0) with the homogeneous kernel of−λ-degree and a best constant factor and its operator expression are given

Banach J. Math. Anal. 4 (2010), no. 2, 100–110

B

J

M

A

www.emis.de/journals/BJMA/

ON A HILBERT-TYPE INTEGRAL INEQUALITY IN THE SUBINTERVAL AND ITS OPERATOR EXPRESSION

BICHENG YANG^1∗ AND THEMISTOCLES M. RASSIAS² Communicated by K. Ciesielski

Abstract. In this paper, by using the methods of real analysis and functional analysis, a Hilbert-type integral inequality in the subinterval (a,∞) (a > 0) with the homogeneous kernel of−λ-degree and a best constant factor and its operator expression are given. As applications, a few improved results, the equivalent forms and some new inequalities with the particular kernels are obtained.

1. Introduction If f, g ≥ 0, f, g ∈ L²(0,∞),||f|| = {R∞

0 f²(x)dx}¹² and ||g|| = {R∞

0 g²(x)dx}¹², then we have the following Hilbert’s integral inequality:

Z ∞

0

Z ∞

0

f(x)g(y)

x+y dxdy ≤π||f|| · ||g||, (1.1) where the constant factor π is the best possible. Inequality (1.1) is important in analysis and its applications (cf. [2, 5]). Define an integral operator T : L²(0,∞)→L²(0,∞) as: for f(≥0)∈L²(0,∞),

T(f)(y) :=

Z ∞

0

f(x)

x+ydx(y ∈(0,∞)). (1.2)

Then inequality (1.1) is rewritten as: (T f, g) ≤ π||f|| · ||g||, where (T f, g) :=

R∞ 0 (R∞

0 f(x)

x+ydx)g(y)dy is the inner product of T f and g. We named of T Hilbert

Date: Received: 4 January 2010; Accepted: 19 February 2010.

∗ Corresponding author.

2000Mathematics Subject Classification. Primary 47A07; Secondary 26D15.

Key words and phrases. Hilbert-type integral inequality, homogenous kernel, operator.

100

integral operator. By (1.1), we can prove the equivalent form that||T f|| ≤π||f||, and conclude that||T||=π; [1].

If we replace _x+y¹ by a bilinear functionk(x, y)(≥0) in (1.1), then the problem is how to make sure the conditions of k(x, y) for giving an integral operator T as (1.2) and the inequality with a best constant factor as (1.1). In recent years, Yang [7, 8] considered the case ofk(x, y) being continuous and symmetric in the function space L^p(0,∞), Yang [9, 10, 11] considered the same case of k(x, y) in the disperse space l^p, and Zhong et al. [18] considered the case of k(x, y) in L^p(Rⁿ₊).But their given conditions are not quite simple.

In 1998, by introducing λ∈(0,1] and the Beta function B(u, v) as[6]:

B(u, v) :=

Z ∞

0

1

(1 +t)^u+vt^−u+1dt(u, v >0), (1.3) Yang [12] gave an extension of (1.1) in the subinterval (a,∞)(a >0) as:

Z ∞

a

Z ∞

a

f(x)g(y)dxdy

(x+y)^λ ≤k_λ{ Z ∞

a

σ(x)x^1−λf²(x)dx Z ∞

a

σ(x)x^1−λg²(x)dx}¹²,(1.4) wherek_λ =B(^λ₂,^λ₂) andσ(x) = 1−¹₂(^a_x)^λ2. When λ= 1, a→0⁺,inequality (1.4) deduces to (1.1). In recent years, a number of papers studied some improvements and extensions of (1.4) (cf. [13, 14, 15, 17]).

In this paper, a simple condition of the homogeneous kernel with −λ-degree (λ > 0) is considered. By using the methods of real analysis and functional analysis, a Hilbert-type integral inequality in the subinterval (a,∞) with the homogeneous kernel and a best constant factor and its operator expression are given. As applications, a few improved results, the equivalent forms and some new inequalities with the particular kernels are obtained.

2. Lemmas and main results

If λ > 0, the function kλ(x, y) is non-negative measurable in (0,∞)× (0,∞), satisfying kλ(ux, uy) = u^−λkλ(x, y) for any u, x, y > 0, then we call kλ(x, y) the homogeneous function of −λ−degree; if for any x, y > 0, k_λ(x, y) = k_λ(y, x), then we call the homogeneous function k_λ(x, y) is symmetric. Assume that r >

1,¹_r + ¹_s = 1. Setting k_λ(r) and ek_λ(s) as kλ(r) :=

Z ∞

0

kλ(u,1)u^λr−1du,ekλ(s) :=

Z ∞

0

kλ(1, u)u^λs−1du, then it follows k_λ(r) = ek_λ(s). In fact, setting v = _u¹,we obtain

ekλ(s) = Z ∞

0

kλ(1,1

v)v^{−λs +1}dv v² =

Z ∞

0

kλ(v,1)v^λr−1dv=kλ(r).

Suppose that k_λ(r) is a positive number. For a > 0, x, y ∈ (a,∞), define the weight functions ω_λ(r, y, a) and$_λ(s, x, a) as:

ωλ(r, y, a) :=

Z ∞

a

kλ(x, y) y^λs

x^1−λr dx, $λ(s, x, a) :=

Z ∞

a

kλ(x, y) x^λr

y^1−λsdy. (2.1)

Settingu= ^y_x in the integral ω_λ(r, y, a), for anyy∈(a,∞), we find ω_λ(r, y, a) =

Z ^y_a

0

k_λ(1, u)u^λs−1du≤ Z ∞

0

k_λ(1, u)u^λs−1du=ek_λ(s).

Similarly, $_λ(s, x, a)≤k_λ(r) (x∈(a,∞)).Setting θ_λ(r) and θe_λ(s) as θ_λ(r) :=

Z 1

0

k_λ(u,1)u^λr−1du,θe_λ(s) :=

Z 1

0

k_λ(1, u)u^λs−1du, if θ_λ(r),θe_λ(s)>0, then for any y > a,we find

ω_λ(r, y, a) = Z ^y_a

0

k_λ(1, u)u^λs−1du≥ Z 1

0

k_λ(1, u)u^λs−1du=eθ_λ(s)>0.

Similarly, $_λ(s, x, a) ≥ θ_λ(r) > 0 (x > a). Hence by (2.1), for fixed y >

a, k_λ(x, y)>0 a.e. in (a,∞), and for fixedx > a, k_λ(x, y)>0 a.e. in (a,∞).

Lemma 2.1. If both k_λ(1, u), k_λ(u,1)≥l_λ >0, u∈(0,1], then we have ω_λ(r, y, a) ≤ k_λ(r)[1− rl_λ

λkλ(r)(a

y)^λr] (y∈(a,∞)); (2.2)

$_λ(s, x, a) ≤ k_λ(r)[1− sl_λ λk_λ(r)(a

x)^λs] (x∈(a,∞)). (2.3) Proof. Setting u= ^x_y, we find

ω_λ(r, y, a) = Z ∞

a y

k_λ(u,1)u^λr−1du=k_λ(r)− Z ^a_y

0

k_λ(u,1)u^λr−1du

≤ k_λ(r)−l_λ Z ^a_y

0

u^λr−1du=k_λ(r)− rl_λ λ (a

y)^λr (y∈(a,∞)).

Hence we have (2.2). Similarly we have (2.3). The lemma is proved.

Lemma 2.2. If both kλ(1, u) and kλ(u,1) are derivable decreasing function in (0,1], then we have

ω_λ(r, y, a) ≤ k_λ(r)[1− θ_λ(r) k_λ(r)(a

y)^λr] (y∈(a,∞)); (2.4)

$_λ(s, x, a) ≤ k_λ(r)[1− θe_λ(s) kλ(r)(a

x)^λs] (x∈(a,∞)). (2.5) In particular, if k_λ(x, y) is symmetric, setting k_λ :=k_λ(2), then

ω_λ(2, y, a)≤k_λ[1−1 2(a

y)^λ2];$_λ(2, x, a)≤k_λ[1− 1 2(a

x)^λ2] (x, y > a). (2.6) Proof. Since k_λ⁰(u,1)≤0, u∈(0,1), for y∈(0,1), we obtain

d dy[y^−λr

Z y

0

k_λ(u,1)u^λr−1du] = −λ r y^−λr

Z y

0

k_λ(u,1)u^λr−1du+k_λ(y,1)y⁻¹

=−y^−λr Z y

0

kλ(u,1)du^λr +kλ(y,1)y⁻¹ =y^−λr Z y

0

k_λ⁰(u,1)u^λrdu≤0,

and y^−λr Ry

0 k_λ(u,1)u^λr−1du≥R1

0 k_λ(u,1)u^λr−1du=θ_λ(r). Hence we find ω_λ(r, y, a) = k_λ(r)−[(a

y)^−λr Z ^a_y

0

k_λ(u,1)u^λr−1du](a y)^λr

≤ k_λ(r)−θ_λ(r)(a

y)^λr (y∈(a,∞)).

Then we obtain (2.4). Similarly, we obtain (2.5). If k_λ(x, y) is symmetric, then we find θ_λ(2) =θe_λ(2) and

k_λ =θ_λ(2) + Z ∞

1

k_λ(1, u)u^λ2−1du=θ_λ(2) + Z 1

0

k_λ(v,1)v^λ2−1dv= 2θ_λ(2).

Then by (2.4) and (2.5), we have (2.6). The lemma is proved.

For the measurable function ϕ(x)>0,set the function spaces as:

L^ρ_ϕ(a,∞) :={h≥0;||h||ρ,ϕ :={ Z ∞

a

ϕ(x)h^ρ(x)dx}^1ρ <∞}(ρ =p, q).

Theorem 2.3. Assume that p, r > 1,¹_p + ¹_q = 1,¹_r + ¹_s = 1, λ > 0, kλ(x, y) is a homogeneous function of −λ−degree in (0,∞)× (0,∞), satisfying k_λ(r), θ_λ(r) and θeλ(s) are positive numbers. For a >0, there exist measurable functions κ(y) and µ(x),e such that 0< κ(y),µ(x)e ≤1 and

ω_λ(r, y, a)≤k_λ(r)κ(y), $_λ(s, x, a)≤k_λ(r)µ(x)(x, ye ∈(a,∞)). (2.7) Ifφ_r(x) := x^{p(1−λr)−1}, ψ_s(x) :=x^{q(1−λs)−1}(x∈(a,∞)), f ∈L^p_φ

r(a,∞), g ∈L^q_ψ

s(a,∞),

||f||_p,φr,||g||_q,ψs >0, then we have the equivalent inequalities as I_λ(a) :=

Z ∞

a

Z ∞

a

k_λ(x, y)f(x)g(y)dxdy < k_λ(r)||f||_p,µ·φ_e r||g||q,κ·ψs; (2.8)

J_λ(a) :=

Z ∞

a

y^{pλs −1} κ^p−1(y)(

Z ∞

a

k_λ(x, y)f(x)dx)^pdy < k_λ^p(r)||f||^p_p,eµ·φ

r, (2.9) where the constant factors k_λ(r) and k_λ^p(r) are the best possible.

In particular, for κ(y) = µ(x) = 1,e we have the equivalent inequalities as:

I_λ(a)< k_λ(r){

Z ∞

a

x^{p(1−λr)−1}f^p(x)dx}^1p{ Z ∞

a

y^{q(1−λs)−1}g^q(y)dy}^1q; (2.10) Z ∞

a

y^{pλs −1}( Z ∞

a

k_λ(x, y)f(x)dx)^pdy < k_λ^p(r) Z ∞

a

x^{p(1−λr)−1}f^p(x)dx. (2.11) Proof. Since 0< θ_λ(r)/k_λ(r)≤κ(y)≤k_λ(r),0<eθ_λ(s)/k_λ(r)≤µ(x)e ≤k_λ(r) (x, y ∈ (a,∞)), it is obvious that the condition 0 < ||f||_p,φr,||g||_q,ψs < ∞ is equivalent to the condition 0<||f||_p,eµ·φr,||g||q,κ·ψs <∞.

By H¨older’s inequality [3], in view of (2.1) and Fubini’s theorem [4], we find Iλ(a) =

Z ∞

a

Z ∞

a

kλ(x, y)[x^(1−λr)/q

y^(1−λs)/pf(x)][y^(1−λs)/p

x^(1−λr)/qg(y)]dxdy

≤ { Z ∞

a

Z ∞

a

k_λ(x, y)x^{(1−λr)(p−1)}

y^1−λs f^p(x)dxdy}^1p

×{

Z ∞

a

Z ∞

a

k_λ(x, y)y^{(1−λs)(q−1)}

x^1−λr g^q(y)dxdy}^1q

= { Z ∞

a

$λ(s, x, a)φr(x)f^p(x)dx}^1p{ Z ∞

a

ωλ(r, y, a)ψs(y)g^q(y)dy}^1q.(2.12) If inequality (2.12) keeps the form of equality, then [3] there exist constantsA and B, such that they are not all zero and

Ax^{(1−λr)(p−1)}

y^1−λs f^p(x) =By^{(1−λs)(q−1)}

x^1−λr g^q(y) a.e. in (a,∞)×(a,∞).

It followsAx^p(1−λr)f^p(x) = By^q(1−λs)g^q(y) a.e. in (a,∞)×(a,∞). Assuming that A6= 0, there exists y > a, x^{p(1−λr)−1}f^p(x) = [By^q(1−λs)g^q(y)]_Ax¹ a.e. inx∈(a,∞).

This contradicts the fact that 0<||f||p,φr <∞.Then inequality (2.12) keeps the strict form and inequality (2.12) is valid by using (2.7).

Forx∈(a,∞),setting a bounded measurable function [f(x)]_n as [f(x)]n := min{f(x), n}=

f(x), for f(x)≤n n, forf(x)> n, since ||f||_p,φr >0, there exists n₀ ∈N,such that Rn

a φ_r(x)[f(x)]^p_ndx >0(n≥n₀), and then Rn

a µ(x)φe _r(x)[f(x)]^p_ndx >0. Setting eg_n(y) as eg_n(y) := y^pλs−1

κ^p−1(y)( Z n

a

k_λ(x, y)[f(x)]_ndx)^p−1(y∈(a, n);n ≥n₀), then by (2.8), we have

0 <

Z n

a

κ(y)ψ_s(y)eg_n^q(y)dy= Z n

a

y^pλs−1 κ^p−1(y)(

Z n

a

k_λ(x, y)[f(x)]_ndx)^pdy

= Z n

a

Z n

a

k_λ(x, y)[f(x)]_neg_n(y)dxdy

< k_λ(r){

Z n

a eµ(x)φ_r(x)[f(x)]^p_ndx}^1p{ Z n

a

κ(y)ψ_s(y)eg^q_n(y)dy}^1q <∞;(2.13) 0<

Z n

a

κ(y)ψ_s(y)eg_n^q(y)dy < k^p_λ(r) Z ∞

a µ(x)φe _r(x)f^p(x)dx <∞. (2.14) It follows 0 <||g||_q,κ·ψs < ∞ and 0 < ||g||_q,ψs <∞. For n → ∞, by (2.8), both (2.13) and (2.14) still keep the forms of strict inequality. Hence we have (2.9).

On the other-hand, suppose that (2.9) is valid. By H¨older’s inequality, J_λ(a) =

Z ∞

a

[y^{−1p +λs}κ^−1q (y) Z ∞

a

k_λ(x, y)f(x)dx][κ^1q(y)y^1p−λsg(y)]dy

≤ { Z ∞

a

y^{pλs −1} κ^p−1(y)(

Z ∞

a

k_λ(x, y)f(x)dx)^pdy}^p1{ Z ∞

a

κ(y)ψ_s(y)g^q(y)dy}^1q. (2.15)

In view of (2.9), we have (2.8). Hence (2.8) is equivalent to (2.9).

For n ∈ N, n > max{^λ_r,^λ_s}, setting fn, gn as: fn(x) = x^{λr−np1 −1}, gn(x) = x^{λs−nq1 −1}, for x ∈ (a,∞), if there exists 0 < K ≤ k_λ(r) = ek_λ(s), such that (2.8) is still valid if we replace kλ(r) by K, then we have

I_λ⁽ⁿ⁾(a) :=

Z ∞

a

Z ∞

a

k_λ(x, y)f_n(x)g_n(y)dxdy < K||f_n||p,·φr||g_n||_q,ψs = nK

aⁿ¹ ; (2.16) I_λ⁽ⁿ⁾(a) =

Z ∞

a

[ Z ∞

a

k_λ(x, y)x^{λr−np1 −1}y^{λs−nq1 −1}dx]dy

u=y/x

= Z ∞

a

y⁻¹⁻ⁿ¹[ Z ^y_a

0

k_λ(1, u)u^{λs+np1 −1}du]dy

= n

aⁿ¹ Z 1

0

k_λ(1, u)u^{λs+np1 −1}du+ Z ∞

a

y⁻¹⁻¹ⁿ Z ^y_a

1

k_λ(1, u)u^{λs+np1 −1}dudy

= n

aⁿ¹ Z 1

0

kλ(1, u)u^{λs+np1 −1}du+ Z ∞

1

( Z ∞

au

y⁻¹⁻¹ⁿdy)kλ(1, u)u^{λs+np1 −1}du

= n

aⁿ¹[ Z 1

0

k_λ(1, u)u^{λs+np1 −1}du+ Z ∞

1

k_λ(1, u)u^{λs−nq1 −1}du]. (2.17) Hence by (2.16) and (2.17), we have

Z 1

0

k_λ(1, u)u^{λs+np1 −1}du+ Z ∞

1

k_λ(1, u)u^{λs−nq1 −1}du < K, and by Fatou’s lemma [4], it follows

ek_λ(s) = Z 1

0 lim

n→∞ k_λ(1, u)u^{λs+np1 −1}du+ Z ∞

1 lim

n→∞ k_λ(1, u)u^{λs−nq1 −1}du

≤ _n→∞^lim [ Z 1

0

k_λ(1, u)u^{λs+np1 −1}du+ Z ∞

1

k_λ(1, u)u^{λs−nq1 −1}du]≤K.

Therefore K = k_λ(r) is the best constant factor of (2.8). If the constant factor in (2.9) is not the best possible, then by (2.15), we can get a contradiction that the constant factor in (2.8) is not the best possible. The theorem is proved.

Define an operatorT_a:L^p_φ

r(a,∞)→L^p

ψs^1−p

(a,∞) as: for f ∈L^p_φ

r(a,∞), (T_af)(y) :=

Z ∞

a

k_λ(x, y)f(x)dx(y∈(a,∞)).

In view of (2.11), it follows Taf ∈ L^p

ψs^1−p(a,∞). For g ∈ L^q_ψs(a,∞), define the formal inner of T_af and g as:

(T_af, g) :=

Z ∞

a

Z ∞

a

k_λ(x, y)f(x)g(y)dxdy.

Hence the equivalent inequalities (2.10) and (2.11) may be rewritten as (T_af, g)< k_λ(r)||f||_p,φr||g||_q,ψs;||T_af||_p,ψ^1−p

s < k_λ(r)||f||_p,φr,

where the constant factor is the best possible,T_ais obviously bounded and||T_a||= k_λ(r). We call T_a Hilbert-type integral operator with the homogeneous kernel of

−λ-degree in the subinterval (a,∞).

Corollary 2.4. Let the assumptions of Theorem 2.3 be fulfilled and additionally both k_λ(1, u), k_λ(u,1)≥l_λ >0, u∈ (0,1]. Then we have the following equivalent inequalities:

(T_af, g) < k_λ(r){

Z ∞

a

[1− sl_λ λk_λ(r)(a

x)^λs]φ_r(x)f^p(x)dx}^p1

×{

Z ∞

a

[1− rl_λ λkλ(r)(a

y)^λr]ψ_s(y)g^q(y)dy}^1q; (2.18) Z ∞

a

y^{pλs −1} [1− _λk^rlλ

λ(r)(^a_y)^λr]^p−1( Z ∞

a

k_λ(x, y)f(x)dx)^pdy

< k^p_λ(r) Z ∞

a

[1− sl_λ λk_λ(r)(a

x)^λs]φ_r(x)f^p(x)dx, (2.19) where the constant factors k_λ(r) andk_λ^p(r) are the best possible. We still have the following two pairs of equivalent inequalities:

(T_af, g)< k_λ(r){

Z ∞

a

[1− sl_λ λk_λ(r)(a

x)^λs]φ_r(x)f^p(x)dx}^1p||g||_q,ψs, (2.20)

||T_af||^p

p,ψ^1−ps < k_λ^p(r) Z ∞

a

[1− sl_λ λk_λ(r)(a

x)^λs]φ_r(x)f^p(x)dx; (2.21) (T_af, g)< k_λ(r)||f||_p,φr{

Z ∞

a

[1− rlλ

λk_λ(r)(a

y)^λr]ψ_s(y)g^q(y)dy}^1q, (2.22) Z ∞

a

y^{pλs −1} [1− _λk^rlλ

λ(r)(^a_y)^λr]^p−1( Z ∞

a

k_λ(x, y)f(x)dx)^pdy < k_λ^p(r)||f||^p_p,φ

r. (2.23) Proof. By Lemma 2.1, setting κ(y) = 1−_λk^rlλ

λ(r)(^a_y)^λr andµ(x) = 1e −_λk^slλ

λ(r)(^a_x)^λs in (2.8) and (2.9), we have (2.18) and (2.19). Since κ(y) ≤1, by (2.18) and (2.19), we have (2.20) and (2.21). Similarly, since µ(x)e ≤ 1, we have (2.22) and (2.23).

The corollary is proved.

Corollary 2.5. Let the assumptions of Theorem 2.3 be fulfilled and additionally bothk_λ(1, u)andk_λ(u,1)are derivable decreasing function in(0,1].Then we have the following equivalent inequalities with the best constant factors:

(T_af, g) < k_λ(r){

Z ∞

a

[1− eθ_λ(s) k_λ(r)(a

x)^λs]φ_r(x)f^p(x)dx}^1p

×{

Z ∞

a

[1− θ_λ(r) k_λ(r)(a

y)^λr]ψ_s(y)g^q(y)dy}^1q; (2.24)

Z ∞

a

y^{pλs −1} [1− ^θ_k^λ(r)

λ(r)(^a_y)^λr]^p−1( Z ∞

a

k_λ(x, y)f(x)dx)^pdy

< k_λ^p(r) Z ∞

a

[1− θe_λ(s) k_λ(r)(a

x)^λs]φ_r(x)f^p(x)dx. (2.25) If k_λ(x, y) is symmetric and σ(x) = 1− ¹₂(^a_x)^λ2 as (1.4), then we have the equivalent inequalities as:

I_λ(a)< k_λ{ Z ∞

a

σ(x)φ₂(x)f^p(x)dx}^1p{ Z ∞

a

σ(y)ψ₂(y)g^q(y)dy}^1q; (2.26) Z ∞

a

y^{pλ2 −1} [σ(y)]^p−1(

Z ∞

a

k_λ(x, y)f(x)dx)^pdy < k^p_λ Z ∞

a

σ(x)φ₂(x)f^p(x)dx. (2.27) Proof. By Lemma 2.2, setting κ(y) = 1− ^θ_k^λ(r)

λ(r)(^a_y)^λr and eµ(x) = 1− _k^eθλ(s)

λ(r)(^a_x)^λs in (2.8) and (2.9), we have (2.24) and (2.25). Forr=s= 2,by (2.6), we have (2.26)

and (2.27). The corollary is proved.

3. Applications to some particular kernels

In the following, we assume thata, λ >0, p, r >1,¹_p+¹_q = 1,¹_r+¹_s = 1, φr(x) :=

x^{p(1−λr)−1}, ψ_s(x) := x^{q(1−λs)−1}, σ(x) = 1− ¹₂(^a_x)^λ2, f, g ≥0, 0< ||f||_p,φr,||g||_q,ψs <

∞.Some words that the constants are the best possible are omitted.

Example 3.1. Letk_λ(x, y) = _(xα+y^1α)^λ/α (α >0),which is symmetric. Since both k_λ(1, u) andk_λ(u,1) are derivable decreasing in (0,1], and k_λ(u,1) = _(uα+1)¹ λ/α ≥ l_λ = ₂λ/α¹ (u∈(0,1]), setting v =u^α, by (1.3), we have

ek_λ(r) :=

Z ∞

0

u^λr−1du (u^α+ 1)^λα = 1

α Z ∞

0

v^αrλ−1dv (v+ 1)^λα = 1

αB( λ αr, λ

αs).

By (2.18), (2.19) and (2.26), (2.27), we have two pairs of equivalent inequalities as:

H(a) : = Z ∞

a

Z ∞

a

f(x)g(y)dxdy (x^α+y^α)^αλ

< ek_λ(r){

Z ∞

a

[1− s

2^λαλekλ(r)(a

x)^λs]φ_r(x)f^p(x)dx}^1p

×{

Z ∞

a

[1− r

2^αλλek_λ(r)(a

y)^λr]ψs(y)g^q(y)dy}^1q, Z ∞

a

y^{pλs −1} [1− ^r

2^λ/αλekλ(r)(^a_y)^λr]^p−1[ Z ∞

a

f(x)

(x^α+y^α)^λαdx]^pdy

< ek^p_λ(r) Z ∞

a

[1− s

2^αλλek_λ(r)(a

x)^λs]φr(x)f^p(x)dx;

H(a)<ek_λ(2){

Z ∞

a

σ(x)φ₂(x)f^p(x)dx}^1p{ Z ∞

a

σ(y)ψ₂(y)g^q(y)dy}^1q, (3.1) Z ∞

a

y^{pλ2 −1} σ^p−1(y)[

Z ∞

a

f(x)dx

(x^α+y^α)^αλ]^pdy <ek_λ^p(2) Z ∞

a

σ(x)φ₂(x)f^p(x)dx.

Example 3.2. Let k_λ(x, y) = ^ln(x/y)_xλ−y^λ, which is symmetric. We find that both k_λ(1, u) andk_λ(u,1) are derivable decreasing in (0,1], [16] andk_λ(u,1) = _u^lnλ−1^u ≥ lλ = ¹_λ (u∈(0,1]). Settingv =u^λ, we obtain [2]

k_λ(r) = Z ∞

0

(lnu)u^λr−1 u^λ−1 du=

Z ∞

0

(lnv)v^1r−1

λ²(v−1)dv = [ π λsin(^π_r)]².

By (2.18), (2.19) and (2.26), (2.27), we have two pairs of equivalent inequalities as:

H⁰(a) :=

Z ∞

a

Z ∞

a

ln(^x_y)f(x)g(y) x^λ−y^λ dxdy

< [ π λsin(^π_r)]²{

Z ∞

a

[1−s[sin(^π_r) π ]²(a

x)^λs]φr(x)f^p(x)dx}^1p

×{

Z ∞

a

[1−r[sin(^π_r) π ]²(a

y)^λr]ψ_s(y)g^q(y)dy}^1q, Z ∞

a

y^pλs−1 [1−r[^sin(

π r)

π ]²(^a_y)^λr]^p−1 (

Z ∞

a

ln(^x_y)f(x)

x^λ−y^λ dx)^pdy

< [ π λsin(^π_r)]^2p

Z ∞

a

[1−s[sin(^π_r) π ]²(a

x)^λs]φr(x)f^p(x)dx;

H⁰(a)<(π λ)²{

Z ∞

a

σ(x)φ₂(x)f^p(x)dx}^1p{ Z ∞

a

σ(y)ψ₂(y)g^q(y)dy}^1q, Z ∞

a

y^{pλ2 −1} σ^p−1(y)[

Z ∞

a

ln(^x_y)f(x)

x^λ−y^λ dx]^pdy <(π λ)^2p

Z ∞

a

σ(x)φ₂(x)f^p(x)dx.

Example 3.3. Letk_λ(x, y) = _(max{x,y})¹ λ,which is symmetric.Since bothk_λ(1, u) and kλ(u,1) are derivable decreasing in (0,1], and kλ(u,1) = lλ = 1 (u ∈(0,1]), we have

k_λ(r) = Z ∞

0

u^λr−1

(max{u,1})^λdu= Z 1

0

u^λr−1du+ Z ∞

1

u^λr−1

u^λ du= rs λ. Then by (2.18) and (2.19), we have two equivalent inequalities as follows:

Z ∞

a

Z ∞

a

f(x)g(y)dxdy

(max{x, y})^λ < rs λ{

Z ∞

a

[1− 1 r(a

x)^λs]φ_r(x)f^p(x)dx}^p1

×{

Z ∞

a

[1−1 s(a

y)^λr]ψs(y)g^q(y)dy}^1q,

Z ∞

a

y^{pλs −1} [1− ¹_s(^a_y)^λr]^p−1[

Z ∞

a

f(x)dx

(max{x, y})^λ]^pdy

< (rs λ)^p

Z ∞

a

[1− 1 r(a

x)^λs]φr(x)f^p(x)dx;

Example 3.4. Letk_λ(x, y) = _(max{x,y})^|ln(x/y)|λ,which is symmetric.We find that k_λ(r) =

Z ∞

0

|lnu|u^λr−1du (max{u,1})^λ

= Z 1

0

(−lnu)u^λr−1du+ Z ∞

1

(lnu)u^λr−1

u^λ du = r²+s² λ² ; ω_λ(r, y, a) =

Z ∞

a y

|lnu|u^λr−1

(max{u,1})^λdu= r²+s² λ² −

Z ^a_y

0

(−lnu)u^λr−1du

= r²+s² λ² − r

λ Z ^a_y

0

(−lnu)du^λr ≤ r²+s² λ² κ(y), κ(y) = 1− r²

r²+s²(a y)^λr;

$_λ(s, x, a)≤ r²+s²

λ² µ(x),e µ(x) = 1e − s² r²+s²(a

x)^λs. Then by (2.8) and (2.9), we have two equivalent inequalities as follows:

Z ∞

a

Z ∞

a

|ln(^x_y)|f(x)g(y) (max{x, y})^λ dxdy

< r²+s² λ² {

Z ∞

a

[1− s² r²+s²(a

x)^λs]φ_r(x)f^p(x)dx}^1p

×{

Z ∞

a

[1− r² r²+s²(a

y)^λr]ψ_s(y)g^q(y)dy}^1q; Z ∞

a

y^{pλs −1}

[1− _r2^r+s^{2 2}(^a_y)^λr]^p−1[ Z ∞

a

|ln(^x_y)|f(x)dx (max{x, y})^λ ]^pdy

< (r²+s² λ² )^p

Z ∞

a

[1− s² r²+s²(a

x)^λs]φ_r(x)f^p(x)dx.

Remark 3.5. (i) For α = 1, p = q = 2, inequality (3.1) deduces to (1.2). (ii) Inequality (2.8) is a refinement of (2.10), because of

I_λ(a)< k_λ(r)||f||_p,eµ·φr||g||q,κ·ψs ≤k_λ(r)||f||_p,φr,||g||_q,ψs.

(iii) When a→0⁺, (2.10) deduces to a Hilbert-type integral inequality in (0,∞) with a best constant factor k_λ(r) as:

I_λ(0)≤k_λ(r){

Z ∞

0

φ_r(x)f^p(x)dx}^1p{ Z ∞

0

ψ_s(y)g^q(y)dy}^1q.

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1Department of Mathematics, Guangdong Education Institute and Guangzhou, Guangdong 510303, P. R. China.

E-mail address: [email protected]

2 Department of Mathematics, National Technical University of Athens, Zo- grafou campus, 15780, Athens, Greece.

E-mail address: [email protected]