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A half-discrete Hilbert-type inequality with a homogeneous kernel and an extension

Journal of Inequalities and Applications20112011:124

https://doi.org/10.1186/1029-242X-2011-124

Received: 28 August 2011

Accepted: 30 November 2011

Published: 30 November 2011

Abstract

Using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel is obtained, and a best extension with two interval variables is given. The equivalent forms, the operator expressions, the reverses and some particular cases are considered.

2000 Mathematics Subject Classification: 26D15; 47A07.

Keywords

Hilbert-type inequalityhomogeneous kernelweight functionequivalent formreverse

1 Introduction

Assuming that p > 1, 1 p + 1 q = 1 , f (≥ 0) L p (R+), g(≥ 0) L q (R+), f p = 0 f p ( x ) d x 1 p > 0 , || g || q > 0, we have the following Hardy-Hilbert's integral inequality [1]:
0 0 f ( x ) g ( y ) x + y d x d y < π s i n ( π / p ) | | f | | p | | g | | q ,
(1)
where the constant factor π s i n ( π / p ) is the best possible. If a m , b n ≥ 0, a = { a m } m = 1 l p , b = { b n } n = 1 l q , | | a | | p = m = 1 a m p 1 p > 0 , || b || q > 0, then we still have the following discrete Hardy-Hilbert's inequality with the same best constant factor π s i n ( π / p ) :
m = 1 n = 1 a m b n m + n < π s i n ( π / p ) | | a | | p | | b | | q .
(2)

For p = q = 2, the above two inequalities reduce to the famous Hilbert's inequalities. Inequalities (1) and (2) are important in analysis and its applications [24].

In 1998, by introducing an independent parameter λ (0, 1], Yang [5] gave an extension of (1) for p = q = 2. Refinement and generalizing the results from [5], Yang [6] gave some best extensions of (1) and (2) as follows: If λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative homogeneous function of degree - λ satisfying for any x, y, t > 0, kλ(tx, ty) = t kλ (x, y), k ( λ 1 ) = 0 k λ ( t , 1 ) t λ 1 - 1 d t R + , ϕ ( x ) = x p ( 1 - λ 1 ) - 1 , ψ ( x ) = x q ( 1 - λ 2 ) - 1 , f ( 0 ) L p , ϕ ( R + ) = { f | | | f | | p , ϕ : = { 0 ϕ ( x ) | f ( x ) | p d x } 1 p < } , g(≥ 0) L q , ψ , || f || p , ϕ , || g || q , ψ > 0, then we have
0 0 k λ ( x , y ) f ( x ) g ( y ) d x d y < k ( λ 1 ) | | f | | p , ϕ | | g | | q , ψ ,
(3)
where the constant factor k1) is the best possible. Moreover, if kλ(x, y) is finite and k λ ( x , y ) x λ 1 - 1 ( k λ ( x , y ) y λ 2 - 1 ) is decreasing with respect to x > 0(y > 0), then for a m ,b n ≥ 0, a = { a m } m = 1 l p , ϕ = { a | | | a | | p , ϕ : = { n = 1 ϕ ( n ) | a n | p } 1 p < } , b = { b n } n = 1 l q , ψ , || a || p , ϕ , || b || q , Ψ > 0, we have
m = 1 n = 1 k λ ( m , n ) a m b n < k ( λ 1 ) | | a | | p , ϕ | | b | | q , ψ ,
(4)

with the best constant factor k1). Clearly, for λ = 1, k 1 ( x , y ) = 1 x + y , λ 1 = 1 q , λ 2 = 1 p (3) reduces to (1), and (4) reduces to (2). Some other results about Hilbert-type inequalities are provided by [715].

On half-discrete Hilbert-type inequalities with the non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors are the best possible. And, Yang [16] gave a result by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang [17] gave the following half-discrete Hilbert's inequality with the best constant factor B1, λ2)(λ, λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ):
0 f ( x ) n = 1 a n ( x + n ) λ d x < B ( λ 1 , λ 2 ) | | f | | p , φ | | a | | q , ψ .
(5)
In this article, using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel and a best constant factor is given as follows:
0 f ( x ) n = 1 k λ ( x , n ) a n d x < k ( λ 1 ) | | f | | p , φ | | a | | q , ψ ,
(6)

which is a generalization of (5). A best extension of (6) with two interval variables, some equivalent forms, the operator expressions, the reverses and some particular cases are considered.

2 Some lemmas

We set the following conditions:

Condition (i) v(y)(y [n0 - 1, ∞)) is strictly increasing with v(n0 - 1) ≥ 0 and for any fixed x (b, c), f(x, y) is decreasing for y (n0 - 1, ∞) and strictly decreasing in an interval of (n0 - 1, ∞).

Condition (ii) v ( y ) ( y [ n 0 - 1 2 , ) ) is strictly increasing with v n 0 - 1 2 0 and for any fixed x (b, c), f(x, y) is decreasing and strictly convex for y n 0 - 1 2 , .

Condition (iii) There exists a constant β ≥ 0, such that v(y)(y [n0 - β, ∞)) is strictly increasing with v(n0 - β) ≥ 0, and for any fixed x (b, c), f(x, y) is piecewise smooth satisfying
R ( x ) : = n 0 - β n 0 f ( x , y ) d y - 1 2 f ( x , n 0 ) - n 0 ρ ( y ) f y ( x , y ) d y > 0 ,

where ρ ( y ) ( = y - [ y ] - 1 2 ) is Bernoulli function of the first order.

Lemma 1 If λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree - λ in R + 2 , u ( x ) ( x ( b , c ) , - b < c ) and v(y)(y [n0, ∞), n0 N) are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c - ) = v(∞) = ∞, setting K(x, y) = kλ(u(x), v(y)), then we define weight functions ω(n) and ϖ(x) as follows:
ω ( n ) : = [ v ( n ) ] λ 2 b c K ( x , n ) [ u ( x ) ] λ 1 1 u ( x ) d x , n n 0 ( n N ) ,
(7)
ϖ ( x ) : = [ u ( x ) ] λ 1 n = n 0 K ( x , n ) [ v ( n ) ] λ 2 1 v ( n ) , x ( b , c ) .
(8)
It follows
ω ( n ) = k ( λ 1 ) : = 0 k λ ( t , 1 ) t λ 1 - 1 d t .
(9)
Moreover, setting f ( x , y ) : = [ u ( x ) ] λ 1 K ( x , y ) [ v ( y ) ] λ 2 1 v ( y ) , if k1) R+ and one of the above three conditions is fulfilled, then we still have
ϖ ( x ) < k ( λ 1 ) ( x ( b , c ) ) .
(10)
Proof . Setting t = u ( x ) v ( n ) in (7), by calculation, we have (9).
  1. (i)
    If Condition (i) is fulfilled, then we have
    ϖ ( x ) = n = n 0 f ( x , n ) < [ u ( x ) ] λ 1 n 0 1 K ( x , y ) [ v ( y ) ] λ 2 1 v ( y ) d y = t = u ( x ) / v ( y ) 0 u ( x ) v ( n 0 1 ) k λ ( t , 1 ) t λ 1 1 d t k ( λ 1 ) .
     
  2. (ii)
    If Condition (ii) is fulfilled, then by Hadamard's inequality [18], we have
    ϖ ( x ) = n = n 0 f ( x , n ) < n 0 - 1 2 f ( x , y ) d y = t = u ( x ) / v ( y ) 0 u ( x ) v ( n 0 - 1 2 ) k λ ( t , 1 ) t λ 1 - 1 d t k ( λ 1 ) .
     
  3. (iii)
    If Condition (iii) is fulfilled, then by Euler-Maclaurin summation formula [6], we have
    ϖ ( x ) = n = n 0 f ( x , n ) = n 0 f ( x , y ) d y + 1 2 f ( x , n 0 ) + n 0 ρ ( y ) f y ( x , y ) d y = n 0 - β f ( x , y ) d y - R ( x ) = 0 u ( x ) v ( n 0 - β ) k λ ( t , 1 ) t λ 1 - 1 d t - R ( x ) k ( λ 1 ) - R ( x ) < k ( λ 1 ) .
     

The lemma is proved.    ■

Lemma 2 Let the assumptions of Lemma 1 be fulfilled and additionally, p > 0(p ≠ 1), 1 p + 1 q = 1 , a n ≥ 0, nn0(n N), f (x) is a non-negative measurable function in (b, c). Then, (i) for p > 1, we have the following inequalities:
J 1 : = n = n 0 v ( n ) [ v ( n ) ] 1 - p λ 2 d c K ( x , n ) f ( x ) d x p 1 p [ k ( λ 1 ) ] 1 q d c ϖ ( x ) [ u ( x ) ] p ( 1 - λ 1 ) - 1 [ u ( x ) ] p - 1 f p ( x ) d x 1 p ,
(11)
L 1 : = b c [ ϖ ( x ) ] 1 - q u ( x ) [ u ( x ) ] 1 - q λ 1 n = n 0 K ( x , n ) a n q d x 1 q k ( λ 1 ) n = n 0 [ v ( n ) ] q ( 1 - λ 2 ) - 1 [ v ( n ) ] q - 1 a n q 1 q ;
(12)

(ii) for 0 < p < 1, we have the reverses of (11) and (12).

Proof(i) By Hölder's inequality with weight [18] and (9), it follows
[ b c K ( x , n ) f ( x ) d x ] p = { b c K ( x , n ) [ [ u ( x ) ] ( 1 λ 1 ) / q [ v ( n ) ] ( 1 λ 2 ) / p [ v ( n ) ] 1 / p [ u ( x ) ] 1 / q f ( x ) ] × [ [ v ( n ) ] ( 1 λ 2 ) / p [ u ( x ) ] ( 1 λ 1 ) / q [ u ( x ) ] 1 / q [ v ( n ) ] 1 / p ] d x } p b c K ( x , n ) [ u ( x ) ] ( 1 λ 1 ) ( p 1 ) v ( n ) [ v ( n ) ] 1 λ 2 [ u ( x ) ] p 1 f p ( x ) d x × { b c K ( x , n ) [ v ( n ) ] ( 1 λ 2 ) ( q 1 ) u ( x ) [ u ( x ) ] 1 λ 1 [ v ( n ) ] q 1 d x } p 1 = { ω ( n ) [ v ( n ) ] q ( 1 λ 2 ) 1 [ v ( n ) ] q 1 } p 1 b c K ( x , n ) [ u ( x ) ] ( 1 λ 1 ) ( p 1 ) v ( n ) f p ( x ) [ v ( n ) ] 1 λ 2 [ u ( x ) ] p 1 d x = [ k ( λ 1 ) ] p 1 [ v ( n ) ] p λ 2 1 v ( n ) b c K ( x , n ) [ u ( x ) ] ( 1 λ 1 ) ( p 1 ) v ( n ) [ v ( n ) ] 1 λ 2 [ u ( x ) ] p 1 f p ( x ) d x .
Then, by Lebesgue term-by-term integration theorem [19], we have
J 1 [ k ( λ 1 ) ] 1 q n = n 0 b c K ( x , n ) [ u ( x ) ] ( 1 - λ 1 ) ( p - 1 ) v ( n ) f p ( x ) [ v ( n ) ] 1 - λ 2 [ u ( x ) ] p - 1 d x 1 p = [ k ( λ 1 ) ] 1 q b c n = n 0 K ( x , n ) [ u ( x ) ] ( 1 - λ 1 ) ( p - 1 ) v ( n ) f p ( x ) [ v ( n ) ] 1 - λ 2 [ u ( x ) ] p - 1 d x 1 p = [ k ( λ 1 ) ] 1 q b c ϖ ( x ) [ u ( x ) ] p ( 1 - λ 1 ) - 1 [ u ( x ) ] p - 1 f p ( x ) d x 1 p ,

and (11) follows.

Still by Hölder's inequality, we have
n = n 0 K ( x , n ) a n q = n = n 0 K ( x , n ) [ u ( x ) ] ( 1 - λ 1 ) q [ v ( n ) ] ( 1 - λ 2 ) p [ v ( n ) ] 1 p [ u ( x ) ] 1 q × [ v ( n ) ] ( 1 - λ 2 ) p [ u ( x ) ] ( 1 - λ 1 ) q [ u ( x ) ] 1 q [ v ( n ) ] 1 p a n q n = n 0 K ( x , n ) [ u ( x ) ] ( 1 - λ 1 ) ( p - 1 ) [ v ( n ) ] 1 - λ 2 v ( n ) [ u ( x ) ] p - 1 q - 1 × n = n 0 K ( x , n ) [ v ( n ) ] ( 1 - λ 2 ) ( q - 1 ) [ u ( x ) ] 1 - λ 1 u ( x ) [ v ( n ) ] q - 1 a n q = [ u ( x ) ] 1 - q λ 1 [ ϖ ( x ) ] 1 - q u ( x ) n = n 0 K ( x , n ) u ( x ) [ u ( x ) ] 1 - λ 1 [ v ( n ) ] ( q - 1 ) ( 1 - λ 2 ) [ v ( n ) ] q - 1 a n q .
Then, by Lebesgue term-by-term integration theorem, we have
L 1 b c n = n 0 K ( x , n ) u ( x ) [ u ( x ) ] 1 - λ 1 [ v ( n ) ] ( q - 1 ) ( 1 - λ 2 ) [ v ( n ) ] q - 1 a n q d x 1 q = n = n 0 [ v ( n ) ] λ 2 b c K ( x , n ) u ( x ) d x [ u ( x ) ] 1 - λ 1 [ v ( n ) ] q ( 1 - λ 2 ) - 1 [ v ( n ) ] q - 1 a n q 1 q = n = n 0 ω ( n ) [ v ( n ) ] q ( 1 - λ 2 ) - 1 [ v ( n ) ] q - 1 a n q 1 q ,
and then in view of (9), inequality (12) follows.
  1. (ii)

    By the reverse Hölder's inequality [18] and in the same way, for q < 0, we have the reverses of (11) and (12). ■

     

3 Main results

We set Φ ( x ) : = [ u ( x ) ] p ( 1 - λ 1 ) - 1 [ u ( x ) ] p - 1 ( x ( b , c ) ) , Ψ ( n ) : = [ v ( n ) ] q ( 1 - λ 2 ) - 1 [ v ( n ) ] q - 1 ( n n 0 , n N ) , wherefrom [ Φ ( x ) ] 1 - q = u ( x ) [ u ( x ) ] 1 - q λ 1 , [ Ψ ( n ) ] 1 - p = v ( n ) [ v ( n ) ] 1 - p λ 2 .

Theorem 1 Suppose that λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree - λ¸ in R + 2 , u ( x ) ( x ( b , c ) , - b < c ) and v(y)(y [n0, ∞), n0 N are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c - ) = v(∞) = ∞, ϖ(x) < k1) R+(x (b, c)). If p > 1 , 1 p + 1 q = 1 , f ( x ), a n ≥ 0, f L p Φ ( b , c ), a = { a n } n = n 0 l q , Ψ , || f || p ,Φ > 0 and || a || q ,Ψ > 0, then we have the following equivalent inequalities:
I : = n = n 0 a n b c K ( x , n ) f ( x ) d x = b c f ( x ) n = n 0 K ( x , n ) a n d x < k ( λ 1 ) | | f | | p , Φ | | a | | q , Ψ ,
(13)
J : = n = n 0 [ Ψ ( n ) ] 1 - p b c K ( x , n ) f ( x ) d x p 1 p < k ( λ 1 ) | | f | | p , Φ ,
(14)
L : = b c [ Φ ( x ) ] 1 - q n = n 0 K ( x , n ) a n q d x 1 q < k ( λ 1 ) | | a | | q , Ψ .
(15)

Moreover, if v ( y ) v ( y ) ( y n 0 ) is decreasing and there exist constants δ < λ1 and M > 0, such that k λ ( t , 1 ) M t δ t 0 , 1 v ( n 0 ) , then the constant factor k1) in the above inequalities is the best possible.

Proof By Lebesgue term-by-term integration theorem, there are two expressions for I in (13). In view of (11), for ϖ(x) < k1) R+, we have (14). By Hölder's inequality, we have
I = n = n 0 Ψ - 1 q ( n ) b c K ( x , n ) f ( x ) d x [ Ψ 1 q ( n ) a n ] J | | a | | q , Ψ .
(16)
Then, by (14), we have (13). On the other hand, assuming that (13) is valid, setting
a n : = [ Ψ ( n ) ] 1 - p b c K ( x , n ) f ( x ) d x p - 1 , n n 0 ,
then J p -1 = || a || q . By (11), we find J < ∞. If J = 0, then (14) is naturally valid; if J > 0, then by (13), we have
| | a | | q , Ψ q = J p = I < k ( λ 1 ) | | f | | p , Φ | | a | | q , Ψ , | | a | | q , Ψ q - 1 = J < k ( λ 1 ) | | f | | p , Φ ,

and we have (14), which is equivalent to (13).

In view of (12), for [ϖ(x) ]1-q> [k1)]1-q, we have (15). By Hölder's in equality, we find
I = b c [ Φ 1 p ( x ) f ( x ) ] Φ - 1 p ( x ) n = n 0 K ( x , n ) a n d x | | f | | p , Φ L .
(17)
Then, by (15), we have (13). On the other hand, assuming that (13) is valid, setting
f ( x ) : = [ Φ ( x ) ] 1 - q n = n 0 K ( x , n ) a n q - 1 , x ( b , c ) ,
then L q -1 = || f || p Φ . By (12), we find L < ∞. If L = 0, then (15) is naturally valid; if L > 0, then by (13), we have
| | f | | p , Φ p = L q = I < k ( λ 1 ) | | f | | p , Φ | | a | | q , Ψ , | | f | | p , Φ p - 1 = L < k ( λ 1 ) | | a | | q , Ψ ,

and we have (15) which is equivalent to (13).

Hence, inequalities (13), (14) and (15) are equivalent.

There exists an unified constant d (b, c), satisfying u(d) = 1. For 0 < ε < p1 - δ), setting f ̃ ( x ) = 0 , x (b, d); f ̃ ( x ) = [ u ( x ) ] λ 1 - ε p - 1 u ( x ) , x (d, c), and ã n = [ v ( n ) ] λ 2 - ε q - 1 v ( n ) , nn0, if there exists a positive number k(≤ k1)), such that (13) is valid as we replace k1) by k, then in particular, we find
Ĩ : = n = n 0 b c K ( x , n ) ã n f ̃ ( x ) d x < k | | f ̃ | | p , Φ | | ã | | q , Ψ = k d c u ( x ) [ u ( x ) ] ε + 1 d x 1 p v ( n 0 ) [ v ( n 0 ) ] ε + 1 + n = n 0 + 1 v ( n ) [ v ( n ) ] ε + 1 1 q < k 1 ε 1 p v ( n 0 ) [ v ( n 0 ) ] ε + 1 + n 0 [ v ( y ) ] - ε - 1 v ( y ) d y 1 q = k ε ε v ( n 0 ) [ v ( n 0 ) ] ε + 1 + [ v ( n 0 ) ] - ε 1 q
(18)
I ˜ = n = n 0 [ v ( n ) ] λ 2 ε q 1 v ( n ) d c K ( x , n ) [ u ( x ) ] λ 1 ε p 1 u ( x ) d x = t = u ( x ) / v ( n ) n = n 0 [ v ( n ) ] ε 1 v ( n ) 1 / v ( n ) k λ ( t , 1 ) t λ 1 ε p 1 d t = k ( λ 1 ε p ) n = n 0 [ v ( n ) ] ε 1 v ( n ) A ( ε ) > k ( λ 1 ε p ) n 0 [ v ( y ) ] ε 1 v ( y ) d y A ( ε ) = 1 ε k ( λ 1 ε p ) [ v ( n 0 ) ] ε A ( ε ), A ( ε ) : = n = n 0 [ v ( n ) ] ε 1 v ( n ) 0 1 / v ( n ) k λ ( t, 1 ) t λ 1 ε p 1 d t .
(19)
For k λ ( t , 1 ) M 1 t δ ( δ < λ 1 ; t ( 0 , 1 v ( n 0 ) ] ) , we find
0 < A ( ε ) M n = n 0 [ v ( n ) ] - ε - 1 v ( n ) 0 1 v ( n ) t λ 1 - δ - ε p - 1 d t = M λ 1 - δ - ε p n = n 0 [ v ( n ) ] - λ 1 + δ - ε q - 1 v ( n ) = M λ 1 - δ - ε p v ( n 0 ) [ v ( n 0 ) ] λ 1 - δ + ε q + 1 + n = n 0 + 1 v ( n ) [ v ( n ) ] λ 1 - δ + ε q + 1 M λ 1 - δ - ε p v ( n 0 ) [ v ( n 0 ) ] λ 1 - δ + ε q + 1 + n 0 v ( y ) [ v ( y ) ] λ 1 - δ + ε q + 1 d y = M λ 1 - δ - ε p v ( n 0 ) [ v ( n 0 ) ] λ 1 - δ + ε q + 1 + [ v ( n 0 ) ] - λ 1 + δ - ε q λ 1 - δ + ε q < ,
namely A(ε) = O(1)(ε → 0+). Hence, by (18) and (19), it follows
k λ 1 - ε p [ v ( n 0 ) ] - ε - ε O 1 < k ε v ( n 0 ) [ v ( n 0 ) ] ε + 1 + [ v ( n 0 ) ] - ε 1 q .
(20)

By Fatou Lemma [19], we have k ( λ 1 ) lim ε 0 + k λ 1 - ε p , then by (20), it follows k1) ≤ k(ε → 0+). Hence, k = k1) is the best value of (12).

By the equivalence, the constant factor k1) in (14) and (15) is the best possible, otherwise we can imply a contradiction by (16) and (17) that the constant factor in (13) is not the best possible. ■

Remark 1 (i) Define a half-discrete Hilbert's operator T : L p , Φ ( b , c ) l p , Ψ 1 - p as: for f L p Φ (b, c), we define T f l p , Ψ 1 - p , satisfying
T f ( n ) = b c K ( x , n ) f ( x ) d x , n n 0 .
Then, by (14), it follows | | T f | | p . Ψ 1 - p k ( λ 1 ) | | f | | p , Φ and then T is a bounded operator with || T || ≤ k1). Since, by Theorem 1, the constant factor in (14) is the best possible, we have || T || = k1).
  1. (ii)
    Define a half-discrete Hilbert's operator T ̃ : l q , Ψ L q , Φ 1 - q ( b , c ) as: for a l q Ψ , we define T ̃ a L q , Φ 1 - q ( b , c ) , satisfying
    T ̃ a ( x ) = n = n 0 K ( x , n ) a n , x ( b , c ) .
     

Then, by (15), it follows | | T ̃ a | | q , Φ 1 - q k ( λ 1 ) | | a | | q , Ψ and then T ̃ is a bounded operator with | | T ̃ | | k ( λ 1 ) . Since, by Theorem 1, the constant factor in (15) is the best possible, we have | | T ̃ | | = k ( λ 1 ) .

In the following theorem, for 0 < p < 1, or p < 0, we still use the formal symbols of | | f | | p , Φ ̃ and ||a|| q ,Ψ and so on. ■

Theorem 2 Suppose that λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degreein R + 2 , u(x)(x (b, c), -∞ ≤ b < c ≤ ∞) and v(y)(y [n0, ∞), n0 N) are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, k1) R+, θλ(x) (0, 1), k1)(1 - θλ(x)) < ϖ(x) < k1)(x (b, c)). If 0 < p < 1, 1 p + 1 q = 1 , f(x), a n ≥ 0, Φ ̃ ( x ) : = ( 1 - θ λ ( x ) ) Φ ( x ) ( x ( b , c ) ) , 0 < | | f | | q , Φ ̃ < and 0 < ||a|| q ,Ψ < ∞. Then, we have the following equivalent inequalities:
I : = n = n 0 b c K ( x , n ) a n f ( x ) d x = b c n = n 0 K ( x , n ) a n f ( x ) d x > k ( λ 1 ) | | f | | p , Φ ̃ | | a | | q , Ψ ,
(21)
J : = n = n 0 [ Ψ ( n ) ] 1 - p b c K ( x , n ) f ( x ) d x p 1 p > k ( λ 1 ) | | f | | p , Φ ̃ ,
(22)
L ̃ : = b c [ Φ ̃ ( x ) ] 1 - q n = n 0 K ( x , n ) a n q d x 1 q > k ( λ 1 ) | | a | | q , Ψ .
(23)

Moreover, if v ( y ) v ( y ) ( y n 0 ) is decreasing and there exist constants δ, δ0 > 0, such that θ λ ( x ) = O ( 1 [ u ( x ) ] δ ) ( x ( d , c ) ) and k1 - δ0) R+, then the constant factor k1) in the above inequalities is the best possible.

Proof. In view of (9) and the reverse of (11), for ϖ(x) > k1)(1 - θλ(x)), we have (22). By the reverse Hölder's inequality, we have
I = n = n 0 Ψ - 1 q ( n ) b c K ( x , n ) f ( x ) d x [ Ψ 1 q ( n ) a n ] J | | a | | q , Ψ .
(24)
Then, by (22), we have (21). On the other hand, assuming that (21) is valid, setting a n as Theorem 1, then J p -1 = ||a|| q Ψ. By the reverse of (11), we find J > 0. If J = ∞, then (24) is naturally valid; if J < ∞, then by (21), we have
| | a | | q , Ψ q = J p = I > k ( λ 1 ) | | f | | p , Φ ̃ | | a | | q , Ψ , | | a | | q , Ψ q - 1 = J > k ( λ 1 ) | | f | | p , Φ ̃ ,

and we have (22) which is equivalent to (21).

In view of (9) and the reverse of (12), for [ϖ(x)]1-q> [k1)(1 - θ λ (x))]1-q(q < 0), we have (23). By the reverse Hölder's inequality, we have
I = b c [ Φ ̃ 1 p ( x ) f ( x ) ] Φ ̃ - 1 p ( x ) n = n 0 K ( x , n ) a n d x | | f | | p , Φ ̃ L ̃ .
(25)
Then, by (23), we have (21). On the other hand, assuming that (21) is valid, setting
f ( x ) : = [ Φ ̃ ( x ) ] 1 - q n = n 0 K ( x , n ) a n q - 1 , x ( b , c ) ,
then L ̃ q - 1 = | | f | | p , Φ ̃ . By the reverse of (12), we find L ̃ > 0 . If L ̃ = , then (23) is naturally valid; if L ̃ < , then by (21), we have
| | f | | p , Φ ̃ p = L ̃ q = I > k ( λ 1 ) | | f | | p , Φ ̃ | | a | | q , Ψ , | | f | | p , Φ ̃ p - 1 = L ̃ > k ( λ 1 ) | | a | | q , Ψ ,

and we have (23) which is equivalent to (21).    ■

Hence, inequalities (21), (22) and (23) are equivalent.

For 0 < ε < pδ0, setting f ̃ ( x ) and ã n as Theorem 1, if there exists a positive number k(≥ k1)), such that (21) is still valid as we replace k1) by k, then in particular, for q < 0, in view of (9) and the conditions, we have
Ĩ : = b c n = n 0 K ( x , n ) ã n f ̃ ( x ) d x > k | | f ̃ | | p , Φ ̃ | | ã | | q , Ψ = k d c 1 - O 1 [ u ( x ) ] δ u ( x ) d x [ u ( x ) ] ε + 1 1 p n = n 0 v ( n ) [ v ( n ) ] ε + 1 1 q = k 1 ε - O ( 1 ) 1 p v ( n 0 ) [ v ( n 0 ) ] ε + 1 + n = n 0 + 1 v ( n ) [ v ( n ) ] ε + 1 1 q > k 1 ε - O ( 1 ) 1 p v ( n 0 ) [ v ( n 0 ) ] ε + 1 + n 0 v ( y ) [ v ( y ) ] + 1 d y 1 q = k ε 1 - ε O ( 1 ) 1 p ε v ( n 0 ) [ v ( n 0 ) ] ε + 1 + [ v ( n 0 ) ] - ε 1 q ,
(26)
Ĩ = n = n 0 [ v ( n ) ] λ 2 - ε q - 1 v ( n ) d c K ( x , n ) [ u ( x ) ] λ 1 - ε p - 1 u ( x ) d x n = n 0 [ v ( n ) ] λ 2 - ε q - 1 v ( n ) b c K ( x , n ) [ u ( x ) ] λ 1 - ε p - 1 u ( x ) d x = t = u ( x ) v ( n ) n = n 0 [ v ( n ) ] - ε - 1 v ( n ) 0 k λ ( t , 1 ) t ( λ 1 - ε p ) - 1 d t k λ 1 - ε p v ( n 0 ) [ v ( n 0 ) ] ε + 1 + n 0 [ v ( y ) ] - ε - 1 v ( y ) d y = 1 ε k λ 1 - ε p ε v ( n 0 ) [ v ( n 0 ) ] ε + 1 + [ v ( n 0 ) ] - ε .
(27)
Since we have k λ ( t , 1 ) t λ 1 - ε p - 1 k λ ( t , 1 ) t λ 1 - δ 0 - 1 , t (0, 1] and
0 1 k λ ( t , 1 ) t λ 1 - δ 0 - 1 d t k ( λ 1 - δ 0 ) < ,
then by Lebesgue control convergence theorem [19], it follows
k λ 1 - ε p 1 k λ ( t , 1 ) t λ 1 - 1 d t + 0 1 k λ ( t , 1 ) t λ 1 - ε p - 1 d t = k ( λ 1 ) + o ( 1 ) ( ε 0 + ) .
By (26) and (27), we have
( k ( λ 1 ) + o ( 1 ) ) ε v ( n 0 ) [ v ( n 0 ) ] ε + 1 + [ v ( n 0 ) ] - ε > k { 1 - ε O ( 1 ) } 1 p ε v ( n 0 ) [ v ( n 0 ) ] ε + 1 + [ v ( n 0 ) ] - ε 1 q ,

and then k1) ≥ k(ε → 0+). Hence, k = k1) is the best value of (21).

By the equivalence, the constant factor k1) in (22) and (23) is the best possible, otherwise we can imply a contradiction by (24) and (25) that the constant factor in (21) is not the best possible.    ■

In the same way, for p < 0, we also have the following theorem.

Theorem 3 Suppose that λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree¸ in R + 2 , u(x)(x (b, c), -∞ ≤ b < c ≤ ∞) and v(y)(y [n0, ∞), n0 N) are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, ϖ(x) < k1) R + (x (b, c)). If p < 0, 1 p + 1 q = 1 , f(x), a n ≥ 0, 0 < || f || p Φ < ∞ and 0 < ||a|| q ,Ψ < ∞. Then, we have the following equivalent inequalities:
I : = n = n 0 b c K ( x , n ) a n f ( x ) d x = b c n = n 0 K ( x , n ) a n f ( x ) d x > k ( λ 1 ) | | f | | p , Φ | | a | | q , Ψ ,
(28)
J : = n = n 0 [ Ψ ( n ) ] 1 - p b c K ( x , n ) f ( x ) d x p 1 p > k ( λ 1 ) | | f | | p , Φ ,
(29)
L ̃ : = b c [ Φ ( x ) ] 1 - q n = n 0 K ( x , n ) a n q d x 1 q > k ( λ 1 ) | | a | | q , Ψ .
(30)

Moreover, if v ( y ) v ( y ) ( y n 0 ) is decreasing and there exists constant δ0 > 0, such that k1 + δ0) R+, then the constant factor k1) in the above inequalities is the best possible.

Remark 2 (i) For n0 = 1, b = 0, c = ∞, u(x) = v(x) = x, if
ϖ ( x ) = x λ 1 n = 1 k λ ( x , n ) n λ 2 - 1 < k ( λ 1 ) R + ( x ( 0 , ) ) ,
then (13) reduces to (6). In particular, for k λ ( x , n ) = 1 ( x + n ) λ ( λ = λ 1 + λ 2 , λ 1 > 0 , 0 < λ 2 1 ) ,(6) reduces to (5).
  1. (ii)
    For n0 = 1, b = 0, c = ∞, u(x) = v(x) = x α (α > 0), k λ ( x , y ) = 1 ( m a x { x , y } ) λ ( λ , λ 1 > 0 , 0 < α λ 2 1 ) , since
    f ( x , y ) = α x α λ 1 y α λ 2 - 1 ( m a x { x α , y α } ) λ = α x - α λ 2 y α λ 2 - 1 , y x α x α λ 1 y - α λ 1 - 1 , y > x
     
is decreasing for y (0, ∞) and strictly decreasing in an interval of (0, ∞), then by Condition (i), it follows
ϖ ( x ) < α x α λ 1 0 1 ( m a x { x α , y α } ) λ y α ( λ 2 1 ) y α 1 d y = t = ( y / x ) α 0 t λ 2 1 ( m a x { 1 , t } ) λ d t = λ λ 1 λ 2 = k ( λ 1 ) .
Since for δ = λ 1 2 < λ 1 , k λ ( t , 1 ) = 1 1 t δ ( t ( 0 , 1 ] ) , then by (13), we have the following inequality with the best constant factor λ α λ 1 λ 2 :
n = 1 a n 0 1 ( m a x { x α , n α } ) λ f ( x ) d x < λ α λ 1 λ 2 0 x p ( 1 - α λ 1 ) - 1 f p ( x ) d x 1 p n = 1 n q ( 1 - α λ 2 ) - 1 a n q 1 q .
(31)
  1. (iii)
    For n0 = 1, b = β, c = ∞, u ( x ) = v ( x ) = x - β ( 0 β 1 2 ) , k λ ( x , y ) = l n ( x y ) x λ - y λ ( λ , λ 1 > 0 , 0 < λ 2 1 ) , since for any fixed x (β, ∞),
    f ( x , y ) = ( x β ) λ 1 l n [ ( x β ) / ( y β ) ] ( x β ) λ ( y β ) λ ( y β ) λ 2 1
     
is decreasing and strictly convex for y ( 1 2 , ) , then by Condition (ii), it follows
ϖ ( x ) < ( x β ) λ 1 1 2 l n [ ( x β ) / ( y β ) ] ( x β ) λ ( y β ) λ ( y β ) λ 2 1 d y = t = [ ( y β ) / ( x β ) ] λ 1 λ 2 [ 1 2 β x β ] λ l n t t 1 t ( λ 2 / λ ) 1 d t 1 λ 2 0 l n t t 1 t ( λ 2 / λ ) 1 d t = [ π λ s i n ( π λ 1 λ ) ] 2 .
Since for δ = λ 1 2 < λ 1 , k λ ( t , 1 ) = l n t t λ - 1 M t δ ( t ( 0 , 1 ] ) , then by (13), we have the following inequality with the best constant factor [ π λ s i n ( π λ 1 λ ) ] 2 :
n = 1 a n β l n [ ( x - β ) ( n - β ) ] ( x - β ) λ - ( n - β ) λ f ( x ) d x < π λ s i n ( π λ 1 λ ) 2 × β ( x - β ) p ( 1 - λ 1 ) - 1 f p ( x ) d x 1 p n = 1 ( n - β ) q ( 1 - λ 2 ) - 1 a n q 1 q .
(32)
  1. (iv)
    For n0 = 1, b = 1 - β = γ, c = ∞, u(x) = v(x) = (x - γ), γ 1 - 1 8 [ λ + λ ( 3 λ + 4 ) ] , k λ ( x , y ) = 1 x λ + y λ ( 0 < λ 4 ) , λ 1 = λ 2 = λ 2 , we have
    k λ 2 = 0 k λ ( t , 1 ) t λ 2 - 1 d t = 2 λ 0 1 t λ + 1 d t λ 2 = 2 λ arctan t λ 2 | 0 = π λ R +
     
and
f ( x , y ) = ( x - γ ) λ 2 ( y - γ ) λ 2 - 1 ( x - γ ) λ + ( y - γ ) λ ( x , y ( γ , ) ) .
Hence, v(y)(y [γ, ∞)) is strictly increasing with v(1 - β) = v(γ) = 0, and for any fixed x (γ, ∞), f(x, y) is smooth with
f y ( x , y ) = ( 1 + λ 2 ) ( x γ ) λ 2 ( y γ ) λ 2 2 ( x γ ) λ + ( y γ ) λ + λ ( x γ ) 3 λ 2 ( y γ ) λ 2 2 [ ( x γ ) λ + ( y γ λ ] 2 .
We set
R ( x ) : = γ 1 f ( x , y ) d y - 1 2 f ( x , 1 ) - 1 ρ ( y ) f y ( x , y ) d y .
(33)
For x (γ, ∞), 0 < λ ≤ 4, by (33) and the following improved Euler-Maclaurin summation formula [6]:
1 8 g ( 1 ) < 1 ρ ( y ) g ( y ) d y < 0 ( ( 1 ) i g ( i ) ( y ) > 0 , g ( i ) ( ) = 0 , i = 0 , 1 ) ,
we have
R ( x ) = γ 1 ( x γ ) λ 2 ( y γ ) λ 2 1 ( x γ ) λ + ( y γ ) λ d y 1 2 ( x γ ) λ 2 ( 1 γ ) λ 2 1 ( x γ ) λ + ( 1 γ ) λ + ( 1 + λ 2 ) 1 ρ ( y ) ( x γ ) λ 2 ( y γ ) λ 2 2 ( x γ ) λ + ( y γ ) λ d y 1 ρ ( y ) λ ( x γ ) 3 λ 2 ( y γ ) λ 2 2 [ ( x γ ) λ + ( y γ λ ] 2 d y > 2 λ arctan ( 1 γ x γ ) λ 2 ( 1 γ ) λ 2 1 ( x γ ) λ 2 2 [ ( 1 γ ) λ + ( x γ λ ] 1 8 ( 1 + λ 2 ) ( 1 γ ) λ 2 2 ( x γ ) λ 2 ( 1 γ ) λ + ( x γ ) λ + 0 = h ( x ) : = 2 λ arctan ( 1 γ x γ ) λ 2 [ 1 γ 2 + 1 8 ( 1 + λ 2 ) ] ( 1 γ ) λ 2 2 ( x γ ) λ 2 ( 1 γ ) λ + ( x γ ) λ .
Since for γ 1 - 1 8 [ λ + λ ( 3 λ + 4 ) ] , i.e. 1 - γ 1 8 [ λ + λ ( 3 λ + 4 ) ] ( 0 < λ 4 ) ,
h ( x ) = ( 1 γ ) λ 2 ( x γ ) λ 2 1 ( 1