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On a half-discrete Hilbert-type inequality similar to Mulholland’s inequality

Abstract

By using the way of weight functions and Hadamard’s inequality, a half-discrete Hilbert-type inequality similar to Mulholland’s inequality with a best constant factor is given. The extension with multi-parameters, the equivalent forms as well as the operator expressions are also considered.

MSC:26D15.

1 Introduction

Assuming that f,g L 2 ( R + ), f= { 0 f 2 ( x ) d x } 1 2 >0, g>0, we have the following Hilbert integral inequality (cf. [1]):

0 0 f ( x ) g ( y ) x + y dxdy<πfg,
(1)

where the constant factor π is the best possible. If a= { a n } n = 1 , b= { b n } n = 1 l 2 , a= { n = 1 a n 2 } 1 2 >0, b>0, then we still have the following discrete Hilbert inequality:

m = 1 n = 1 a m b n m + n <πab,
(2)

with the same best constant factor π. Inequalities (1) and (2) are important in analysis and its applications (cf. [24]). Also we have the following Mulholland inequality with the same best constant factor (cf. [1, 5]):

m = 2 n = 2 a m b n ln m n <π { m = 2 m a m 2 n = 2 n b n 2 } 1 2 .
(3)

In 1998, by introducing an independent parameter λ(0,1], Yang [6] gave an extension of (1). By generalizing the results from [6], Yang [7] gave some best extensions of (1) and (2) as follows: If p>1, 1 p + 1 q =1, λ 1 + λ 2 =λ, k λ (x,y) is a non-negative homogeneous function of degree −λ with k( λ 1 )= 0 k λ (t,1) t λ 1 1 dt 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 ( R + ), f p , ϕ , g q , ψ >0, then

0 0 k λ (x,y)f(x)g(y)dxdy<k( λ 1 ) f p , ϕ g q , ψ ,
(4)

where the constant factor k( λ 1 ) is the best possible. Moreover, if k λ (x,y) is finite and k λ (x,y) x λ 1 1 ( k λ (x,y) y λ 2 1 ) is decreasing for 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 ,
(5)

with the same best constant factor k( λ 1 ). Clearly, for p=q=2, λ=1, k 1 (x,y)= 1 x + y , λ 1 = λ 2 = 1 2 , (4) reduces to (1), while (5) reduces to (2). Some other results about Hilbert-type inequalities are provided by [5, 816].

On the topic of half-discrete Hilbert-type inequalities with the general non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors in the inequalities are the best possible. Moreover, Yang [17] gave an inequality with the particular kernel 1 ( 1 + n x ) λ and an interval variable, and proved that the constant factor is the best possible. Recently, [18] and [19] gave the following half-discrete Hilbert inequality with the best constant factor π:

0 f(x) n = 1 a n ( x + n ) λ dx<πfa.
(6)

In this paper, by using the way of weight functions and Hadamard’s inequality, a half-discrete Hilbert-type inequality similar to (3) and (6) with the best constant factor is given as follows:

0 f(x) n = 1 a n ln e ( n + 1 2 ) x dx<πf { n = 1 ( n + 1 2 ) a n 2 } 1 2 .
(7)

Moreover, the best extension of (7) with multi-parameters, some equivalent forms as well as the operator expressions are considered.

2 Some lemmas

Lemma 1 If 0<λ2, α 1 2 , setting weight functions ω(n) and ϖ(x) as follows:

ω(n):= ln λ 2 (n+α) 0 x λ 2 1 ln λ e ( n + α ) x dx,n N ,
(8)
ϖ(x):= x λ 2 n = 1 ln λ 2 1 ( n + α ) ( n + α ) ln λ e ( n + α ) x ,x(0,),
(9)

we have

ϖ(x)<ω(n)=B ( λ 2 , λ 2 ) .
(10)

Proof Substitution of t=xln(n+α) in (8), by calculation, yields

ω(n)= 0 1 ( 1 + t ) λ t λ 2 1 dt=B ( λ 2 , λ 2 ) .

Since, for fixed x>0 and in view of the conditions,

h ( x , y ) : = ln λ 2 1 ( y + α ) ( y + α ) ln λ e ( y + α ) x = ln λ 2 1 ( y + α ) ( y + α ) [ 1 + x ln ( y + α ) ] λ

is decreasing and strictly convex for y( 1 2 ,), then by Hadamard’s inequality (cf. [20]), we find

ϖ ( x ) < x λ 2 1 2 ln λ 2 1 ( y + α ) ( y + α ) [ 1 + x ln ( y + α ) ] λ d y = t = x ln ( y + α ) x ln ( 1 2 + α ) t λ 2 1 ( 1 + t ) λ d t B ( λ 2 , λ 2 ) ,

namely, (10) follows. □

Lemma 2 Let the assumptions of Lemma  1 be fulfilled and, additionally, let p>1, 1 p + 1 q =1, a n 0, nN, f(x) be a non-negative measurable function in (0,). Then we have the following inequalities:

J : = { n = 1 ln p λ 2 1 ( n + α ) n + α [ 0 f ( x ) ln λ e ( n + α ) x d x ] p } 1 p J [ B ( λ 2 , λ 2 ) ] 1 q { 0 ϖ ( x ) x p ( 1 λ 2 ) 1 f p ( x ) d x } 1 p ,
(11)
L 1 : = { 0 x q λ 2 1 [ ϖ ( x ) ] q 1 [ n = 1 a n ln λ e ( n + α ) x ] q d x } 1 q L 1 { B ( λ 2 , λ 2 ) n = 1 ( n + α ) q 1 ln q ( 1 λ 2 ) 1 ( n + α ) a n q } 1 q .
(12)

Proof By Hölder’s inequality (cf. [20]) and (10), it follows

[ 0 f ( x ) d x ln λ e ( n + α ) x ] p = { 0 1 ln λ e ( n + α ) x [ x ( 1 λ 2 ) / q ln ( 1 λ 2 ) / p ( n + α ) f ( x ) ( n + α ) 1 p ] [ ln ( 1 λ 2 ) / p ( n + α ) x ( 1 λ 2 ) / q ( n + α ) 1 p ] d x } p 0 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α { 0 ( n + α ) q 1 ln λ e ( n + α ) x ln ( 1 λ 2 ) ( q 1 ) ( n + α ) x 1 λ 2 d x } p 1 = { ω ( n ) ( n + α ) q 1 ln q ( λ 2 1 ) + 1 ( n + α ) } p 1 0 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α = [ B ( λ 2 , λ 2 ) ] p 1 n + α ln p λ 2 1 ( n + α ) 0 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α .

Then by the Lebesgue term-by-term integration theorem (cf. [21]), we have

J [ B ( λ 2 , λ 2 ) ] 1 q { n = 1 0 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α } 1 p = [ B ( λ 2 , λ 2 ) ] 1 q { 0 n = 1 ln λ 2 1 ( n + α ) ln λ e ( ( n + α ) ) x x ( 1 λ 2 ) ( p 1 ) f p ( x ) d x n + α } 1 p = [ B ( λ 2 , λ 2 ) ] 1 q { 0 ϖ ( x ) x p ( 1 λ 2 ) 1 f p ( x ) d x } 1 p ,

and (11) follows. Still by Hölder’s inequality, we have

[ n = 1 a n ln λ e ( n + α ) x ] q = { n = 1 1 ln λ e ( n + α ) x [ x ( 1 λ 2 ) / q ln ( 1 λ 2 ) / p ( n + α ) 1 ( n + α ) 1 p ] [ ln ( 1 λ 2 ) / p ( n + α ) x ( 1 λ 2 ) / q ( n + α ) 1 p a n ] } q { n = 1 ln λ 2 1 ( n + α ) ln λ e ( n + α ) x x ( 1 λ 2 ) ( p 1 ) ( n + α ) } q 1 n = 1 ( n + α ) q 1 ln λ e ( n + α ) x ln ( 1 λ 2 ) ( q 1 ) ( n + α ) x 1 λ 2 a n q = [ ϖ ( x ) ] q 1 x q λ 2 1 n = 1 ( n + α ) q 1 ln λ e ( n + α ) x x λ 2 1 ln ( 1 λ 2 ) ( q 1 ) ( n + α ) a n q .

Then by the Lebesgue term-by-term integration theorem, we have

L 1 { 0 n = 1 ( n + α ) q 1 ln λ e ( n + α ) x x λ 2 1 ln ( 1 λ 2 ) ( q 1 ) ( n + α ) a n q d x } 1 q = { n = 1 [ ln λ 2 ( n + α ) 0 x λ 2 1 d x ln λ e ( n + α ) x ] ( n + α ) q 1 ln q ( 1 λ 2 ) 1 ( n + α ) a n q } 1 q = { n = 1 ω ( n ) ( n + α ) q 1 ln q ( 1 λ 2 ) 1 ( n + α ) a n q } 1 q ,

and then in view of (10), inequality (12) follows. □

3 Main results

We introduce two functions

Φ(x):= x p ( 1 λ 2 ) 1 (x>0)andΨ(n):= ( n + α ) q 1 ln q ( 1 λ 2 ) 1 (n+α)(nN),

wherefrom, [ Φ ( x ) ] 1 q = x q λ 2 1 , and [ Ψ ( n ) ] 1 p = ln p λ 2 1 ( n + α ) n + α .

Theorem 1 If  0<λ2, α 1 2 , p>1, 1 p + 1 q =1, f(x), a n 0, f L p , Φ ( R + ), a= { a n } n = 1 l q , Ψ , f p , Φ >0 and a q , Ψ >0, then we have the following equivalent inequalities:

I : = n = 1 0 a n f ( x ) d x ln λ e ( n + α ) x I = 0 n = 1 a n f ( x ) d x ln λ e ( n + α ) x < B ( λ 2 , λ 2 ) f p , Φ a q , Ψ ,
(13)
J = { n = 1 [ Ψ ( n ) ] 1 p [ 0 f ( x ) ln λ e ( n + α ) x d x ] p } 1 p J < B ( λ 2 , λ 2 ) f p , Φ ,
(14)
L : = { 0 [ Φ ( x ) ] 1 q [ n = 1 a n ln λ e ( n + α ) x ] q d x } 1 q L < B ( λ 2 , λ 2 ) a q , Ψ ,
(15)

where the constant B( λ 2 , λ 2 ) is the best possible in the above inequalities.

Proof By the Lebesgue term-by-term integration theorem, there are two expressions for I in (13). In view of (11), for ϖ(x)<B( λ 2 , λ 2 ), we have (14). By Hölder’s inequality, we have

I= n = 1 [ Ψ 1 q ( n ) 0 1 ln λ e ( n + α ) x 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 [ 0 1 ln λ e ( n + α ) x f ( x ) d x ] p 1 ,nN,

then J p 1 = a q , Ψ . By (11), we find J<. If J=0, then (14) is trivially valid; if J>0, then by (13), we have

a q , Ψ q = J p = I < B ( λ 2 , λ 2 ) f p , Φ a q , Ψ , i.e. a q , Ψ q 1 = J < B ( λ 2 , λ 2 ) f p , Φ ,

that is, (14) is equivalent to (13). In view of (12), for [ ϖ ( x ) ] 1 q > [ B ( λ 2 , λ 2 ) ] 1 q , we have (15). By Hölder’s inequality, we find

I= 0 [ Φ 1 p ( x ) f ( x ) ] [ Φ 1 p ( x ) n = 1 1 ln λ e ( n + α ) x a n ] dx 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 = 1 1 ln λ e ( n + α ) x a n ] q 1 ,x(0,),

then L q 1 = f p , Φ . By (12), we find L<. If L=0, then (15) is trivially valid; if L>0, then by (13), we have

f p , Φ p = L q = I < B ( λ 2 , λ 2 ) f p , Φ a q , Ψ , i.e. , f p , Φ p 1 = L < B ( λ 2 , λ 2 ) a q , Ψ ,

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

For 0<ε< p λ 2 , setting f ˜ (x)= x λ 2 + ε p 1 , x(0,1); f ˜ (x)=0, x[1,), and a ˜ n = 1 n + α ln λ 2 ε q 1 (n+α), nN, if there exists a positive number k (B( λ 2 , λ 2 )) such that (13) is valid as we replace B( λ 2 , λ 2 ) with k, then, in particular, it follows

I ˜ : = n = 1 0 1 ln λ e ( n + α ) x a ˜ n f ˜ ( x ) d x < k f ˜ p , Φ a ˜ q , Ψ I ˜ = k { 0 1 d x x ε + 1 } 1 p { 1 ( 1 + α ) ln ε + 1 ( 1 + α ) + n = 2 1 ( n + α ) ln ε + 1 ( n + α ) } 1 q I ˜ < k ( 1 ε ) 1 p { 1 ( 1 + α ) ln ε + 1 ( 1 + α ) + 1 1 ( x + α ) ln ε + 1 ( x + α ) d x } 1 q I ˜ = k ε { ε ( 1 + α ) ln ε + 1 ( 1 + α ) + 1 ln ε ( 1 + α ) } 1 q ,
(18)
I ˜ = n = 1 1 n + α ln λ 2 ε q 1 ( n + α ) 0 1 1 ln λ e ( n + α ) x x λ 2 + ε p 1 d x = t = x ln ( n + α ) n = 1 1 ( n + α ) ln ε + 1 ( n + α ) 0 ln ( n + α ) 1 ( t + 1 ) λ t λ 2 + ε p 1 d t = B ( λ 2 + ε p , λ 2 ε p ) n = 1 1 ( n + α ) ln ε + 1 ( n + α ) A ( ε ) > B ( λ 2 + ε p , λ 2 ε p ) 1 1 ( y + α ) ln ε + 1 ( y + α ) d y A ( ε ) = 1 ε ln ε ( 1 + α ) B ( λ 2 + ε p , λ 2 ε p ) A ( ε ) ,
A(ε):= n = 1 1 ( n + α ) ln ε + 1 ( n + α ) ln ( n + α ) 1 ( t + 1 ) λ t λ 2 + ε p 1 dt.
(19)

We find

0 < A ( ε ) n = 1 1 ( n + α ) ln ε + 1 ( n + α ) ln ( n + α ) 1 t λ t λ 2 + ε p 1 d t = 1 λ 2 ε p n = 1 1 ( n + α ) ln λ 2 + ε q + 1 ( n + α ) < ,

and then A(ε)=O(1) (ε 0 + ). Hence by (18) and (19), it follows

1 ln ε ( 1 + α ) B ( λ 2 + ε p , λ 2 ε p ) ε O ( 1 ) < k { ε ( 1 + α ) ln ε + 1 ( 1 + α ) + 1 ln ε ( 1 + α ) } 1 q ,
(20)

and B( λ 2 , λ 2 )k (ε 0 + ). Hence k=B( λ 2 , λ 2 ) is the best value of (13).

By equivalence, the constant factor B( λ 2 , λ 2 ) 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 the first type half-discrete Hilbert-type operator T 1 : L p , Φ ( R + ) l p , Ψ 1 p as follows: For f L p , Φ ( R + ), we define T 1 f l p , Ψ 1 p , satisfying

T 1 f(n)= 0 1 ln λ e ( n + α ) x f(x)dx,nN.

Then by (14) it follows T 1 f p . Ψ 1 p B( λ 2 , λ 2 ) f p , Φ , and then T 1 is a bounded operator with T 1 B( λ 2 , λ 2 ). Since by Theorem 1 the constant factor in (14) is the best possible, we have T 1 =B( λ 2 , λ 2 ).

  1. (ii)

    Define the second type half-discrete Hilbert-type operator T 2 : l q , Ψ L q , Φ 1 q ( R + ) as follows: For a l q , Ψ , we define T 2 a L q , Φ 1 q ( R + ), satisfying

    T 2 a(x)= n = 1 1 ln λ e ( n + α ) x a n ,x(0,).

Then by (15) it follows T 2 a q , Φ 1 q B( λ 2 , λ 2 ) a q , Ψ , and then T 2 is a bounded operator with T 2 B( λ 2 , λ 2 ). Since by Theorem 1 the constant factor in (15) is the best possible, we have T 2 =B( λ 2 , λ 2 ).

Remark 2 For p=q=2, λ=1, λ 1 = λ 2 = 1 2 , α= 1 2 in (13), (14) and (15), we have (7) and the following equivalent inequalities:

{ n = 1 1 n + 1 2 [ 0 f ( x ) ln e ( n + 1 2 ) x d x ] 2 } 1 2 <πf,
(21)
{ 0 [ n = 1 a n ln e ( n + 1 2 ) x ] 2 d x } 1 2 <π { n = 1 ( n + 1 2 ) a n 2 } 1 2 .
(22)

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Acknowledgements

This work is supported by 2012 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2012KJCX0079).

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ZH wrote and reformed the article. BY conceived of the study, and participated in its design and coordination. All authors read and approved the final manuscript.

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Huang, Z., Yang, B. On a half-discrete Hilbert-type inequality similar to Mulholland’s inequality. J Inequal Appl 2013, 290 (2013). https://doi.org/10.1186/1029-242X-2013-290

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Keywords

  • Hilbert-type inequality
  • weight function
  • equivalent form