Skip to content

Advertisement

Journal of Inequalities and Applications
Journal of Inequalities and Applications Cover Image
  • Research
  • Open Access

On a more accurate half-discrete mulholland's inequality and an extension

Journal of Inequalities and Applications20122012:70

https://doi.org/10.1186/1029-242X-2012-70

©  Chen and Yang; licensee Springer. 2012

  • Received: 14 November 2011
  • Accepted: 26 March 2012
  • Published:

Abstract

By using the way of weight functions and Jensen-Hadamard's inequality, a more accurate half-discrete 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 considered.

Mathematics Subject Classication 2000: 26D15; 47A07.

Keywords

  • Mulholland's inequality
  • weight function
  • equivalent form
  • operator expression

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's integral inequality (cf. [1]):
∫ 0 ∞ ∫ 0 ∞ f x g y x + y d x d y < π f g ,
(1)
where the constant factor π is the best possible. Moreover, for a = a m m = 1 ∞ ∈ l 2 , b = b n n = 1 ∞ ∈ l 2 , a = ∑ m = 1 ∞ a m 2 1 2 < 0 , b > 0 , we still have the following discrete Hilbert's inequality
∑ m = 1 ∞ ∑ n = 1 ∞ a m b n m + n < π a b ,
(2)
with the same best constant factor π. Inequalities (1) and (2) are important in analysis and its applications (cf. [2–4]) and they still represent the field of interest to numerous mathematicians. Also we have the following Mulholland's 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 satisfying 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 ( R + ) , ‖ f ‖ p , ϕ , ‖ g ‖ q , ψ > 0 , then
∫ 0 ∞ ∫ 0 ∞ k λ x , y f x g y d x d y < k λ 1 f p , ϕ g q , ψ ,
(4)
where the constant factor k(λ1) 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 = 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)

where, k(λ1) is still the best value. Clearly, for p = q = 2 , λ = 1 , k 1 x , y = 1 x + y , λ 1 = λ 2 = 1 2 , inequality (4) reduces to (1), while (5) reduces to (2). Some other results about Hilbert-type inequalities are provided by [8–16].

On 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 the constant factors in the inequalities are the best possible. However Yang [17] gave a result with the kernel 1 1 + n x λ by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang [18] gave the following half-discrete Hilbert's inequality with the best constant factor B(λ1, λ2)(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 .
(6)
In this article, by using the way of weight functions and Jensen-Hadamard's inequality, a more accurate half-discrete Mulholland's inequality with a best constant factor similar to (6) is given as follows:
∫ 3 2 ∞ f x ∑ n = 2 ∞ a n ln 4 9 x n d x < π ∫ 3 2 ∞ x f 2 ( x ) d x ∑ n = 2 ∞ n a n 2 1 2 .
(7)

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

2 Some lemmas

Lemma 1 If λ 1 > 0 , 0 < λ 2 ≤ 1 , λ 1 + λ 2 = λ , α ≥ 4 9 , setting weight functions ω(n) and ϖ(x) as follows:
ω n : = ln α n λ 2 ∫ 1 α ∞ ln α x λ 1 - 1 x ln α x n λ d x , n ∈ N \ 1 ,
(8)
ϖ x : = ln α x λ 1 ∑ n = 2 ∞ ln α n λ 2 - 1 n ln α x n λ , x ∈ 1 α , ∞ ,
(9)
then we have
ϖ x < ω n = B λ 1 , λ 2 .
(10)
Proof. Applying the substitution t = ln α x ln α n to (8), we obtain
ω n = ∫ 0 ∞ 1 1 + t λ t λ 1 - 1 d t = B λ 1 , λ 2 .
Since by the conditions and for fixed x ≥ 1 α ,
h x , y : = ln α y λ 2 - 1 y ln α x y λ = 1 y ln α x + ln α y λ ln α y 1 - λ 2
is decreasing and strictly convex in y ∈ 3 2 , ∞ , then by Jensen-Hadamard's inequality (cf. [1]), we find
ϖ x < α ln x λ 1 ∫ 3 2 ∞ 1 y ln α x y λ ln α y λ 2 - 1 d y t = ( ln α y )/( ln α x ) = ∫ ln 3 α / 2 ln α x ∞ t λ 2 - 1 d t 1 + t λ ≤ B λ 2 , λ 1 B λ 1 , λ 2 ,

namely, (10) follows. □ ▪

Lemma 2 Let the assumptions of Lemma 1 be fulfilled and additionally, p > 1 , 1 p + 1 q = 1 , a n ≥ 0 , ∈ N \ 1 , f x is a non-negative measurable function in 1 α , ∞ . Then we have the following inequalities:
J : = ∑ n = 2 ∞ ln α n p λ 2 - 1 n ∫ 1 α ∞ f x ln α x n λ d x p 1 p ≤ B λ 1 , λ 2 1 q ∫ 1 α ∞ ϖ x x p - 1 ln α x p 1 - λ 1 f p x d x 1 p ,
(11)
L 1 : = ∫ 1 α ∞ ln α x q λ 1 - 1 x ϖ x q - 1 ∑ n = 2 ∞ a n ln α x n λ q d x 1 q ≤ B λ 1 , λ 2 ∑ n = 2 ∞ n q - 1 ln α n q 1 - λ 2 - 1 a n q 1 q .
(12)
Proof. By Hälder's inequality cf. [1] and (10), it follows
∫ 1 α ∞ f x d x ln α x n λ p = ∫ 1 α ∞ 1 ln α x n λ ln α x 1 - λ 1 / q x 1 / q ln α n 1 - λ 2 / p n 1 / p f x × ln α n 1 - λ 2 / p n 1 / p ln α x 1 - λ 1 / q x 1 / q d x p ≤ ∫ 1 α ∞ x p - 1 ln α x n λ ln α x 1 - λ 1 p - 1 n ln α n 1 - λ 2 f p x d x × ∫ 1 α ∞ 1 ln α x n λ n q - 1 ln α n 1 - λ 2 q - 1 x ln α x 1 - λ 1 d x p - 1 = ω n ln α n q 1 - λ 2 - 1 n 1 - q p - 1 ∫ 1 α ∞ x p - 1 ln α x 1 - λ 1 p - 1 n ln α x n λ ln α n 1 - λ 2 f p x d x = B λ 1 , λ 2 p - 1 n ln α n p λ 2 - 1 ∫ 1 α ∞ x p - 1 ln α x 1 - λ 1 p - 1 n ln α x n λ ln α n 1 - λ 2 f p x d x .
Then by Beppo Levi's theorem (cf. [19]), we have
J ≤ B λ 1 , λ 2 1 q ∑ n = 2 ∞ ∫ 1 α ∞ x p - 1 ln α x 1 - λ 1 p - 1 n ln α x n λ ln α n 1 - λ 2 f p x d x 1 p = B λ 1 , λ 2 1 q ∫ 1 α ∞ ∑ n = 2 ∞ x p - 1 ln α x 1 - λ 1 p - 1 n ln α x n λ ln α n 1 - λ 2 f p x d x 1 p = B λ 1 , λ 2 1 q ∫ 1 α ∞ ϖ x x p - 1 ln α x p 1 - λ 1 - 1 f p x d x 1 p ,
that is, (11) follows. Still by Hölder's inequality, we have
∑ n = 2 ∞ a ln α x n λ q = ∑ n = 2 ∞ 1 ln α x n λ ln α x 1 - λ 1 / q x 1 / q ln α n 1 - λ 2 / p n 1 / p × ln α n 1 - λ 2 / p n 1 / p ln α x 1 - λ 1 / q x 1 / p a n q ≤ ∑ n = 2 ∞ x p - 1 ln α x n λ ln α x 1 - λ 1 p - 1 n ln α n 1 - λ 2 q - 1 × ∑ n = 2 ∞ 1 ln α x n λ n q - 1 ln α n 1 - λ 2 q - 1 x ln α x 1 - λ 1 a n q = x ϖ x q - 1 ln α x q λ 1 - 1 ∑ n = 2 ∞ ln α x λ 1 - 1 x ln α x n λ n q - 1 ln α n q - 1 1 - λ 2 a n q .
Then by Beppo Levi's theorem, we have
L 1 ≤ ∫ 1 α ∞ ∑ n = 2 ∞ ln α x λ 1 - 1 x ln α x n λ n q - 1 ln α n q - 1 1 - λ 2 a n q d x 1 q = ∑ n = 2 ∞ ln α n λ 2 ∫ 1 α ∞ ln α x λ 1 - 1 x ln α x n λ d x n q - 1 ln α n q 1 - λ 2 - 1 a n q 1 q = ∑ n = 2 ∞ ω n n q - 1 ln α n q 1 - λ 2 - 1 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 ln α x p 1 - λ 1 - 1 x ∈ 1 α , ∞ , and Ψ n : = n q - 1 ln α n q 1 - λ 2 - 1 n ∈ N \ 1 ,

wherefrom, Φ x 1 - q = 1 x ln α x q λ 1 - 1 , and Ψ n 1 - p = 1 n ln α n p λ 2 - 1 .

Theorem 3 If p > 1 , 1 p + 1 q = 1 , λ 1 > 0 , 0 < λ 2 ≤ 1 , λ 1 + λ 2 = λ , α ≥ 4 9 , f x , a n ≥ 0 , f ∈ L p , Φ 1 α , ∞ , a = a n n = 2 ∞ ∈ l q , Ψ , f p , Φ > 0 , then we have the following equivalent inequalities:
I : = ∑ n = 2 ∞ ∫ 1 α ∞ a n f x d x ln α x n λ = ∫ 1 α ∞ ∑ n = 2 ∞ f x a n d x ln α x n λ < B λ 1 , λ 2 f p , Φ a q , Ψ ,
(13)
J = ∑ n = 2 ∞ Ψ n 1 - p ∫ 1 α ∞ f x d x ln α x n λ p 1 p < B λ 1 , λ 2 f p , Φ ,
(14)
L : = ∫ 1 α ∞ Φ x 1 - q ∑ n = 2 ∞ a n ln α x n λ q d x 1 q < B λ 1 , λ 2 a q , Ψ ,
(15)

where the constant B(λ1, λ2) is the best possible in the above inequalities.

Proof. By Beppo Levi's theorem (cf. [19]), there are two expressions for I in (13). In view of (11), for ϖ(x) < B(λ1, λ2), we have (14). By Hälder's inequality, we have
I = ∑ n = 2 ∞ Ψ - 1 q n ∫ 1 α ∞ 1 ln α x n λ 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 ∫ 1 α ∞ 1 ln α x n λ f x d x p - 1 , n ∈ N \ 1 ,
then J p -1 = ||a|| q , Ψ. By (11), we find J < ∝. If J = 0, then (14) is valid trivially; if J > 0, then by (13), we have
a q , Ψ q = J p = I < B λ 1 , λ 2 f p , Ψ a q , Ψ , i .e . a q , Ψ q - 1 = J < B λ 1 , λ 2 f p , Φ ,
that is, (14) is equivalent to (13). By (12), since [ϖ(x)]1-q>[B(λ1, λ2)]1-q, we have (15). By Hälder's inequality, we find
I = ∫ 1 α ∞ Φ 1 p x f x Φ - 1 p x ∑ n = 2 ∞ a n ln α x n λ 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 = 2 ∞ a n ln α x n λ q - 1 , x ∈ 1 α , ∞ ,
then L q -1 = ║f║ p , Φ.. By (12), we find L < ∝. If L = 0, then (15) is valid trivially; if L > 0, then by (13), we have
f p , Φ p = L q = I < B λ 1 , λ 2 f p , Φ a q , Ψ , i .e . f p , Φ p - 1 = L < B λ 1 , λ 2 a q , Ψ ,

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

For 0 < ε <pλ1, setting ã n = 1 n ln α n λ 2 - ∈ q - 1 , n ∈ N \ { 1 } , and
f ̃ x : = 0 , x ∈ 1 α , e α 1 x ln α x λ 1 - ∈ p - 1 , x ∈ e α , ∞ ,
if there exists a positive number k(≤ B(λ1, λ2)), such that (13) is valid as we replace B(λ1, λ2) with k, then in particular, it follows
Ĩ : = ∑ n = 2 ∞ ∫ 1 α ∞ 1 ln α x n λ ã n f ̃ x d x < k f ̃ p , Φ ã q , Ψ = k ∫ e α ∞ d x x ln α x ε + 1 1 p 1 2 ln 2 α ε + 1 + ∑ n = 3 ∞ 1 n ln α n ε + 1 1 q < k 1 ε 1 p 1 2 ln 2 α ε + 1 + ∫ 2 ∞ 1 x ln α x ε + 1 d x 1 q = k ε ε 2 ln 2 α ε + 1 + 1 ln 2 α ε 1 q ,
(18)
Ĩ = ∑ n = 2 ∞ ln α n λ 2 - ε q - 1 1 n ∫ e α ∞ 1 x ln α x n λ ln α x λ 1 - ε p - 1 d x t = ln α x / ln α n = ∑ n = 2 ∞ 1 n ln α n ε + 1 ∫ 1 / ln α n ∞ t λ 1 - ε p - 1 t + 1 λ d t = B λ 1 - ε p , λ 2 + ε p ∑ n = 2 ∞ 1 n ln α n ε + 1 - A ε > B λ 1 - ε p , λ 2 + ε p ∫ 2 ∞ 1 y ln α y ε + 1 d y - A ε = 1 ε ln 2 α ε B λ 1 - ε p , λ 2 + ε p - A ε , A ε : = ∑ n = 2 ∞ 1 n ln α n ε + 1 ∫ 0 1 / ln α n 1 t + 1 λ t λ 1 - ε p - 1 d t .
(19)
W find
0 < A ε ≤ ∑ n = 2 ∞ 1 n ln α n ε + 1 ∫ 0 1 / ln α n t λ 1 - ε p - 1 d t = 1 λ 1 - ε p ∑ n = 2 ∞ 1 n ln α n λ 1 + ε q + 1 < ∞ ,
that is, A(ε) = O(1) (ε → 0+). Hence by (18) and (19), it follows
B λ 1 - ε p , λ 2 + ε p ln 2 α ε - ε O 1 < k ε 2 ln 2 α ε + 1 + 1 ln 2 α ε 1 q ,
(20)

and B(λ1, λ2) ≤ k(ε → 0+). Hence, k = B(λ1, λ2) is the best value of (13).

Due to the equivalence, the constant factor B(λ1, λ2) 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 Mulholland's operator T : L p , Φ 1 α , ∞ → l p , Ψ 1 - p as follows: for f ∈ L p , Φ 1 α , ∞ , we define T f ∈ l p , Ψ 1 - p as
T f ( n ) = ∫ 1 α ∞ 1 ( ln α x n ) λ f ( x ) d x , n ∈ N \ { 1 } .

Then by (14), it follows T f p , Ψ 1 - p ≤ B ( λ 1 , λ 2 ) f p , Φ and then T is a bounded operator with ║T║ ≤ B(λ1, λ2). Since by Theorem 1, the constant factor in (14) is the best possible, we have ║T║ = B(λ1, λ2).

(ii) Define the second type half-discrete Mulholland's operator T ̃ : l q , Ψ → L q , Φ 1 - q 1 α , ∞ as follows: For a ∈ l q, ψ, define T ̃ a ∈ L q , Φ 1 - q 1 α , ∞ as
T ̃ a ( x ) = ∑ n = 2 ∞ 1 ( ln α x n ) λ a n , x ∈ 1 α , ∞ .

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

Remark 2 We set p = q = 2 , λ = 1 , λ 1 = λ 2 = 1 2 in (13), (14) and (15). (i) if α = 4 9 , then we deduce (7) and the following equivalent inequalities:
∑ n = 2 ∞ 1 n ∫ 3 2 ∞ f ( x ) ln 4 9 x n d x 2 < π 2 ∫ 3 2 ∞ x f 2 ( x ) d x ,
(21)
∫ 3 2 ∞ 1 x ∑ n = 2 ∞ a n ln 4 9 x n 2 d x < π 2 ∑ n = 2 ∞ n a n 2 ;
(22)
(ii) if α = 1, then we have the following half-discrete Mulholland's inequality and its equivalent forms:
∫ 1 ∞ f ( x ) ∑ n = 2 ∞ a n ln x n d x < π ∫ 1 ∞ x f 2 ( x ) d x ∑ n = 2 ∞ n a n 2 1 2 ,
(23)
∑ n = 2 ∞ 1 n ∫ 1 ∞ f ( x ) ln x n d x 2 < π 2 ∫ 1 ∞ x f 2 ( x ) d x ,
(24)
∫ 1 ∞ 1 x ∑ n = 2 ∞ a n ln x n 2 d x < π 2 ∑ n = 2 ∞ n a n 2 .
(25)

Declarations

Acknowledgements

This work is supported by the Guangdong Science and Technology Plan Item (No. 2010B010600018), and the Guangdong Modern Information Service industry Develop Particularly item 2011 (No. 13090).

Authors’ Affiliations

(1)
Department of Computer Science, Guangdong University of Education, Guangzhou, Guangdong, 510303, PR China
(2)
Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong, 510303, PR China

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 Acaremic Publishers, Boston; 1991.View ArticleGoogle Scholar
  3. Yang B: Hilbert-Type Integral Inequalities. Bentham Science Publishers Ltd., dubai; 2009.Google Scholar
  4. Yang B: Discrete Hilbert-Type Inequalities. Bentham Science Publishers Ltd., dubai; 2011.Google Scholar
  5. Yang B: An extension of Mulholand'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.1997.5877MathSciNetView ArticleGoogle Scholar
  7. Yang B: The Norm of Operator and Hilbert-type Inequalities. Science Press, Beijin; 2009.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.MathSciNetGoogle Scholar
  9. Krnić M, Pečarić J: Hilbert's inequalities and their reverses. Publ Math Debrecen 2005, 67(3–4):315–331.MathSciNetGoogle 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.2010.06.002MathSciNetView ArticleGoogle Scholar
  11. Azar L: On some extensions of Hardy-Hilbert's inequality and applications. J Inequal Appl 2009, 2009: 14. Article ID 546829Google 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.MathSciNetGoogle Scholar
  13. Arpad B, Choonghong O: Best constant for certain multilinear integral operator. J Inequal Appl 2006, 2006: 12. Article ID 28582Google 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.MathSciNetGoogle Scholar
  15. Zhong W: The Hilbert-type integral inequality with a homogeneous kernel of Lambda-degree. J Inequal Appl 2008, 2008: 13. Article ID 917392View ArticleGoogle Scholar
  16. Li Y, He B: On inequalities of Hilbert's type. Bull Aust Math Soc 2007, 76(1):1–13. 10.1017/S0004972700039423View ArticleGoogle Scholar
  17. Yang B: A mixed Hilbert-type inequality with a best constant factor. Int J Pure Appl Math 2005, 20(3):319–328.MathSciNetGoogle Scholar
  18. Yang B: A half-discrete Hilbert's inequality. J Guangdong Univ Edu 2011, 31(3):1–7.Google Scholar
  19. Donald LC: Measure Theorey. Birkhäuser, Boston; 1980.Google Scholar

Copyright

© Chen and Yang; licensee Springer. 2012

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.

Advertisement