) covers condition (i). By condition (ii), it follows that  1+ε ∞   y p (9) k(y, x) dx = kε (p) +

0(1) a → 0+ . x a

For any a, ε > 0, set fε as: fε (x) = 0, x ∈ (0, a); fε (x) = (εa ε )1/p x −(1+ε)/p , x ∈ [a, ∞), then one has fε p = 1, and by (9),  ∞ ∞ p  1 p

T p  Tfε p =

k(x, y)fε (x) dx 0



 εa

 ε 1/p

0

p

 ∞ ∞ k(x, y)x a

a

dy

− 1+ε p

dx

1

p

dy

B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

 1 = εa ε p

 ∞

y −1−ε

 ∞

a

 1 = εa ε p

185

 1+ε p  p1 y p k(y, x) dx dy x

a

 ∞ y

−1−ε

 p kε (p) +

0(1) dy

1

p

= kε (p) +

0(1).

a

In virtue of condition (ii), it follows that T p  k0 (p) (for a, ε → 0+ ). Hence, combining with T p  k0 (p) in (i), one has T p = k0 (p). By the same way, one has T q = k0 (p). The theorem is proved. 2 Theorem 2. Let p > 1, p1 + q1 = 1, k ε (r, x) (r = p, q; x > 0) satisfy condition (i) of Theorem 1. If f, g  0, and f ∈ Lp (0, ∞), g ∈ Lq (0, ∞), then one has the following two equivalent inequalities: ∞∞ k(x, y)f (x)g(y) dx dy  k0 (p)f p gq , 0 0

p

 ∞ ∞ k(x, y)f (x) dx 0

(10)

1

p

 k0 (p)f p ,

dy

(11)

0

where the constant factor k0 (p)(=

∞ 0

1

k(x, t)( xt ) p dt) is independent of x.

Proof. By Hölder’s inequality with weight and condition (i), one has ∞∞ k(x, y)f (x)g(y) dx dy 0 0

∞∞ =

 1   1  x pq y pq k(x, y) f (x) g(y) dx dy y x

0 0

 ∞∞ 

 1  ∞∞ 1  1  1 p q x q p y p q k(x, y) f (x) dx dy k(x, y) g (y) dx dy y x

0 0

 ∞ ∞ = 0

0 0

 1  ∞ ∞ 1  1   1  p q p x q y k(x, y) dy f p (x) dx k(x, y) dx g q (y) dy y x

0

 ∞

= k0 (p)

 1  ∞ p

f p (x) dx 0

0

1

0

q

g q (y) dy

,

0

and (10) is valid. ∞ Set g(y) = ( 0 k(x, y)f (x) dx)p−1 (y ∈ (0, ∞)) and use (10) to obtain

(12)

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B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

g (y) dy = q

0<

p

∞ ∞

∞ 0

k(x, y)f (x) dx 0

dy

0

k(x, y)f (x)g(y) dx dy  k0 (p)f p

=  ∞

0 0

1/p

g q (y) dy

 ∞ ∞ =

0

1

 ∞

∞∞

p k(x, y)f (x) dx

0

p

q

g (y) dy 0

,

(13)

1

p

 k0 (p)f p .

dy

(14)

0

Hence (11) is valid, and one shows that (10) implies (11). If (11) is valid, by Hölder’s inequality, one has ∞∞ ∞ ∞ k(x, y)f (x)g(y) dx dy = 0 0



k(x, y)f (x) dx g(y) dy 0

0

p

 ∞ ∞ 

k(x, y)f (x) dx 0

1

p

dy

gq .

(15)

0

Then by (11), one has (10). It follows that (10) is equivalent to (11). The theorem is proved.

2

Note 1. Since T q  k0 (p), by the same way, one still can show that q  1  ∞ ∞ q

k(x, y)g(y) dy 0

dx

 k0 (p)gq ,

(16)

0

and (16) is equivalent to (10). It follows that (10), (11) and (16) are equivalent. Theorem 3. Let p > 1, p1 + q1 = 1, and k ε (r, x) (r = p, q; x > 0) satisfy condition (ii) of Theorem 1. If f, g  0, f ∈ Lp (0, ∞), g ∈ Lq (0, ∞), and f p , gq > 0, T is defined by (6) (or (7)), and the formal inner product of Tf and g is defined by ∞∞ k(x, y)f (x)g(y) dx dy, (Tf, g) := 0 0

then one has the following two equivalent inequalities: (Tf, g) < T p f p gq , Tf p < T p f p , where the constant factor T p = possible.

(17) (18) ∞ 0

1 p

k(x, t)( xt ) dt (> 0) in the above inequalities is the best

Proof. If (12) takes the form of equality, then there exist real numbers A and B such that they are not all zero, and (see [8])  1  1 y p q x q p f (x) = B g (y) a.e. in (0, ∞) × (0, ∞). A y x

B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

187

It follows that Axf p (x) = Byg q (y) a.e. in (0, ∞) × (0, ∞). Then there exists a constant C, such that Axf p (x) = C

a.e. in (0, ∞),

Byg q (y) = C

a.e. in (0, ∞).

Assume that A = 0 and then one has a.e. in (0, ∞), which contradicts the fact that f ∈ Lp (0, ∞). Hence (12) takes the form of strict inequality, and in view of T p = k0 (p) in the result of (ii) in Theorem 1, one has (17). Since f p > 0, by (13) and (14), one has g ∈ Lq (0, ∞) and gq > 0. Hence by using (17), (13) takes the form of strict inequality and k0 (p) = T p ; so does (14), and then (18) is valid. By the same way of Theorem 2, (17) and (18) are obviously equivalent. In view of the fact that the constant factor T p in (18) is the best possible, one can conclude that the constant factor T p in (17) is the best possible. Otherwise, by (13) and (14), one can get a contradiction that the constant factor in (18) is not the best possible. The theorem is proved. 2 C f p (x) = Ax

Note 2. By the same way and in view of T p = T q , one has T gq < T p gq ,

(19)

where the constant factor T p is the best possible, and (19), (17) and (18) are equivalent. For giving some particular cases of Theorems 3 and 2, one needs the formula of the Beta function B(u, v) as (see [10]): 1 (1 − t)u−1 t v−1 dt = B(v, u)

B(u, v) =

(u, v > 0).

(20)

0

(a) Setting k(x, y) = finds that k ε (r, x) =

∞ 0

1 x+y ,

for any 0  ε < min{p, q} − 1 and fixed x > 0, putting u = yx , one

 1+ε ∞ ∞ r−1−ε r−1 x r 1 1 1 −1 r u u r −1 du dy = du → x +y y 1+u 1+u 0

  ε → 0+ ; r = p, q .

π = k0 (p) = sin(π/p)

0

π (T is defined by (2)), and by Theorem 3 Hence by Theorem 1, one has T p = k0 (p) = sin(π/p) and Note 2, one has (5) and the following two equivalent inequalities:

 ∞ ∞ 0 0  ∞ ∞

0

0

1 f (x) dx x+y 1 g(y) dy x+y

p

1

p

dy

<

π f p , sin(π/p)

(21)

<

π gq , sin(π/p)

(22)

1

q

q

dx

π sin(π/p) is the best possible. |x λ−1 −y λ−1 | (λ > 0, λ = 1, 1r , 2r (1 − 1r )), (max{x,y})λ

where the constant factor (b) Setting k(x, y) = one obtains that

for any 0  ε < min{p, q} − 1,

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B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

(i) if 0 < λ < 1, λ = 1r , 2r (1 − 1r ) (r = p, q), then k ε (r, x) =

∞

 1+ε |x λ−1 − y λ−1 | x r dy (max{x, y})λ y

0

 1+ε  1+ε ∞ λ−1 y λ−1 − x λ−1 x r x − y λ−1 x r dy + dy xλ y yλ y

x =

x

0

1 =

 λ−1  1+ε u − 1 u− r du +

1

0

 λ−1  1+ε u − 1 u r −1 du

0

1 →

 λ−1   1 1 u − 1 u− r + u r −1 du

0

=

(λpq − 2)(1 − λ)pq = k0 (p) (λp − 1)(λq − 1)

  ε → 0+ ; r = p, q ;

(ii) if λ > 1, then k ε (r, x) =

∞

 1+ε |x λ−1 − y λ−1 | x r dy (max{x, y})λ y

0

→−

(λpq − 2)(1 − λ)pq = k0 (p) (λp − 1)(λq − 1)

  ε → 0+ ; r = p, q .

Hence by Theorem 1, one has T p = k0 (p) = | (λpq−2)(1−λ)pq (λp−1)(λq−1) | > 0, and by Theorem 3, one has Corollary 1. If p > 1,

f ∈ Lp (0, ∞), g ∈ Lq (0, ∞), and f p , gq > 0,

then for λ > 0, λ =

following two equivalent inequalities:

∞∞ 0 0

   (λpq − 2)(1 − λ)pq  |x λ−1 − y λ−1 | f p gq ,  f (x)g(y) dx dy <  (max{x, y})λ (λp − 1)(λq − 1) 

 ∞ ∞ 0

1 1 p + q = 1, f, g  0, 1 1 2 p , q , pq , 1, one has the

|x λ−1 − y λ−1 | f (x) dx (max{x, y})λ

p

1

p

dy

0

   (λpq − 2)(1 − λ)pq  f p , <  (λp − 1)(λq − 1) 

(23)

(24)

where the constant factor | (λpq−2)(1−λ)pq (λp−1)(λq−1) | is the best possible. |x−y| 1 1 (c) Setting k(x, y) = (min{x,y}) λ (0 < λ < min{ p , q }), then for any 0  ε < min{p, q} × (1 − λ) − 1, one obtains from (20) that λ−1

k ε (r, x) =

∞ 0

 1+ε |x − y|λ−1 x r dy (min{x, y})λ y

B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

x =

 1+ε  1+ε ∞ (x − y)λ−1 x r (y − x)λ−1 x r dy + dy yλ y xλ y x

0

1

λ−1 (1− 1+ε r −λ)−1

(1 − u)

=

189

u

1 du +

0

(1 − u)λ−1 u(

1+ε r −λ)−1

du

0

  1 1 → B λ, − λ + B λ, − λ = k0 (p) q p

  ε → 0+ ; r = p, q .

Hence by Theorem 1, one has T p = k0 (p) = B(λ, q1 − λ) + B(λ, p1 − λ), and by Theorem 3, one has 1 1 p q p + q = 1, f, g  0, f ∈ L (0, ∞), g ∈ L (0, ∞), and f p , gq 0 < λ < min{ p1 , q1 }, one has the following two equivalent inequalities:

Corollary 2. If p > 1, then for

∞∞

 

 1 |x − y|λ−1 1 − λ + B λ, − λ f p gq , (25) f (x)g(y) dx dy < B λ, (min{x, y})λ q p

0 0

 ∞ ∞ 0

> 0,

|x − y|λ−1 f (x) dx (min{x, y})λ

p

1

   1 1 < B λ, − λ + B λ, − λ f p , (26) q p

p

dy

0

where the constant factor [B(λ, q1 − λ) + B(λ, p1 − λ)] is the best possible. λ/2

(d) Setting k(x, y) = (min{(x/y),(y/x)}) (λ  0), then for any 0  ε < min{p, q} − 1, one max{x,y} obtains  1+ε ∞ (min{(x/y), (y/x)})λ/2 x r dy k ε (r, x) = max{x, y} y 0

x =

 λ  1+ε ∞  λ2  1+ε x r 1 y 2 x r 1 x dy + dy x x y y y y x

0

=

1

1

+ λ 1+ε 1 + λ2 − 1+ε r 2 + r 4pq(λ + 1) = k0 (p) → (pλ + 2)(qλ + 2)

  ε → 0+ ; r = p, q .

Hence by Theorem 1, one has T p = k0 (p) =

4pq(λ+1) (pλ+2)(qλ+2) ,

and by Theorem 3, one has

Corollary 3. If p > 1, p1 + q1 = 1, f, g  0, f ∈ Lp (0, ∞), g ∈ Lq (0, ∞), and f p , gq > 0, then for λ  0, one has the following two equivalent inequalities: ∞∞ 0 0

4pq(λ + 1) (min{(x/y), (y/x)})λ/2 f (x)g(y) dx dy < f p gq , max{x, y} (pλ + 2)(qλ + 2)

(27)

190

B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

 ∞ ∞ 0

(min{(x/y), (y/x)})λ/2 f (x) dx max{x, y}

0 0

4pq(λ+1) (pλ+2)(qλ+2)

<

4pq(λ + 1) f p , (pλ + 2)(qλ + 2)

min{(x/y), (y/x)} f (x) dx max{x, y}

p

1

p

dy

0

(e) Setting k(x, y) = k ε (r, x) =

∞

|x−y|λ−1 (max{x,y})λ

(28)

is the best possible. In particular, for λ = 2, one has

min{(x/y), (y/x)} 3pq f (x)g(y) dx dy < f p gq , max{x, y} 2pq + 1

 ∞ ∞ 0

p

dy

0

where the constant factor ∞∞

1

p

<

3pq f p . 2pq + 1

(29)

(30)

(λ > 0), then for any 0  ε < min{p, q} − 1, one obtains that

 1+ε |x − y|λ−1 x r dy (max{x, y})λ y

0

x =

 1+ε  1+ε ∞ (x − y)λ−1 x r (y − x)λ−1 x r dy + dy xλ y yλ y x

0

1 =

λ−1 (1− 1+ε r )−1

(1 − u)

u

1 du +

0

1

1

(1 − u)λ−1 u(1− r )−1 du + 0

1+ε r −1

du

0

1 →

(1 − u)λ−1 u



1 = B λ, p





1 + B λ, q

1

(1 − u)λ−1 u r −1 du

0



= k0 (p)

  ε → 0+ ; r = p, q .

Hence by Theorem 1, one has T p = k0 (p) = B(λ, p1 ) + B(λ, q1 ), and by Theorem 3, one has Corollary 4. If p > 1, p1 + q1 = 1, f, g  0, f ∈ Lp (0, ∞), g ∈ Lq (0, ∞), and f p , gq > 0, then for λ > 0, one has the following two equivalent inequalities: ∞∞

 

 1 |x − y|λ−1 1 + B λ, f p gq , f (x)g(y) dx dy < B λ, (max{x, y})λ p q

0 0

 ∞ ∞ 0

|x − y|λ−1 f (x) dx (max{x, y})λ

p

1

p

dy

 

 1 1 + B λ, f p , < B λ, p q

0

where the constant factor [B(λ, p1 ) + B(λ, q1 )] is the best possible.

(31)

(32)

B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

(f) Setting k(x, y) = one obtains k ε (r, x) =

∞

(xy)(λ−1)/2 (max{x,y})λ

191

(λ > 2 max{ p1 , q1 }−1), then for any 0  ε < min{p, q} λ+1 2 −1,

 1+ε (xy)(λ−1)/2 x r dy (max{x, y})λ y

0

x =

 1+ε  1+ε ∞ (xy)(λ−1)/2 x r (xy)(λ−1)/2 x r dy + dy xλ y yλ y x

0

= =

1 λ+1 2

−

1+ε r

+

1 λ−1 2

+

1+ε r

1

→

λ+1 2

−

1 p

+

1 λ+1 2

−

1 q

  4pqλ = k0 (p) ε → 0+ ; r = p, q . [p(λ + 1) − 2][q(λ + 1) − 2]

Hence by Theorem 1, one has T p = k0 (p) =

4pqλ [p(λ+1)−2][q(λ+1)−2] ,

and by Theorem 3, one has

1 1 p q p + q = 1, f, g  0, f ∈ L (0, ∞), g ∈ L (0, ∞), and f p , gq λ > 2 max{ p1 , q1 } − 1, one has the following two equivalent inequalities:

Corollary 5. If p > 1, then for

∞∞ 0 0

(xy)(λ−1)/2 4pqλ f (x)g(y) dx dy < f p gq , λ (max{x, y}) [p(λ + 1) − 2][q(λ + 1) − 2]

 ∞ ∞ 0

(xy)(λ−1)/2 f (x) dx (max{x, y})λ

p

1

p

dy

<

0

where the constant factor

4pqλ [p(λ+1)−2][q(λ+1)−2]

4pqλ f p , [p(λ + 1) − 2][q(λ + 1) − 2]

> 0,

(33)

(34)

is the best possible.

Remarks. For λ = 0 in (27), or for λ = 1 in (31) and (33), one has ∞∞

1 f (x)g(y) dx dy < pqf p gq ; max{x, y}

(35)

0 0

and for λ = 0 in (28), or for λ = 1 in (32) and (34), one has  ∞ ∞ 0

1 f (x) dx max{x, y}

p

1

p

dy

< pqf p .

(36)

0

(35) is the classical Hilbert’s type integral inequality and (36) is the equivalent form (see [3]). All of the above given results are new Hilbert’s type inequalities and the equivalent forms with the best constant factors. In the following, one gives a particular case of Theorem 2 (for p = q = 2):

192

B. Yang / J. Math. Anal. Appl. 321 (2006) 182–192

(f) Setting k(x, y) = k ε (2, x) =

∞ 0

Since k 0 (2, x) = rem 2, one has

π λ

(xy)(λ−1)/2 1+(xy)λ

(λ > 0), then for any 0  ε < λ, one obtains

 1+ε ∞ ( 1 − ε )−1 (xy)(λ−1)/2 x 2 u 2 2λ xε dy = du = kε (2) 1 + (xy)λ y λ 1+u

(constant).

0

= k0 (2), by Theorem 1 (for p = q = 2), one has T 2 

π λ

and by Theo-

Corollary 6. If f, g  0, f ∈ L2 (0, ∞), g ∈ L2 (0, ∞), then for λ > 0, one has the following two equivalent inequalities: ∞∞ 0 0

(xy)(λ−1)/2 π f (x)g(y) dx dy  f 2 g2 , λ 1 + (xy) λ

 ∞ ∞ 0

0

(xy)(λ−1)/2 f (x) dx 1 + (xy)λ

2

1 2

dy



π f 2 . λ

(37)

(38)

Note 3. By (1), since T 2  πλ , one guides that we can obtain the sharper form of (37) as ∞∞ 0 0

1 (xy)(λ−1)/2 π  f (x)g(y) dx dy  √ f 2 g2 + (f, g)2 2 , λ 1 + (xy) λ 2

(39)

but one cannot conform that the constant factors in (37)–(39) are the best possible. References [1] K. Zhang, A bilinear inequality, J. Math. Anal. Appl. 271 (2002) 288–296. [2] G.H. Hardy, Note on a theorem of Hilbert concerning series of positive term, Proc. London Math. Soc. 23 (1925), Records of Proc. xlv-xlvi. [3] G.H. Hardy, J.E. Littlewood, G. Polya, Inequalities, Cambridge Univ. Press, Cambridge, 1952. [4] D.S. Mitrinovic, J.E. Pecaric, A.M. Kink, Inequalities Involving Functions and Their Integrals and Derivatives, Kluwer Academic, Boston, 1991. [5] B. Yang, On Hilbert’s integral inequality, J. Math. Anal. Appl. 220 (1998) 778–785. [6] B. Yang, L. Debnath, On the extended Hardy–Hilbert’s inequality, J. Math. Anal. Appl. 272 (2002) 187–199. [7] B. Yang, T.M. Rassias, On the way of weight coefficient and research for Hilbert-type inequalities, Math. Inequal. Appl. 6 (2003) 625–658. [8] J. Kuang, Applied Inequalities, Shandong Science Press, Jinan, 2003. [9] A.E. Taylor, D.C. Lay, Introduction to Functional Analysis, Wiley, New York, 1976. [10] Z. Wang, D. Gua, An Introduction to Special Functions, Science Press, Beijing, 1979.