Skip to main content

On a half-discrete reverse Mulholland-type inequality and an extension

Abstract

By using the way of weight functions and the Hermite-Hadamard inequality, a half-discrete reverse Mulholland-type inequality with a best constant factor is given. The extension with multi-parameters, the equivalent forms as well as the relating homogeneous inequalities 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 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 π (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). For 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, 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 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 strict 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 including the reverse Hilbert-type inequalities are provided by [816]. 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 in the inequalities are best possible. However, Yang [17] gave a result by introducing an interval variable and proved that the constant factor is best possible. Recently, Yang [18] gave a half-discrete Hilbert inequality with multi-parameters, and [19] gave the following half-discrete reverse Hilbert-type inequality with the best constant factor 4: For 0<p<1, 1 p + 1 q =1, we have θ 1 (x)(0,1), and

0 f ( x ) n = 1 min { x , n } a n d x > 4 { 0 1 θ 1 ( x ) x 1 3 p 2 f p ( x ) d x } 1 p { n = 1 a n q n 1 3 q 2 } 1 q .
(6)

In this paper, by using the way of weight functions and the Hermite-Hadamard inequality, a half-discrete reverse Mulholland-type inequality similar to (6) is given as follows:

0 f ( x ) n = 1 a n ln e ( n + 1 2 ) x d x > π { 0 1 θ 1 ( x ) x 1 p 2 f p ( x ) d x } 1 p { n = 1 ( n + 1 2 ) q 1 a n q ln 1 q 2 ( n + 1 2 ) } 1 q .
(7)

Moreover, a best extension of (7) with multi-parameters, the equivalent forms and the relating homogeneous inequalities 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 1 ln λ e ( n + α ) x x λ 2 1 dx,nN,
(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

B ( λ 2 , λ 2 ) ( 1 θ λ ( x ) ) <ϖ(x)<ω(n)=B ( λ 2 , λ 2 ) ,
(10)

where

θ λ (x)= 1 B ( λ 2 , λ 2 ) 0 x ln ( 1 + α ) t ( λ / 2 ) 1 ( 1 + t ) λ dt(0,1),

satisfying θ λ (x)=O( x λ 2 ).

Proof Substituting of t=xln(n+α) in (8), by calculation, we have

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

Since by the conditions and for fixed x>0

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 strictly decreasing and strictly convex in y( 1 2 ,), then by the Hermite-Hadamard 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 ) , ϖ ( x ) > x λ 2 1 ln λ 2 1 ( y + α ) ( y + α ) [ 1 + x ln ( y + α ) ] λ d y = t = x ln ( y + α ) x ln ( 1 + α ) t λ 2 1 d t ( 1 + t ) λ = B ( λ 2 , λ 2 ) ( 1 θ λ ( x ) ) > 0 , 0 < θ λ ( x ) : = 1 B ( λ 2 , λ 2 ) 0 x ln ( 1 + α ) t λ 2 1 ( 1 + t ) λ d t < 1 B ( λ 2 , λ 2 ) 0 x ln ( 1 + α ) t λ 2 1 d t = 2 [ x ln ( 1 + α ) ] λ 2 λ B ( λ 2 , λ 2 ) ,

that is, (10) is valid. □

Lemma 2 Let the assumptions of Lemma 1 be fulfilled and, additionally, 0<p<1, 1 p + 1 q =1, a n 0, nN, f(x) is 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 [ 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 { 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 the reverse Hölder inequality (cf. [20]) and (10), it follows that

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

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

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

In this paper, for 0<p<1 (q<0), we still use the normal expressions f p , Φ and a q , Ψ . We also introduce two functions

Φ ( x ) : = ( 1 θ λ ( x ) ) x p ( 1 λ 2 ) 1 ( x > 0 ) and Ψ ( n ) : = ( n + α ) q 1 ln q ( 1 λ 2 ) 1 ( n + α ) ( n N ) ,

wherefrom [ Φ ( x ) ] 1 q = ( 1 θ λ ( 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 , 0<p<1, 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 = 1 a n 0 f ( x ) ln λ e ( n + α ) x d x = 0 f ( x ) n = 1 a n ln λ e ( n + α ) x d 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 >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 >B ( λ 2 , λ 2 ) a q , Ψ ,
(15)

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

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 )(1 θ λ (x)), we have (14). By the reverse Hölder 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,

we obtain that J p 1 = a q , Ψ . By (11), we find J>0. If J=, then (14) is trivially valid; if J<, 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 θ λ ( x ) ) ] 1 q ,

we have (15). By the reverse Hölder 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,),

we obtain that L q 1 = f p , Φ . By (12), we find L>0. If L=, then (15) is trivially valid; if L<, 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 , set 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 when replacing B( λ 2 , λ 2 ) with k, then, in particular, it follows that

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

Hence by (18) and (19) it follows that

B ( λ 2 + ε p , λ 2 ε p ) [ ε ( 1 + α ) ln ε + 1 ( 1 + α ) + 1 ln ε ( 1 + α ) ] > k { 1 ε O ( 1 ) } 1 p { ε ( 1 + α ) ln ε + 1 ( 1 + α ) + 1 ln ε ( 1 + α ) } 1 q ,

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

By the equivalence, the constant factor B( λ 2 , λ 2 ) in (14) and (15) is best possible. Otherwise we would reach a contradiction by (16) and (17) that the constant factor in (13) is not best possible. □

Remark 1 (i) For 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 ln p 2 1 ( n + 1 2 ) n + 1 2 [ 0 f ( x ) ln e ( n + 1 2 ) x d x ] p > π p 0 1 θ 1 ( x ) x 1 p 2 f p ( x ) d x ,
(20)
0 ( 1 θ 1 ( x ) ) 1 q x 1 q 2 [ n = 1 a n ln e ( n + 1 2 ) x ] q d x < π q n = 1 ( n + 1 2 ) q 1 ln 1 q 2 ( n + 1 2 ) a n q .
(21)

(ii) Setting x= 1 ln y , g(y):= 1 y ( ln y ) λ 2 f( 1 ln y ) and

ϕ(y):= ( 1 θ λ ( 1 ln y ) ) y p 1 ( ln y ) p ( 1 λ 2 ) 1 ( y ( 1 , ) )

in (13), by simplifications, we find the following inequality with the homogeneous kernel:

n = 1 a n 1 g ( y ) ln λ y ( n + α ) d y = 1 g ( y ) n = 1 a n ln λ y ( n + α ) d x > B ( λ 2 , λ 2 ) g p , ϕ a q , Ψ .
(22)

It is evident that (22) is equivalent to (13), and then the constant factor B( λ 2 , λ 2 ) in (22) is still best possible. In the same way as in (14) and (15), we have the following inequalities equivalent to (13) with the best constant factor B( λ 2 , λ 2 ):

{ n = 1 [ Ψ ( n ) ] 1 p [ 1 g ( y ) ln λ y ( n + α ) d y ] p } 1 p >B ( λ 2 , λ 2 ) g p , ϕ ,
(23)
{ 1 [ ϕ ( y ) ] 1 q [ n = 1 a n ln λ y ( n + α ) ] q d y } 1 q >B ( λ 2 , λ 2 ) a q , Ψ .
(24)

References

  1. Hardy GH, Littlewood JE, Pólya G: Inequalities. Cambridge University Press, Cambridge; 1934.

    Google Scholar 

  2. Mitrinović DS, Pečarić JE, Fink AM: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Academic, Boston; 1991.

    Book  Google Scholar 

  3. Yang B: Hilbert-Type Integral Inequalities. Bentham Science Publishers, Sharjah; 2009.

    Google Scholar 

  4. Yang B: Discrete Hilbert-Type Inequalities. Bentham Science Publishers, Sharjah; 2011.

    Google Scholar 

  5. Yang B: An extension of Mulholland’s inequality. Jordan J. Math. Stat. 2010,3(3):151-157.

    Google Scholar 

  6. Yang B: On Hilbert’s integral inequality. J. Math. Anal. Appl. 1998, 220: 778-785. 10.1006/jmaa.1997.5877

    Article  MathSciNet  Google Scholar 

  7. Yang B: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing; 2009. (in Chinese)

    Google Scholar 

  8. Yang B, Brnetić I, Krnić M, Pečarić J: Generalization of Hilbert and Hardy-Hilbert integral inequalities. Math. Inequal. Appl. 2005,8(2):259-272.

    MathSciNet  Google Scholar 

  9. Krnić M, Pečarić J: Hilbert’s inequalities and their reverses. Publ. Math. (Debr.) 2005,67(3-4):315-331.

    Google Scholar 

  10. Jin J, Debnath L: On a Hilbert-type linear series operator and its applications. J. Math. Anal. Appl. 2010, 371: 691-704. 10.1016/j.jmaa.2010.06.002

    Article  MathSciNet  Google Scholar 

  11. Azar L: On some extensions of Hardy-Hilbert’s inequality and applications. J. Inequal. Appl. 2009., 2009: Article ID 546829

    Google Scholar 

  12. Yang B, Rassias T: On the way of weight coefficient and research for Hilbert-type inequalities. Math. Inequal. Appl. 2003,6(4):625-658.

    MathSciNet  Google Scholar 

  13. Arpad B, Choonghong O: Best constant for certain multilinear integral operator. J. Inequal. Appl. 2006., 2006: Article ID 28582

    Google Scholar 

  14. Kuang J, Debnath L: On Hilbert’s type inequalities on the weighted Orlicz spaces. Pac. J. Appl. Math. 2007,1(1):95-103.

    MathSciNet  Google Scholar 

  15. Zhong W: The Hilbert-type integral inequality with a homogeneous kernel of λ -degree. J. Inequal. Appl. 2008., 2008: Article ID 917392

    Google Scholar 

  16. Li Y, He B: On inequalities of Hilbert’s type. Bull. Aust. Math. Soc. 2007,76(1):1-13. 10.1017/S0004972700039423

    Article  Google Scholar 

  17. Yang B: A mixed Hilbert-type inequality with a best constant factor. Int. J. Pure Appl. Math. 2005,20(3):319-328.

    MathSciNet  Google Scholar 

  18. Yang B: A half-discrete Hilbert’s inequality. J. Guangdong Univ. Educ. 2011,31(3):1-7.

    Google Scholar 

  19. Yang B: A half-discrete reverse Hilbert-type inequality with a homogeneous kernel of positive degree. J. Zhanjiang Norm. Coll. 2011,32(3):5-9.

    Google Scholar 

  20. Kuang J: Applied Inequalities. Shangdong Science Technic Press, Jinan; 2004. (in Chinese)

    Google Scholar 

  21. Kuang J: Introduction to Real Analysis. Hunan Education Press, Chansha; 1996. (in Chinese)

    Google Scholar 

Download references

Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 61370186).

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Leping He.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

TL participated in the design of the study and performed the numerical analysis. BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. LH conceived of the study, and participated in its design and coordination. All authors read and approved the final manuscript.

Rights and permissions

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

Reprints and Permissions

About this article

Cite this article

Liu, T., Yang, B. & He, L. On a half-discrete reverse Mulholland-type inequality and an extension. J Inequal Appl 2014, 103 (2014). https://doi.org/10.1186/1029-242X-2014-103

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1029-242X-2014-103

Keywords

  • Mulholland-type inequality
  • weight function
  • equivalent form
  • reverse