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A multiple Hilbert-type integral inequality with a non-homogeneous kernel

Journal of Inequalities and Applications20132013:73

https://doi.org/10.1186/1029-242X-2013-73

  • Received: 3 October 2012
  • Accepted: 11 February 2013
  • Published:

Abstract

By using the way of weight functions and the technic of real analysis, a multiple Hilbert-type integral inequality with a non-homogeneous kernel is given. The operator expression with the norm, the reverses and some examples with the particular kernels are also considered.

MSC:47A07, 26D15.

Keywords

  • multiple Hilbert-type integral inequality
  • kernel
  • weight function
  • norm
  • operator

1 Introduction

If p > 1 , 1 p + 1 q = 1 , f ( 0 ) L p ( 0 , ) , g ( 0 ) L q ( 0 , ) , f p , g q > 0 , then we have the following equivalent inequalities (cf. [1]):
(1)
(2)

where the constant factor π sin ( π / p ) is the best possible. (1) is the well-known Hardy-Hilbert integral inequality. Define the Hardy-Hilbert integral operator T : L p ( 0 , ) L p ( 0 , ) as follows: for f L p ( 0 , ) , T f ( y ) : = 0 1 x + y f ( x ) d x ( y ( 0 , ) ).

Then in view of (2), it follows T f p < π sin ( π / p ) f p and T π sin ( π / p ) . Since the constant is the best possible, we find T = π sin ( π / p ) .

Inequalities (1) and (2) and the operator are important in analysis and its applications (cf. [2, 3]). In 2002, [4] considered the property of the Hardy-Hilbert integral operator and gave an improvement of (1) (for p = q = 2 ). In 2004, by introducing another pair of conjugate exponents ( r , s ) ( r > 1 , 1 r + 1 s = 1 ) and an independent parameter λ > 0 , [5] gave the best extensions of (1) as follows:
0 0 f ( x ) g ( y ) x λ + y λ d x d y < π λ sin ( π / r ) f p , ϕ g q , ψ ,
(3)
where ϕ ( x ) = x p ( 1 λ r ) 1 , ψ ( x ) = x q ( 1 λ s ) 1 , f p , ϕ = { 0 ϕ ( x ) f p ( x ) d x } 1 p > 0 , g q , ψ > 0 . In 2007, [6] gave the following inequality with the best constant B ( λ 2 , λ 2 ) ( λ > 0 ; B ( u , v ) is the beta function):
0 0 f ( x ) g ( y ) ( 1 + x y ) λ d x d y < B ( λ 2 , λ 2 ) { 0 x 1 λ f 2 ( x ) d x 0 x 1 λ g 2 ( x ) d x } 1 2 .
(4)
In 2009, [7] gave an extension of (4) in R 2 with the kernel 1 | 1 + x y | λ ( 0 < λ < 1 ); [8] gave another extension of (4) to the general kernel k λ ( 1 , x y ) ( λ > 0 ) with a pair of conjugate exponents ( p , q ) and obtained the following multiple Hilbert-type integral inequality. Suppose that n N { 1 } , p i > 1 , i = 1 n 1 p i = 1 , λ > 0 , k λ ( x 1 , , x n ) 0 is a measurable function of −λ-degree in R + n , and for any ( r 1 , , r n ) ( r i > 1 ), satisfies i = 1 n 1 r i = 1 and
k λ = R + n 1 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ r j 1 d u 1 d u n 1 > 0 .
If ϕ i ( x ) = x p i ( 1 λ r i ) 1 , f i ( 0 ) L ϕ i p i ( 0 , ) , f p i , ϕ i > 0 ( i = 1 , , n ), then we have the following inequality:
R + n k λ ( x 1 , , x n ) i = 1 n f i ( x i ) d x 1 d x n < k λ i = 1 n f i p i , ϕ i ,
(5)

where the constant factor k λ is the best possible. For n = 2 , k λ ( x , y ) = 1 x λ + y λ in (5), we obtain (3). Inequality (5) is an extension of the results in [912] and [13]. In recent years, [14] and [15] considered some Hilbert-type operators relating (1)-(3); [16] also considered a multiple Hilbert-type integral operator with the homogeneous kernel of n + 1 -degree and the relating particular case of (5) (for λ = n 1 , 1 r i = 1 n 1 ( 1 1 p i ) ).

In this paper, by using the way of weight functions and the technic of real analysis, a multiple Hilbert-type integral inequality with a non-homogeneous kernel is given. The operator expression with the norm, the reverses and some examples with the particular kernels are considered.

2 Some lemmas

Lemma 1 If n N { 1 } , λ i R ( i = 1 , , n ), i = 1 n 1 p i = 1 , then we have
A : = i = 1 n [ x i ( λ i 1 ) ( 1 p i ) j = 1 ( j i ) n x j λ j 1 ] 1 p i = 1 .
(6)

Proof

We find
A : = i = 1 n [ x i ( λ i 1 ) ( 1 p i ) + 1 λ i j = 1 n x j λ j 1 ] 1 p i = i = 1 n [ x i ( 1 λ i ) p i j = 1 n x j λ j 1 ] 1 p i = i = 1 n x i 1 λ i ( j = 1 n x j λ j 1 ) i = 1 n 1 p i ,

and then (6) is valid. □

Definition 1 If n N , R + n : = { ( x 1 , , x n ) | x i > 0 ( i = 1 , , n ) } , λ R , k λ ( x 1 , , x n ) is a measurable function in R + n such that for any u > 0 and ( x 1 , , x n ) R + n , k λ ( u x 1 , , u x n ) = u λ k λ ( x 1 , , x n ) , then we call k λ ( x 1 , , x n ) the homogeneous function of −λ-degree in R + n .

Lemma 2 Suppose n N { 1 } , λ i R ( i = 1 , , n ), λ n = i = 1 n 1 λ i = λ 2 , k λ ( x 1 , , x n ) 0 is a homogeneous function ofλ-degree. If
H ( i ) : = R + n 1 k λ ( u 1 , , u i 1 , 1 , u i + 1 , , u n ) × j = 1 ( j i ) n u j λ j 1 d u 1 d u i 1 d u i + 1 d u n ( i = 1 , , n )
satisfying k λ : = H ( n ) R , then each H ( i ) = H ( n ) = k λ and for any i = 1 , , n ,
ω i ( x i ) : = x i λ i R + n 1 k λ ( x 1 x n , , x n 1 x n , 1 ) × j = 1 ( j i ) n x j λ j 1 d x 1 d x i 1 d x i + 1 d x n = k λ ( R ) .
(7)
Proof Setting u j = u n v j ( j i , n ) in the integral H ( i ) , we find
H ( i ) = R + n 1 k λ ( v 1 , , v i 1 , u n 1 , v i + 1 , , v n 1 , 1 ) j = 1 ( j i ) n 1 v j λ j 1 × u n 1 λ i d v 1 d v i 1 d v i + 1 d v n 1 d u n .
Setting v i = u n 1 in the above integral, we obtain H ( i ) = H ( n ) . Setting x n = x n 1 in (7), since λ λ n = λ n , we find
ω i ( x i ) = x i λ i R + n 1 k λ ( x 1 , , x n 1 , x n 1 ) x n λ n 1 × j = 1 ( j i ) n 1 x j λ j 1 d x 1 d x i 1 d x i + 1 d x n = x i λ i R + n 1 k λ ( x 1 , , x n 1 , x n ) ( x n ) λ n + 1 × j = 1 ( j i ) n 1 x j λ j 1 d x 1 d x i 1 d x i + 1 d x n 1 ( x n ) 2 d x n = x i λ i R + n 1 k λ ( x 1 , , x n 1 , x n ) ( x n ) λ n 1 × j = 1 ( j i ) n 1 x j λ j 1 d x 1 d x i 1 d x i + 1 d x n 1 d x n .

Setting u j = x j / x i ( j i , n ) and u n = x n / x i in the above integral, we find ω i ( x i ) = H ( i ) = H ( n ) = k λ . □

Lemma 3 With the assumptions given in Lemma 2, then
k ( λ ˜ 1 , , λ ˜ n 1 ) : = R + n 1 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ ˜ j 1 d u 1 d u n 1

is finite in a neighborhood of ( λ 1 , , λ n 1 ) if any only if k ( λ ˜ 1 , , λ ˜ n 1 ) is continuous at ( λ 1 , , λ n 1 ) .

Proof The sufficiency property is obvious. We prove the necessary property of the condition by mathematical induction in the following. For n = 2 , there exists I : = { λ ˜ 1 | λ ˜ 1 = λ 1 + δ 1 , | δ 1 | δ 0 , δ 0 > 0 } such that for any λ ˜ 1 I , k ( λ ˜ 1 ) R . Since for λ ˜ 1 = λ 1 + δ 1 I ( δ 1 0 ),
and k ( λ 1 δ 0 ) + k ( λ 1 + δ 0 ) < , then by the Lebesgue control convergence theorem (cf. [17]), it follows k ( λ 1 + δ 1 ) = k ( λ 1 ) + o ( 1 ) ( δ 1 0 ). Assuming that for n (≥2), k ( λ ˜ 1 , , λ ˜ n 1 ) is continuous at ( λ 1 , , λ n 1 ) , then for n + 1 , by the result of n = 2 , since k ( λ 1 + δ 1 , , λ n + δ n ) is finite in a neighborhood of ( λ 1 , , λ n ) , we find
then by the assumption for n, it follows
lim δ n 0 k ( λ 1 + δ 1 , , λ n + δ n ) = k ( λ 1 , , λ n ) + o ( 1 ) ( δ i 0 , i = 1 , , n 1 ) .

By mathematical induction, we prove that for n N { 1 } , k ( λ ˜ 1 , , λ ˜ n 1 ) is continuous at ( λ 1 , , λ n 1 ) . □

Lemma 4 With the assumptions given in Lemma 2, if there exists δ > 0 such that for max 1 i n 1 { | δ i | } < δ , k ( λ 1 + δ 1 , , λ n 1 + δ n 1 ) R , p i R { 0 , 1 } ( i = 1 , , n ), 0 < ε < min 1 i n { | p i | } δ , then we have
I ε : = ε 1 1 [ 0 1 x n λ n + ε p n 1 k λ ( x 1 x n , , x n 1 x n , 1 ) d x n ] × j = 1 n 1 x j λ j ε p j 1 d x 1 d x n 1 = k λ + o ( 1 ) ( ε 0 + ) .
(8)
Proof Setting x n = x n 1 in (8), we find
I ε : = ε 1 1 [ 1 ( x n ) λ n ε p n 1 k λ ( x 1 x n , , x 1 x n , 1 ) d x n ] × j = 1 n 1 x j λ j ε p j 1 d x 1 d x n 1 = k λ + o ( 1 ) .
Setting u j = x j / x n ( j = 1 , , n 1 ) in the above integral, since λ λ n = λ n , we find (replacing x n by x n )
I ε = ε 1 x n 1 ε [ x n 1 x n 1 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j ε p j 1 d u 1 d u n 1 ] d x n .
(9)
Setting D j : = { ( u 1 , , u n 1 ) | u j ( 0 , x n 1 ) , u k ( 0 , ) ( k j ) } and
A j ( x n ) : = D j k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j ε p j 1 d u 1 d u n 1 ,
by (9), it follows
I ε R + n 1 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j ε p j 1 d u 1 d u n 1 ε j = 1 n 1 1 x n 1 A j ( x n ) d x n .
(10)

Without loss of generality, we estimate the case of j = n , e.t.

1 x n 1 A n 1 ( x n ) d x n = O ( 1 ) . In fact, setting α > 0 , such that | ε p n 1 + α | < δ , since u n 1 α ln u n 1 0 ( u n 1 0 + ), there exists M > 0 such that u n 1 α ln u n 1 M ( u n 1 ( 0 , 1 ] ), and then by the Fubini theorem, it follows
0 1 x n 1 A n 1 ( x n ) d x n = 1 x n 1 [ R + n 2 0 x n 1 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j ε p j 1 d u n 1 d u 1 d u n 2 ] d x n = 0 1 R + n 2 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j ε p j 1 ( 1 u n 1 1 x n 1 d x n ) d u 1 d u n 1 = 0 1 R + n 2 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j ε p j 1 ( ln u n 1 ) d u 1 d u n 1 M 0 1 R + n 2 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 2 u j λ j ε p j 1 u n 1 λ n 1 ( ε p n 1 + α ) 1 d u 1 d u n 1 M R + n 1 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 2 u j λ j ε p j 1 u n 1 λ n 1 ( ε p n 1 + α ) 1 d u 1 d u n 1 = M k ( λ 1 ε p 1 , , λ n 2 ε p n 2 , λ n 1 ( ε p n 1 + α ) ) < .
Hence by (10), we have
I ε R + n 1 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j ε p j 1 d u 1 d u n 1 o 1 ( 1 ) .
(11)
Since by Lemma 3 we find
I ε ε 1 x n 1 ε [ 0 0 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j ε p j 1 d u 1 d u n 1 ] d x n = 0 0 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j ε p j 1 d u 1 d u n 1 = k ( λ 1 ε p 1 , , λ n 1 ε p n 1 ) = k λ + o 2 ( 1 ) ,

then combining with (11), we have (8). □

Lemma 5 Suppose that n N { 1 } , p i R { 0 , 1 } ( i = 1 , , n ), i = 1 n 1 p i = 1 , 1 q n = 1 1 p n , ( λ 1 , , λ n ) R n , λ n = i = 1 n 1 λ i = λ 2 , k λ ( x 1 , , x n ) (≥0) is a measurable function ofλ-degree in R + n such that
k λ = R + n 1 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j 1 d u 1 d u n 1 R .
If f i 0 are measurable functions in R + ( i = 1 , , n 1 ), k ˜ λ ( x 1 , , x n ) : = k λ ( x 1 x n , , x n 1 x n , 1 ) , then (1) for p i > 1 ( i = 1 , , n ), we have
J : = { 0 x n λ q n 2 1 [ R + n 1 k ˜ λ ( x 1 , , x n ) i = 1 n 1 f i ( x i ) d x 1 d x n 1 ] q n d x n } 1 q n k λ i = 1 n 1 { 0 x p i ( 1 λ i ) 1 f p i ( x ) d x } 1 p i ;
(12)

(2) for 0 < p 1 < 1 , p i < 0 ( i = 2 , , n ), we have the reverse of (12).

Proof (1) For p i > 1 ( i = 1 , , n ), by the Hölder inequality (cf. [18]) and (7), it follows
(13)
For n 3 , by the Hölder inequality again, it follows
J ( k λ ) 1 p n { i = 1 n 1 [ R + n 1 ( 0 k ˜ λ ( x 1 , , x n ) x n λ n 1 d x n ) × x i ( λ i 1 ) ( 1 p i ) j = 1 ( j i ) n 1 x j λ j 1 f i p i ( x i ) d x 1 d x n 1 ] q n p i } 1 q n = ( k λ ) 1 p n i = 1 n 1 { 0 [ R + n 1 k ˜ λ ( x 1 , , x n ) × x i λ i j = 1 ( j i ) n x j λ j 1 d x 1 d x i 1 d x i + 1 d x n ] x i p i ( 1 λ i ) 1 f i p i ( x i ) d x i } 1 p i = ( k λ ) 1 p n i = 1 n 1 { 0 ω i ( x i ) x i p i ( 1 λ i ) 1 f i p i ( x i ) d x i } 1 p i .
(14)

Then by (7), we have (12). (Note: for n = 2 , we do not use the Hölder inequality again in the above.) (2) For 0 < p 1 < 1 , p i < 0 ( i = 2 , , n ), by the reverse Hölder inequality and in the same way, we obtain the reverses of (12). □

3 Main results and applications

With the assumptions given in Lemma 5, setting ϕ i ( x ) : = x p ( i 1 λ i ) 1 ( x ( 0 , ) ; i = 1 , , n ), then we find ϕ n 1 / ( 1 p n ) ( x ) = x q λ n n 1 . If p i > 1 ( i = 1 , , n ), define the following real function spaces:
and a multiple Hilbert-type integral operator T : i = 1 n 1 L ϕ i p i ( 0 , ) L ϕ n 1 / ( 1 p n ) q n as follows: for f = ( f 1 , , f n 1 ) i = 1 n 1 L ϕ i p i ( 0 , ) ,
( T f ) ( x n ) : = R + n 1 k ˜ λ ( x 1 , , x n ) i = 1 n 1 f i ( x i ) d x 1 d x n 1 , x n ( 0 , ) .
(15)
Then by (12), it follows T f L ϕ n 1 / ( 1 p n ) q n , T is bounded, T f q n , ϕ n 1 / ( 1 p n ) k λ i = 1 n 1 f i p i , ϕ i and T k λ , where
T : = sup f i = 1 n 1 L ϕ i p i ( 0 , ) ( f i θ , i = 1 , , n 1 ) T f q n , ϕ n 1 / ( 1 p n ) i = 1 n 1 f i p i , ϕ i .
(16)
Define the formal inner product of T ( f 1 , , f n 1 ) and f n as
( T ( f 1 , , f n 1 ) , f n ) : = R + n k ˜ λ ( x 1 , , x n ) i = 1 n f i ( x i ) d x 1 d x n .
(17)
Theorem 1 With the assumptions given in Lemma 5, suppose that for any ( λ 1 , , λ n ) R n , it satisfies λ n = i = 1 n 1 λ i = λ 2 , and
0 < k λ = R + n 1 k λ ( u 1 , , u n 1 , 1 ) j = 1 n 1 u j λ j 1 d u 1 d u n 1 < .
(18)
If f i ( 0 ) L ϕ i p i ( 0 , ) , f p i , ϕ i > 0 ( i = 1 , , n ), then (i) for p i > 1 ( i = 1 , , n ), we have T = k λ and the following equivalent inequalities:
(19)
(20)

where the constant factor k λ is the best possible; (ii) for 0 < p 1 < 1 , p i < 0 ( i = 2 , , n ), using the formal symbols in the case of (i), we have the equivalent reverses of (19) and (20) with the same best constant factor.

Proof (i) For all p i > 1 , if (12) takes the form of equality, then for n 3 in (14), there exist C i and C k ( i k ) such that they are not all zero and
e.t. C i x i p i ( 1 λ i ) f i p i ( x i ) = C k x k p k ( 1 λ k ) f k p k ( x k ) = C a.e. in R + n . Assuming that C i > 0 , then x i p i ( 1 λ i ) 1 f i p i ( x i ) = C / ( C i x i ) , which contradicts f p i , ϕ i > 0 . (Note: for n = 2 , we consider (13) for f k p i ( x k ) = 1 in the above.) Hence we have (19). By the Hölder inequality, it follows
( T f , f n ) = 0 ( x n λ n 1 q n R + n 1 k ˜ λ ( x 1 , , x n ) i = 1 n 1 f i ( x i ) d x 1 d x n 1 ) ( x n 1 q n λ n f n ( x n ) ) d x n T ( f 1 , , f n 1 ) q n , ϕ n 1 / ( 1 p n ) f n p n , ϕ n ,
(21)
and then by (19), we have (20). Assuming that (20) is valid, setting
f n ( x n ) : = x n q n λ n 1 [ R + n 1 k ˜ λ ( x 1 , , x n ) i = 1 n 1 f i ( x i ) d x 1 d x n 1 ] q n 1 ,
then J = { 0 x n p n ( 1 λ n ) 1 f n p n ( x n ) d x n } 1 q n . By (12), it follows J < . If J = 0 , then (19) is naturally valid. Assuming that 0 < J < , by (20), it follows

and then (19) is valid, which is equivalent to (20).

For ε > 0 small enough, setting f ˜ i ( x ) as: f ˜ i ( x ) = 0 , x ( 0 , 1 ) ; f ˜ i ( x ) = x λ i ε p i 1 , x [ 1 , ) ( i = 1 , , n 1 ), f ˜ n ( x ) = x λ n + ε p n 1 , x ( 0 , 1 ) ; f ˜ n ( x ) = 0 , x [ 1 , ) , if there exists k k λ such that (20) is still valid as we replace k λ by k, then in particular, by Lemma 4, we have
k λ + o ( 1 ) = I ε = ε ( T ( f ˜ 1 , , f ˜ n 1 ) , f ˜ n ) < ε k i = 1 n f ˜ i p i , ϕ i = k
and k λ k ( ε 0 + ) . Hence k = k λ is the best value of (20). We confirm that the constant factor k λ in (19) is the best possible, otherwise we can get a contradiction by (21) that the constant factor in (20) is not the best possible. Therefore T = k λ .
  1. (ii)

    For 0 < p 1 < 1 , p i < 0 ( i = 2 , , n ), by using the reverse Hölder inequality and in the same way, we have the equivalent reverses of (19) and (20) with the same best constant factor. □

     
Example 1 For λ > 0 , λ i = λ r i ( i = 1 , , n ), r n = 2 , i = 1 n 1 r i = 1 , k λ ( x 1 , , x n ) = 1 ( i = 1 n x i ) λ , by mathematical induction, we can show
k λ = R + n 1 1 ( i = 1 n 1 u i + 1 ) λ j = 1 n 1 u j λ r j 1 d u 1 d u n 1 = 1 Γ ( λ ) i = 1 n Γ ( λ r i ) .
(22)
In fact, for n = 2 , we obtain
k λ = R + 1 ( u 1 + 1 ) λ u 1 λ r 1 1 d u 1 = 1 Γ ( λ ) Γ ( λ r 1 ) Γ ( λ r 2 ) .
Assuming that for n (≥2) (22) is valid, then for n + 1 , it follows
k λ = R + n 1 ( i = 1 n u i + 1 ) λ j = 1 n u j λ r j 1 d u 1 d u n = R + n 1 j = 2 n u j λ r j 1 [ R + 1 [ u 1 + ( i = 2 n u i + 1 ) ] λ u 1 λ r 1 1 d u 1 ] d u 2 d u n = R + n 1 1 ( i = 2 n u i + 1 ) λ j = 2 n u j λ r j 1 [ R + 1 ( v 1 + 1 ) λ v 1 λ r 1 1 d v 1 ] d u 2 d u n = Γ ( λ r 1 ) Γ ( λ λ r 1 ) Γ ( λ ) R + n 1 1 ( i = 2 n u i + 1 ) λ ( 1 1 r 1 ) j = 2 n u j λ r j 1 d u 2 d u n = Γ ( λ r 1 ) Γ ( λ λ r 1 ) Γ ( λ ) 1 Γ ( λ λ r 1 ) i = 2 n + 1 Γ ( λ r i ) = 1 Γ ( λ ) i = 1 n + 1 Γ ( λ r i ) .

Then by mathematical induction, (22) is valid for n N { 1 } .

Example 2 For λ > 0 , λ i = λ r i ( i = 1 , , n ), r n = 2 , i = 1 n 1 r i = 1 , k λ ( x 1 , , x n ) = 1 ( max 1 i n { x i } ) λ , by mathematical induction, we can show
k λ = R + n 1 1 ( max 1 i n 1 { u i } + 1 ) λ j = 1 n 1 u j λ r j 1 d u 1 d u n 1 = i = 1 n r i λ n 1 .
(23)
In fact, for n = 2 , we obtain
k λ = R + u 1 λ r 1 1 ( max { u 1 , 1 } ) λ d u 1 = 0 1 u 1 λ r 1 1 d u 1 + 1 u 1 λ r 2 1 d u 1 = 1 λ r 1 r 2 .
Assuming that for n (≥2), (22) is valid, then for n + 1 , it follows
k λ = R + n 1 j = 2 n u j λ r j 1 [ 0 1 ( max 1 i n { u i , 1 } ) λ u 1 λ r 1 1 d u 1 ] d u 2 d u n = R + n 1 j = 2 n u j λ r j 1 [ 0 max { u 2 , , u n , 1 } 1 ( max 2 i n { u i , 1 } ) λ u 1 λ r 1 1 d u 1 + max { u 2 , , u n , 1 } 1 u 1 λ u 1 λ r 1 1 d u 1 ] d u 2 d u n = r 1 2 λ ( r 1 1 ) R + n 1 1 ( max 2 i n { u i , 1 } ) λ ( 1 1 r 1 ) j = 2 n u j λ r j 1 d u 2 d u n = r 1 2 λ ( r 1 1 ) ( r 1 r 1 1 ) n 1 R + n 1 1 ( max 2 i n { v i , 1 } ) λ j = 2 n v j λ r j r 1 r 1 1 1 d v 2 d v n = r 1 2 λ ( r 1 1 ) ( r 1 r 1 1 ) n 1 1 λ n 1 i = 2 n + 1 r 1 1 r 1 r i = 1 λ n i = 1 n + 1 r i .

Then by mathematical induction, (23) is valid for n N { 1 } .

Example 3 For λ > 0 , λ i = λ r i ( i = 1 , , n ), r n = 2 , i = 1 n 1 r i = 1 , k λ ( x 1 , , x n ) = ( min 1 i n { x i } ) λ , by mathematical induction, we can show
k λ = R + n 1 ( min { u 1 , , u n 1 , 1 } ) λ j = 1 n 1 u j λ r j 1 d u 1 d u n 1 = i = 1 n r i λ n 1 .
(24)
In fact, for n = 2 , we obtain
k λ = 0 1 u 1 λ r 2 1 d u 1 + 1 u 1 λ r 1 1 d u 1 = 1 λ r 1 r 2 .
Assuming that for n (≥2), (24) is valid, then for n + 1 , it follows
k λ = R + n 1 j = 2 n u j λ r j 1 [ 0 ( min { u 1 , , u n , 1 } ) λ u 1 λ r 1 1 d u 1 ] d u 2 d u n = R + n 1 j = 2 n u j λ r j 1 [ 0 min { u 2 , , u n , 1 } u 1 λ u 1 λ r 1 1 d u 1 + min { u 2 , , u n , 1 } ( min { u 2 , , u n , 1 } ) λ u 1 λ r 1 1 d u 1 ] d u 2 d u n = r 1 2 λ ( r 1 1 ) R + n 1 ( min { u 2 , , u n , 1 } ) λ ( 1 1 r 1 ) j = 2 n u j λ ( 1 1 r 1 ) ( 1 1 r 1 ) r j 1 d u 2 d u n = r 1 2 λ ( r 1 1 ) 1 [ λ ( 1 1 r 1 ) ] n 1 i = 2 n + 1 ( 1 1 r 1 ) r i = 1 λ n i = 1 n + 1 r i .

Then, by mathematical induction, (24) is valid for n N { 1 } .

Remarks (i) In particular, for n = 2 in (20), we have
(25)
where k λ = 0 k λ ( u , 1 ) u λ 2 1 d u > 0 ( λ R ) is the best possible. Inequality (25) is an extension of (4) and (8.1.7) in [8].
  1. (ii)

    In Examples 1 and 2, by Theorem 1, since for any ( λ 1 , , λ n ) R n ( λ n = i = 1 n λ i = λ 2 ) , we obtain 0 < k λ < , then we have T = k λ and the equivalent inequalities (19) and (20) with the particular kernels and some equivalent reverses. In Example 3, still using Theorem 1, we find 0 < T = k λ < and the relating particular inequalities.

     

Declarations

Acknowledgements

This work is supported by Guangdong Modern Information Service industry Develop Particularly item 2011 (No. 13090).

Authors’ Affiliations

(1)
Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong, 510303, P.R. China

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© Huang and Yang; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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