Next Article in Journal
Head Waves in Modified Weiskopf Sandy Medium
Next Article in Special Issue
Robust Stability of Switched-Interval Positive Linear Systems with All Modes Unstable Using the Φ-Dependent Dwell Time Technique
Previous Article in Journal
An Equivalent Form Related to a Hilbert-Type Integral Inequality
Previous Article in Special Issue
Forecasting Water Consumption in the Yangtze River Delta Based on Deformable Cumulative Multivariable Grey Model

Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article

A New Reverse Extended Hardy–Hilbert’s Inequality with Two Partial Sums and Parameters

by
Jianquan Liao
and
Bicheng Yang
*
School of Mathematics, Guangdong University of Education, Guangzhou 510303, China
*
Author to whom correspondence should be addressed.
Axioms 2023, 12(7), 678; https://doi.org/10.3390/axioms12070678
Submission received: 14 June 2023 / Revised: 30 June 2023 / Accepted: 7 July 2023 / Published: 10 July 2023
(This article belongs to the Special Issue Advances in Analysis and Control of Systems with Uncertainties II)

Abstract

:
By using the methods of real analysis and the mid-value theorem, we introduce some lemmas and obtain a new reverse extended Hardy–Hilbert’s inequality with two partial sums and multi-parameters. We also give a few equivalent conditions of the best possible constant factor related to several parameters in the new inequality. Some particular inequalities are deduced.
MSC:
26D15

1. Introduction

If and $0 < ∑ n = 1 ∞ b n q < ∞$, then we have the following well-known Hardy–Hilbert’s inequality (cf. [1], Theorem 315):
$∑ m = 1 ∞ ∑ n = 1 ∞ a m b n m + n < π sin ( π / p ) ( ∑ m = 1 ∞ a m p ) 1 p ( ∑ n = 1 ∞ b n q ) 1 q ,$
where the constant factor $π sin ( π / p )$ is best possible.
In 2006, Krnic et al. [2] obtained the following inequality, which is a generalization of (1): for
$∑ m = 1 ∞ ∑ n = 1 ∞ a m b n ( m + n ) λ < B ( λ 1 , λ 2 ) [ ∑ m = 1 ∞ m p ( 1 − λ 1 ) − 1 a m p ] 1 p [ ∑ n = 1 ∞ n q ( 1 − λ 2 ) − 1 b n q ] 1 q ,$
with the best possible constant factor $B ( λ 1 , λ 2 )$, in which
is the beta function. In particular, for $p = q = 2 ,$ $λ 1 = λ 2 = λ 2 ,$ (2) deduces to Yang’s inequality in [3]. In 2019, following the way of (2), by using Abel’s summation by parts formula, Adiyasuren et al. [4] provided the following extension of (2) involving two partial sums and some parameters, for
$∑ m = 1 ∞ ∑ n = 1 ∞ a m b n ( m + n ) λ < λ 1 λ 2 B ( λ 1 , λ 2 ) ( ∑ m = 1 ∞ m − p λ 1 − 1 A m p ) 1 p ( ∑ n = 1 ∞ n − q λ 2 − 1 B n q ) 1 q ,$
where $λ 1 λ 2 B ( λ 1 , λ 2 )$ is the best possible constant factor, and $A m : = ∑ i = 1 m a i$ and $B n : = ∑ k = 1 n b k$ $( m , n ∈ N = { 1 , 2 , ⋯ } )$, satisfy the following inequalities:
Both (1) and (2) with their integral analogues played an important role in real analysis, in which some generalizations of (1) are given and a relation between (1) and the other Hilbert-type inequality is obtained (cf. [5,6,7,8,9,10,11,12,13,14,15,16]). In 2021, Gu et. al. [17] provided a generalization of (4) with $1 ( m α + n β ) λ ( α , β ∈ ( 0 , 1 ] )$ as the kernel of inequality. In 2016, by using the weight coefficients and the techniques of real analysis, Hong et al. [18] considered a few equivalent statements of the generalization of (1) with the best possible constant factor related to multi-parameters. The other further results were obtained by [19,20,21,22,23,24,25,26,27,28,29].
In this article, based on the idea of [17,18], by using the techniques of real analysis and the mid-value theorem, we introduce some preserving lemmas and then give a reverse of (2) with two partial sums and multi-parameters, which is a new reverse version of the inequality in [16]. We also consider a few equivalent conditions of the best possible constant factor in the reverse inequality related to multi-parameters. Furthermore, several inequalities are deduced by setting some particular parameters.

2. Some Lemmas

In what follows, we assume that  $λ ∈ ( 0 , 6 ] ,$ $α , β ∈ ( 0 , 1 ] ,$  $λ ^ 1 : = λ − λ 2 p + λ 1 q ,$ $λ ^ 2 : = λ − λ 1 q + λ 2 p$. We also assume that $a m , b n ≥ 0 ,$ $A m : = ∑ j = 1 m a j ,$ $B n : = ∑ k = 1 n b k$ $( m , n ∈ N ) ,$ satisfy $A m = o ( e t m α )$, $B n = o ( e t n β )$ $( t > 0 ; m , n → ∞ ) ,$ and the following inequalities:
$0 < ∑ m = 1 ∞ m p ( 1 − α λ ^ 1 ) − 1 a m p < ∞ , 0 < ∑ n = 1 ∞ n q ( 1 − β λ ^ 2 ) − 1 b n q < ∞ .$
Lemma 1.
For $t > 0$, the following inequalities on the partial sums are valid:
$∑ m = 1 ∞ e − t m α m α − 1 A m ≥ 1 t α ∑ m = 1 ∞ e − t m α a m ,$
$∑ n = 1 ∞ e − t n β n β − 1 B n ≥ 1 t β ∑ n = 1 ∞ e − t n β b n .$
Proof.
Since $A m e − t m α = o ( 1 ) ( m → ∞ )$, by Abel’s summation by parts formula, it follows that
$∑ m = 1 ∞ e − t m α a m = lim m → ∞ A m e − t m α + ∑ m = 1 ∞ A m [ e − t m α − e − t ( m + 1 ) α ] = ∑ m = 1 ∞ A m [ e − t m α − e − t ( m + 1 ) α ] .$
We set function $g ( x ) = e − t x α , x ∈ [ m , m + 1 ] .$ Then, we find $g ′ ( x ) = − t α x α − 1 e − t x α ,$ and for is decreasing in $[ m , m + 1 ] .$ By the mid-value theorem, we have
$∑ m = 1 ∞ e − t m α a m = − ∑ m = 1 ∞ A m ( g ( m + 1 ) − g ( m ) ) = − ∑ m = 1 ∞ A m g ′ ( m + θ ) = t α ∑ m = 1 ∞ ( m + θ ) α − 1 e − t ( m + θ ) α A m ≤ t α ∑ m = 1 ∞ m α − 1 e − t m α A m ( θ ∈ ( 0 , 1 ) ) .$
Hence, we have (6). In the same way, inequality (7) follows.
The lemma is proved. □
In the following lemma, for estimating the weight coefficient in Lemma 3, we introduce some results related to the Bernoulli functions and the related formulas.
Lemma 2.
(Ref. [5]’s section 2.2.3, [30]) (i) If $( − 1 ) i d i d t i g ( t ) > 0 ,$ $t ∈ [ m , ∞ ) ( m ∈ N )$ with $g ( i ) ( ∞ ) = 0$ $( i = 0 , 1 , 2 , 3 )$, are Bernoulli functions and Bernoulli numbers of i-order, then
$∫ m ∞ P 2 k − 1 ( t ) g ( t ) d t = − ε k B 2 k 2 k g ( m ) ( 0 < ε k < 1 ; k = 1 , 2 , ⋯ ) .$
In particular, for $k = 1 ,$ since $B 2 = 1 6$, we find
$− 1 12 g ( m ) < ∫ m ∞ P 1 ( t ) g ( t ) d t < 0 ;$
For $k = 2 ,$ based on $B 4 = − 1 30$, it follows that
$0 < ∫ m ∞ P 3 ( t ) g ( t ) d t < 1 120 g ( m ) .$
(ii) (Ref. [5]’s section 2.2.3, [30]) Suppose that $( i = 0 , 1 , 2 , 3 )$. We have the following Euler–Maclaurin summation formula:
$∑ k = m ∞ f ( k ) = ∫ m ∞ f ( t ) d t + 1 2 f ( m ) + ∫ m ∞ P 1 ( t ) f ′ ( t ) d t ,$
$∫ m ∞ P 1 ( t ) f ′ ( t ) d t = − 1 12 f ′ ( m ) + 1 6 ∫ m ∞ P 3 ( t ) f ‴ ( t ) d t .$
Lemma 3.
For $s ∈ ( 0 , 6 ] ,$ $s 2 ∈ ( 0 , 2 β ] ∩ ( 0 , s ) ,$ $k s ( s 2 ) : = B ( s 2 , s − s 2 )$, the weight coefficient is defined as follows:
$ϖ s ( s 2 , m ) : = m α ( s − s 2 ) ∑ n = 1 ∞ β n β s 2 − 1 ( m α + n β ) s ( m ∈ N ) .$
Then, the following inequalities are valid:
$0 < k s ( s 2 ) ( 1 − O ( 1 m α s 2 ) ) < ϖ s ( s 2 , m ) < k s ( s 2 ) ( m ∈ N ) .$
where we indicate that $O ( 1 m α s 2 ) : = 1 k s ( s 2 ) ∫ 0 1 m α u s 2 − 1 ( 1 + u ) s d u > 0 .$
Proof.
For any $m ∈ N$, the function $g ( m , t )$ is defined by: $g ( m , t ) : = β t β s 2 − 1 ( m α + t β ) s ( t > 0 ) .$ In view of (11), we have
$∑ n = 1 ∞ g ( m , n ) = ∫ 1 ∞ g ( m , t ) d t + 1 2 g ( m , 1 ) + ∫ 1 ∞ P 1 ( t ) g ′ ( m , t ) d t = ∫ 0 ∞ g ( m , t ) d t − h ( m ) ,$
where we set $h ( m ) : = ∫ 0 1 g ( m , t ) d t − 1 2 g ( m , 1 ) − ∫ 1 ∞ P 1 ( t ) g ′ ( m , t ) d t$.
We obtain $− 1 2 g ( m , 1 ) = − β 2 ( m α + 1 ) s$. By integration by parts, it follows that
$∫ 0 1 g ( m , t ) d t = β ∫ 0 1 t β s 2 − 1 ( m α + t β ) s d t = u = t β ∫ 0 1 u s 2 − 1 ( m α + u ) s d u = 1 s 2 ∫ 0 1 d u s 2 ( m α + u ) s = 1 s 2 u s 2 ( m α + u ) s | 0 1 + s s 2 ∫ 0 1 u s 2 ( m α + u ) s + 1 d u = 1 s 2 1 ( m α + 1 ) s + s s 2 ( s 2 + 1 ) ∫ 0 1 d u s 2 + 1 ( m α + u ) s + 1 > 1 s 2 1 ( m α + 1 ) s + s s 2 ( s 2 + 1 ) [ u s 2 + 1 ( m α + u ) s + 1 ] 0 1 + s ( s + 1 ) s 2 ( s 2 + 1 ) ( m α + 1 ) s + 2 ∫ 0 1 u s 2 + 1 d u = 1 s 2 1 ( m α + 1 ) s + λ s 2 ( s 2 + 1 ) 1 ( m α + 1 ) s + 1 + s ( s + 1 ) s 2 ( s 2 + 1 ) ( s 2 + 2 ) 1 ( m α + 1 ) s + 2 ,$
$− g ′ ( m , t ) = − β ( β s 2 − 1 ) t β s 2 − 2 ( m α + t β ) s + β 2 s t β + β s 2 − 2 ( m α + t β ) s + 1 = − β ( β s 2 − 1 ) t β s 2 − 2 ( m α + t β ) s + β 2 s ( m α + t β − m α ) t β s 2 − 2 ( m α + t β ) s + 1 = β ( β s − β s 2 + 1 ) t β s 2 − 2 ( m α + t β ) s − β 2 s m α t β s 2 − 2 ( m α + t β ) s + 1 ,$
and for , it follows that
By (9), (10), (11) and (12), we obtain
$β ( β s − β s 2 + 1 ) ∫ 1 ∞ P 1 ( t ) t β s 2 − 2 ( m α + t β ) s d t > − β ( β s − β s 2 + 1 ) 12 ( m α + 1 ) s ,$
$− β 2 m α s ∫ 1 ∞ P 1 ( t ) t β s 2 − 2 ( m α + t β ) s + 1 d t = β 2 m α s 12 ( m α + 1 ) s + 1 − β 2 m α s 6 ∫ 1 ∞ P 3 ( t ) [ t β s 2 − 2 ( m α + t β ) s + 1 ] ″ d t > β 2 m α s 12 ( m α + 1 ) s + 1 − β 2 m α s 720 [ t β s 2 − 2 ( m α + t β ) s + 1 ] t = 1 ″ > β 2 ( m α + 1 − 1 ) s 12 ( m α + 1 ) s + 1 − β 2 ( m α + 1 ) s 720 [ ( s + 1 ) ( s + 2 ) β 2 ( m α + 1 ) s + 3 + β ( s + 1 ) ( 5 − β − 2 β s 2 ) ( m α + 1 ) s + 2 + ( 2 − β s 2 ) ( 3 − β s 2 ) ( m α + 1 ) s + 1 ] = β 2 s 12 ( m α + 1 ) s − β 2 s 12 ( m α + 1 ) s + 1 − β 2 s 720 [ ( s + 1 ) ( s + 2 ) β 2 ( m α + 1 ) s + 2 + β ( s + 1 ) ( 5 − β − 2 β s 2 ) ( m α + 1 ) s + 1 + ( 2 − β s 2 ) ( 3 − β s 2 ) ( m α + 1 ) s ] .$
Hence, we have $h ( m ) > 1 ( m α + 1 ) s h 1 + λ ( m α + 1 ) s + 1 h 2 + s ( s + 1 ) ( m α + 1 ) s + 2 h 3 ,$ where
$h 1 : = 1 s 2 − β 2 − β − β 2 s 2 12 − β 2 s ( 2 − β s 2 ) ( 3 − β s 2 ) 720 ,$
$h 2 : = 1 s 2 ( s 2 + 1 ) − β 2 12 − β 3 ( s + 1 ) ( 5 − β − 2 β s 2 ) 720 ,$
and $h 3 : = 1 s 2 ( s 2 + 1 ) ( s 2 + 2 ) − β 4 ( s + 2 ) 720 .$ We can find
$h 1 ≥ 1 s 2 − β 2 − β − β 2 s 2 12 − s β 2 ( 2 − β s 2 ) ( 3 − β s 2 ) 720 = g ( s 2 ) 720 s 2 ,$
where the function $g ( σ ) ( σ ∈ ( 0 , 2 β ] )$ is indicated by
$g ( σ ) : = 720 − ( 420 β + 6 s β 2 ) σ + ( 60 β 2 + 5 s β 3 ) σ 2 − s β 4 σ 3 .$
For , we obtain
$g ′ ( σ ) = − ( 420 β + 6 s β 2 ) + 2 ( 60 β 2 + 5 s β 3 ) σ − 3 β 4 σ 2 ≤ − 420 β − 6 s β 2 + 2 ( 60 β 2 + 5 s β 3 ) 2 β = ( 14 s β − 180 ) β < 0 ,$
and then it follows that . For $s 2 ∈ ( 0 , 2 β ]$, we still find
$h 2 > β 2 6 − β 2 12 − 5 ( s + 1 ) β 2 720 = ( 1 12 − s + 1 140 ) β 2 > 0 ,$
and $h 3 ≥ ( 1 24 − s + 2 720 ) β 3 > 0 ( s ∈ ( 0 , 6 ] )$.
Hence, it follows that $h ( m ) > 0$. Setting $t = m α β u 1 β ,$ we have
$ϖ s ( s 2 , m ) = m α ( s − s 2 ) ∑ n = 1 ∞ g ( m , n ) < m α ( s − s 2 ) ∫ 0 ∞ g ( m , t ) d t = β m α ( s − s 2 ) ∫ 0 ∞ t β s 2 − 1 d t ( m α + t β ) s = ∫ 0 ∞ u s 2 − 1 d u ( 1 + u ) s = B ( s 2 , s − s 2 ) = k s ( s 2 ) .$
On the other hand, in view of (11), we find
$∑ n = 1 ∞ g ( m , n ) = ∫ 1 ∞ g ( m , t ) d t + 1 2 g ( m , 1 ) + ∫ 1 ∞ P 1 ( t ) g ′ ( m , t ) d t = ∫ 1 ∞ g ( m , t ) d t + H ( m ) ,$
where we set $H ( m ) : = 1 2 g ( m , 1 ) + ∫ 1 ∞ P 1 ( t ) g ′ ( m , t ) d t$.
We find $1 2 g ( m , 1 ) = β 2 ( m α + 1 ) s$, and
$g ′ ( m , t ) = − β ( β s − β s 2 + 1 ) t β s 2 − 2 ( m α + t β ) s + β 2 s m α t β s 2 − 2 ( m α + t β ) s + 1 .$
For , by (7), we find
$− β ( β s − β s 2 + 1 ) ∫ 1 ∞ P 1 ( t ) t β s 2 − 2 ( m α + t β ) s d t > 0 ,$
and
$β 2 m α s ∫ 1 ∞ P 1 ( t ) t β s 2 − 2 ( m α + t β ) s + 1 d t > − β 2 m α s 12 ( m α + 1 ) s + 1 > − β 2 s 12 ( m α + 1 ) s .$
Hence, it follows that
$H ( m ) > β 2 ( m α + 1 ) s − β 2 s 12 ( m α + 1 ) s ≥ β 2 ( m α + 1 ) s − 6 β 12 ( m α + 1 ) s = 0 .$
Then, we have
$ϖ s ( λ 2 , m ) = m α ( s − s 2 ) ∑ n = 1 ∞ g ( m , n ) > m α ( s − s 2 ) ∫ 1 ∞ g ( m , t ) d t = m α ( s − s 2 ) ∫ 0 ∞ g ( m , t ) d t − m α ( s − s 2 ) ∫ 0 1 g ( m , t ) d t = k s ( s 2 ) [ 1 − 1 k s ( s 2 ) ∫ 0 1 m α u s 2 − 1 ( 1 + u ) s d u ] > 0 ,$
where we indicate that $O ( 1 m α s 2 ) = 1 k s ( s 2 ) ∫ 0 1 m α u s 2 − 1 ( 1 + u ) s d u ,$ satisfying
$0 < ∫ 0 1 m α u s 2 − 1 ( 1 + u ) s d u < ∫ 0 1 m α u s 2 − 1 d u = 1 s 2 m α s 2 .$
Therefore, inequalities (14) are valid.
The lemma is proved. □
In view of Lemma 3, the key inequality is obtained as follows:
Lemma 4.
We have the reverse inequality, as follows:
$I λ : = ∑ n = 1 ∞ ∑ m = 1 ∞ a m b n ( m α + n β ) λ > ( 1 β k λ ( λ 2 ) ) 1 p ( 1 α k λ ( λ 1 ) ) 1 q × [ ∑ m = 1 ∞ ( 1 − O ( 1 m α λ 2 ) ) m p ( 1 − α λ ^ 1 ) − 1 a m p ] 1 p [ ∑ n = 1 ∞ n q ( 1 − β λ ^ 2 ) − 1 b n q ] 1 q .$
Proof.
By the symmetry, for $s 1 ∈ ( 0 , 2 α ] ∩ ( 0 , s ) , s ∈ ( 0 , 6 ] ,$ we obtain the inequalities of the next weight coefficient, as follows:
$0 < k s ( s 1 ) ( 1 − O ( 1 n β s 1 ) ) < ω s ( s 1 , n ) : = n β ( s − s 1 ) ∑ m = 1 ∞ α m α s 1 − 1 ( m α + n β ) s < k s ( s 1 ) ( n ∈ N ) ,$
where we indicate $O ( 1 n β s 1 ) : = 1 k s ( s 1 ) ∫ 0 1 n β u s 1 − 1 ( 1 + u ) s d u > 0$.
Using the reverse Hölder’s inequality (cf. [31]), it follows that
$I λ = ∑ n = 1 ∞ ∑ m = 1 ∞ 1 ( m α + n β ) λ [ m α ( 1 − λ 1 ) / q ( β n β − 1 ) 1 / p n β ( 1 − λ 2 ) / p ( α m α − 1 ) 1 / q a m ] [ n β ( 1 − λ 2 ) / p ( α m α − 1 ) 1 / q m α ( 1 − λ 1 ) / q ( β n β − 1 ) 1 / p b n ] ≥ [ ∑ m = 1 ∞ ∑ n = 1 ∞ β ( m α + n β ) λ m α ( 1 − λ 1 ) ( p − 1 ) n β − 1 a m p n β ( 1 − λ 2 ) ( α m α − 1 ) p − 1 ] 1 p [ ∑ n = 1 ∞ ∑ m = 1 ∞ α ( m α + n β ) λ n β ( 1 − λ 2 ) ( q − 1 ) m α − 1 b n q m α ( 1 − λ 1 ) ( β n β − 1 ) q − 1 ] 1 q = ( 1 α ) 1 q ( 1 β ) 1 p [ ∑ m = 1 ∞ ϖ λ ( λ 2 , m ) m p ( 1 − α λ ^ 1 ) − 1 a m p ] 1 p × [ ∑ n = 1 ∞ ω λ ( λ 1 , n ) n q ( 1 − β λ ^ 2 ) − 1 b n q ] 1 q .$
By (14), (16) and (5) (for $s = λ , s i = λ i ( i = 1 , 2 )$), since $0 < p < 1 ( q < 0 )$ and the assumptions, we obtain (15).
The lemma is proved. □

3. Main Results

By Lemma 1 and Lemma 4, the following theorem follows:
Theorem 1.
The following reverse inequality with two partial sums and parameters is valid:
$I : = ∑ m = 1 ∞ ∑ n = 1 ∞ m α − 1 n β − 1 ( m α + n β ) λ + 2 A m B n > Γ ( λ ) Γ ( λ + 2 ) α β ( 1 β k λ ( λ 2 ) ) 1 p ( 1 α k λ ( λ 1 ) ) 1 q × [ ∑ m = 1 ∞ ( 1 − O ( 1 m α λ 2 ) ) m p ( 1 − α λ ^ 1 ) − 1 a m p ] 1 p [ ∑ n = 1 ∞ n q ( 1 − β λ ^ 2 ) − 1 b n q ] 1 q .$
In particular, for $λ 1 + λ 2 = λ$, we have
as well as:
$∑ m = 1 ∞ ∑ n = 1 ∞ m α − 1 n β − 1 A m B n ( m α + n β ) λ + 2 > ( 1 α ) 1 + 1 q ( 1 β ) 1 + 1 p Γ ( λ ) Γ ( λ + 2 ) B ( λ 1 , λ 2 ) × [ ∑ m = 1 ∞ ( 1 − O ( 1 m α λ 2 ) ) m p ( 1 − α λ 1 ) − 1 a m p ] 1 p [ ∑ n = 1 ∞ n q ( 1 − β λ 2 ) − 1 b n q ] 1 q .$
Proof.
Based on the expression as follows
$1 ( m α + n β ) λ + 2 = 1 Γ ( λ + 2 ) ∫ 0 ∞ t ( λ + 2 ) − 1 e − ( m α + n β ) t d t ,$
by (6) and (7), we have
$I = 1 Γ ( λ + 2 ) ∑ m = 1 ∞ ∑ n = 1 ∞ ( m α − 1 A m ) ( n β − 1 B n ) ∫ 0 ∞ t λ + 1 e − ( m α + n β ) t d t = 1 Γ ( λ + 2 ) ∫ 0 ∞ t λ + 1 ( ∑ m = 1 ∞ e − m α t m α − 1 A m ) ( ∑ n = 1 ∞ e − n β t n β − 1 B n ) d t ≥ 1 Γ ( λ + 2 ) ∫ 0 ∞ t λ + 1 ( 1 t α ∑ m = 1 ∞ e − m α t a m ) ( 1 t β ∑ n = 1 ∞ e − n β t b n ) d t = 1 Γ ( λ + 2 ) α β ∑ m = 1 ∞ ∑ n = 1 ∞ a m b n ∫ 0 ∞ t λ − 1 e − ( m α + n β ) t d t = Γ ( λ ) Γ ( λ + 2 ) α β I λ .$
Then, by (15), inequality (17) follows.
The theorem is proved. □
In the following two theorems, we provide a few equivalent conditions on (17).
Theorem 2.
Assume that  If $λ 1 + λ 2 = λ ,$ then $Γ ( λ ) Γ ( λ + 2 ) α β$$( 1 β k λ ( λ 2 ) ) 1 p ( 1 α k λ ( λ 1 ) ) 1 q$ in (17) is the best possible. constant factor.
Proof.
We now prove that the constant factor $Γ ( λ ) Γ ( λ + 2 ) ( 1 α ) 1 + 1 q ( 1 β ) 1 + 1 p B ( λ 1 , λ 2 )$ in (18) is the best possible. For any $0 < ε < min { p λ 1 , | q | ( 2 β − λ 2 ) }$, we set
Since by (2.2.24) (cf. [5]), we have
In the same way, for $0 < β ( λ 2 − ε q ) < 2 ,$ we have
$B ˜ n : = ∑ k = 1 n b ˜ k = 1 β ( λ 2 − ε q ) ( n β ( λ 2 − ε q ) + c 2 + O 2 ( n β ( λ 2 − ε q ) − 1 ) ) ( n ∈ N ; n → ∞ ) .$
We observe that $A ˜ m = o ( e t m α ) , B ˜ n = o ( e t n β ) ( t > 0 ; m , n → ∞ )$.
If there exists a constant $M ≥ ( 1 α ) 1 + 1 q ( 1 β ) 1 + 1 p Γ ( λ ) Γ ( λ + 2 ) B ( λ 1 , λ 2 )$, such that (18) is valid when we replace $( 1 α ) 1 + 1 q ( 1 β ) 1 + 1 p Γ ( λ ) Γ ( λ + 2 ) B ( λ 1 , λ 2 )$ by $M$, then in particular, we have
By (19) and using the decreasingness property of series, it follows that
$I ˜ > M [ ∑ m = 1 ∞ m − α ε − 1 − ∑ m = 1 ∞ m − α ε − 1 O ( 1 m ε λ 2 ) ] 1 p ( 1 + ∑ n = 2 ∞ n − β ε − 1 ) 1 q > M ( ∫ 1 ∞ x − α ε − 1 d x − O ( 1 ) ) 1 p ( 1 + ∫ 1 ∞ y − β ε − 1 d y ) 1 q = M ε ( 1 α − ε O ( 1 ) ) 1 p ( ε + 1 β ) 1 q .$
We still find that
$I ˜ < 1 α ( λ 1 − ε p ) 1 β ( λ 2 − ε q ) ∑ n = 1 ∞ ∑ m = 1 ∞ 1 ( m α + n β ) λ + 2 [ m α − 1 ( m α ( λ 1 − ε p ) + | c 1 | + | O 1 ( m α ( λ 1 − ε p ) − 1 ) | ) ] × [ n β − 1 ( n β ( λ 2 − ε q ) + | c 2 | + | O 2 ( n β ( λ 2 − ε q ) − 1 ) | ) ] = 1 α ( λ 1 − ε p ) 1 β ( λ 2 − ε q ) × ∑ n = 1 ∞ ∑ m = 1 ∞ 1 ( m α + n β ) λ + 2 [ m α ( λ 1 − ε p + 1 ) − 1 + | c 1 | m α − 1 + | O 1 ( m α ( λ 1 − ε p + 1 ) − 2 ) | ] × [ n β ( λ 2 − ε q + 1 ) − 1 + | c 2 | n β − 1 + | O 2 ( n β ( λ 2 − ε q + 1 ) − 2 ) | ] = 1 α ( λ 1 − ε p ) 1 β ( λ 2 − ε q ) ( I 0 + I 1 ) ,$
where we indicate that $I 0 : = ∑ n = 1 ∞ [ n β ( λ 2 − ε q + 1 ) − ∑ m = 1 ∞ m α ( λ 1 − ε p + 1 ) − 1 ( m α + n β ) λ + 2 ] ,$ and
$I 1 : = ∑ n = 1 ∞ ∑ m = 1 ∞ 1 ( m α + n β ) λ + 2 [ ( | c 1 | m α − 1 + | O 1 ( m α ( λ 1 − ε p + 1 ) − 2 ) ) n β ( λ 2 − ε q + 1 ) − 1 | + ( | c 1 | m α − 1 + | O 1 ( m α ( λ 1 − ε p + 1 ) − 2 ) ) ( | c 2 | n β − 1 + | O 2 ( n β ( λ 2 − ε q + 1 ) − 2 ) | ) + m α ( λ 1 − ε p + 1 ) − 1 ( | c 2 | n β − 1 + | O 2 ( n β ( λ 2 − ε q + 1 ) − 2 ) | ) ] ≤ ∑ n = 1 ∞ n β ( λ 2 − ε q + 1 ) − 1 ( n β ) λ 2 + 1 ∑ m = 1 ∞ 1 ( m α ) λ 1 + 1 ( | c 1 | m α − 1 + | O 1 ( m α ( λ 1 − ε p + 1 ) − 2 ) | ) + ∑ n = 1 ∞ ( | c 2 | n β − 1 + | O 2 ( n β ( λ 2 − ε q + 1 ) − 2 ) | ) ( n β ) λ 2 + 1 ∑ m = 1 ∞ 1 ( m α ) λ 1 + 1 ( | c 1 | m α − 1 + | O 1 ( m α ( λ 1 − ε p + 1 ) − 2 | ) + ∑ n = 1 ∞ ( | c 2 | n β − 1 + | O 2 ( n β ( λ 2 − ε q + 1 ) − 2 ) | ) ( n β ) λ 2 + 1 ∑ m = 1 ∞ 1 ( m α ) λ 1 + 1 m α ( λ 1 − ε p + 1 ) − 1 ≤ M 1 < ∞ .$
By (14), for $s = λ + 2 ∈ ( 0 , 6 ] , s 1 = λ 1 + 1 − ε p$ $( ∈ ( 0 , 2 α ] ∩ ( 0 , λ + 2 ) ) ,$ we have
$I 0 = 1 α ∑ n = 1 ∞ [ n β ( λ 2 + 1 + ε p ) ∑ m = 1 ∞ α m α ( λ 1 + 1 − ε p ) − 1 ( m + n ) λ + 2 ] n − β ε − 1 = 1 α ∑ n = 1 ∞ ω λ + 2 ( λ 1 + 1 − ε p , n ) n − β ε − 1 < 1 α k λ + 2 ( λ 1 + 1 − ε p ) ( 1 + ∑ n = 2 ∞ n − β ε − 1 ) < 1 α k λ + 2 ( λ 1 + 1 − ε p ) ( 1 + ∫ 1 ∞ y − β ε − 1 d y ) = 1 ε 1 α β B ( λ 1 + 1 − ε p , λ 2 + 1 + ε p ) ( 1 + β ε ) .$
Based on the above results, we have
$1 α 2 ( λ 1 − ε p ) 1 β 2 ( λ 2 − ε q ) [ B ( λ 1 + 1 − ε p , λ 2 + 1 + ε p ) ( 1 + β ε ) + ε M 1 ] > ε I ˜ > M ( 1 α − ε O ( 1 ) ) 1 p ( ε + 1 β ) 1 q .$
Setting $ε → 0 +$ in the above inequality, in virtue of the continuity of the beta function, we find
$( 1 α ) 1 + 1 q ( 1 β ) 1 + 1 p Γ ( λ ) Γ ( λ + 2 ) B ( λ 1 , λ 2 ) = ( 1 α ) 1 + 1 q ( 1 β ) 1 + 1 p 1 λ 1 λ 2 B ( λ 1 + 1 , λ 2 + 1 ) ≥ M .$
Hence, $M = ( 1 α ) 1 + 1 q ( 1 β ) 1 + 1 p Γ ( λ ) Γ ( λ + 2 ) B ( λ 1 , λ 2 )$ is the best possible constant factor in (18).
The theorem is proved. □
Theorem 3.
Suppose that . If the constant factor $Γ ( λ ) Γ ( λ + 2 ) α β$$( 1 β k λ ( λ 2 ) ) 1 p ( 1 α k λ ( λ 1 ) ) 1 q$ in (17) is the best possible, then for
$λ − λ 1 − λ 2 ∈ ( − p λ 1 , p ( λ − λ 1 ) ) ∩ [ q ( 2 β − λ 2 ) , p ( 2 α − λ 1 ) ] ,$
we have $λ 1 + λ 2 = λ .$
Proof.
For , we find $λ ^ 1 + λ ^ 2 = λ .$ For $λ − λ 1 − λ 2 ∈ ( − p λ 1 , p ( λ − λ 1 ) ) ,$ we have $λ ^ 1 ∈ ( 0 , λ ) ,$ and then $λ ^ 2 = λ − λ ^ 1 ∈ ( 0 , λ ) ;$ for $λ − λ 1 − λ 2 ∈ [ q ( 2 β − λ 2 ) , p ( 2 α − λ 1 ) ] ,$ we still have $λ ^ 1 ≤ 2 α ,$ $λ ^ 2 ≤ 2 β .$ Then, for $λ 1 + λ 2 = λ$ in (17), substitution of we still have
$∑ m = 1 ∞ ∑ n = 1 ∞ A m B n ( m α + n β ) λ + 2 > ( 1 α ) 1 + 1 q ( 1 β ) 1 + 1 p Γ ( λ ) Γ ( λ + 2 ) B ( λ ^ 1 , λ ^ 2 ) × [ ∑ m = 1 ∞ ( 1 − O ( 1 m α λ ^ ^ 2 ) ) m p ( 1 − α λ ^ 1 ) − 1 a m p ] 1 p [ ∑ n = 1 ∞ n q ( 1 − β λ ^ 2 ) − 1 b n q ] 1 q .$
By using the reverse Hölder’s inequality (cf. [31]), we still obtain
$B ( λ ^ 1 , λ ^ 2 ) = k λ ( λ − λ 2 p + λ 1 q ) = ∫ 0 ∞ 1 ( 1 + u ) λ u λ − λ 2 p + λ 1 q − 1 d u = ∫ 0 ∞ 1 ( 1 + u ) λ ( u λ − λ 2 − 1 p ) ( u λ 1 − 1 q ) d u ≥ [ ∫ 0 ∞ 1 ( 1 + u ) λ u λ − λ 2 − 1 d u ] 1 p [ ∫ 0 ∞ 1 ( 1 + u ) λ u λ 1 − 1 d u ] 1 q = [ ∫ 0 ∞ 1 ( 1 + v ) λ v λ 2 − 1 d v ] 1 p [ ∫ 0 ∞ 1 ( 1 + u ) λ u λ 1 − 1 d u ] 1 q = ( k λ ( λ 2 ) ) 1 p ( k λ ( λ 1 ) ) 1 q .$
If $Γ ( λ ) Γ ( λ + 2 ) α β$$( 1 β k λ ( λ 2 ) ) 1 p ( 1 α k λ ( λ 1 ) ) 1 q$ in (17) is the best possible constant factor, then compare it with the constant factors in (17) and (20), and we have the following inequality:
$Γ ( λ ) Γ ( λ + 2 ) α β ( 1 β k λ ( λ 2 ) ) 1 p ( 1 α k λ ( λ 1 ) ) 1 q ≥ ( 1 α ) 1 + 1 q ( 1 β ) 1 + 1 p Γ ( λ ) Γ ( λ + 2 ) B ( λ ^ 1 , λ ^ 2 ) ( ∈ R + ) ,$
namely, $B ( λ ^ 1 , λ ^ 2 ) ≤$$( k λ ( λ 2 ) ) 1 p ( k λ ( λ 1 ) ) 1 q$. Then, by (21), we have
$B ( λ ^ 1 , λ ^ 21 ) = ( k λ ( λ 2 ) ) 1 p ( k λ ( λ 1 ) ) 1 q ,$
which follows that (21) protains the form of equality.
We observe that (21) protains the form of equality if and only if there exist $A$ and $B$, such that they are not both zero and (cf. [31]) $A u λ − λ 2 − 1 = B u λ 1 − 1 a . e .$ in $R +$. Assume that $A ≠ 0$. It follows that $u λ − λ 2 − λ 1 = B A a . e .$ in $R +$, and then $λ − λ 2 − λ 1 = 0$. Hence, we have $λ 1 + λ 2 = λ$.
The theorem is proved. □
Remark 1.
(i) For $α = β = 1 , λ ∈ ( 0 , 4 ] , λ 1 ∈ ( 0 , 1 ] ∩ ( 0 , λ ) , λ 2 ∈ ( 0 , 2 ) ∩ ( 0 , λ )$ in (18), we have the following reverse inequality with $1 λ ( λ + 1 ) B ( λ 1 , λ 2 )$ as the best possible constant factor:
$∑ m = 1 ∞ ∑ n = 1 ∞ A m B n ( m + n ) λ + 2 > 1 λ ( λ + 1 ) B ( λ 1 , λ 2 ) × [ ∑ m = 1 ∞ ( 1 − O ( 1 m λ 2 ) ) m p ( 1 − λ 1 ) − 1 a m p ] 1 p [ ∑ n = 1 ∞ n q ( 1 − λ 2 ) − 1 b n q ] 1 q .$
(ii) For $α = β = 1 2 , λ ∈ ( 0 , 4 ] , λ 1 ∈ ( 0 , 3 ] ∩ ( 0 , λ ) , λ 2 ∈ ( 0 , λ )$ in (18), we have the following reverse inequality with $8 λ ( λ + 1 ) B ( λ 1 , λ 2 )$ as the best possible constant factor:
$∑ m = 1 ∞ ∑ n = 1 ∞ A m B n ( m + n ) λ + 2 m n > 8 λ ( λ + 1 ) B ( λ 1 , λ 2 ) × [ ∑ m = 1 ∞ ( 1 − O ( 1 m λ 2 / 2 ) ) m p ( 1 − λ 1 2 ) − 1 a m p ] 1 p [ ∑ n = 1 ∞ n q ( 1 − λ 2 2 ) − 1 b n q ] 1 q .$

4. Conclusions

In this article, by using the techniques of real analysis, the way of weight coefficients and the idea of introduced parameters, applying the mid-value theorem. We estimate some lemmas and obtain a new reverse extended inequality (2) with two partial sums and multi-parameters in Theorem 1. We consider a few equivalent statements of the best possible constant factor related to several parameters in Theorems 2 and 3. We also deduce some inequalities for setting particular parameters in Remark 1. The theorems and lemmas in this paper provide a useful extensive account of this type of inequality. Further studies should be using the idea of this article to build some other kinds of Hilbert-type inequalities with partial sums and parameters.

Author Contributions

B.Y. carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. J.L. participated in the design of the study and performed the numerical analysis. All authors have read and agreed to the published version of the manuscript.

Funding

This work is supported by the National Natural Science Foundation (No. 61772140), the Key Construction Discipline Scientific Research Ability Promotion Project of Guangdong Province (No 2021ZDJS056) and the Guangzhou Base Applied Research Project (No. 20220101181-7).

Not applicable.

Not applicable.

Not applicable.

Acknowledgments

The authors thank the referees for their useful proposals to revise this paper.

Conflicts of Interest

The authors have no conflict of interest.

References

1. Hardy, G.H.; Littlewood, J.E.; Polya, G. Inequalities; Cambridge University Press: Cambridge, UK, 1934. [Google Scholar]
2. Krnić, M.; Pečarić, J. Extension of Hilbert’s inequality. J. Math. Anal. Appl. 2006, 324, 150–160. [Google Scholar] [CrossRef] [Green Version]
3. Yang, B.C. On a generalization of Hilbert double series theorem. J. Nanjing Univ. Math. Biquarterly 2001, 18, 145–152. [Google Scholar]
4. Adiyasuren, V.; Batbold, T.; Azar, L.E. A new discrete Hilbert-type inequality involving partial sums. J. Inequalities Appl. 2019, 127, 2019. [Google Scholar] [CrossRef]
5. Yang, B.C. The Norm of Operator and Hilbert-Type Inequalities; Science Press: Beijing, China, 2009. [Google Scholar]
6. Yang, B.C.; Debnath, L. On the extended Hardy–Hilbert’s inequality. J. Math. Anal. Appl. 2002, 272, 187–199. [Google Scholar] [CrossRef]
7. Krnić, M.; Pečarić, J. General Hilbert’s and Hardy’s inequalities. Math. Inequalities Appl. 2005, 8, 29–51. [Google Scholar] [CrossRef] [Green Version]
8. Perić, I.; Vuković, P. Multiple Hilbert’s type inequalities with a homogeneous kernel. Banach J. Math. Anal. 2011, 5, 33–43. [Google Scholar] [CrossRef]
9. Huang, Q.L. A new extension of Hardy-Hilbert-type inequality. J. Inequalities Appl. 2015, 2015, 397. [Google Scholar] [CrossRef]
10. He, B. A multiple Hilbert-type discrete inequality with a new kernel and best possible constant factor. J. Math. Anal. Appl. 2013, 431, 889–902. [Google Scholar] [CrossRef]
11. Xu, J.S. Hardy-Hilbert’s inequalities with two parameters. Adv. Math. 2007, 36, 63–76. [Google Scholar]
12. Xie, Z.T.; Zeng, Z.; Sun, Y.F. A new Hilbert-type inequality with the homogeneous kernel of degree-2. Adv. Appl. Math. Sci. 2013, 12, 391–401. [Google Scholar]
13. Zeng, Z.; Raja Rama Gandhi, K.; Xie, Z.T. A new Hilbert-type inequality with the homogeneous kernel of degree-2 and with the integral. Bull. Math. Sci. Appl. 2014, 3, 11–20. [Google Scholar]
14. Xin, D.M. A Hilbert-type integral inequality with the homogeneous kernel of zero degree. Math. Theory Appl. 2010, 30, 70–74. [Google Scholar]
15. Azar, L.E. The connection between Hilbert and Hardy inequalities. J. Inequalities Appl. 2013, 452, 2013. [Google Scholar] [CrossRef] [Green Version]
16. Adiyasuren, V.; Batbold, T.; Krnić, M. Hilbert–type inequalities involving differential operators, the best constants and applications. Math. Inequal. Appl. 2015, 18, 111–124. [Google Scholar] [CrossRef] [Green Version]
17. Gu, Z.H.; Yang, B.C. An extended Hardy-Hilbert’s inequality with parameters and applications. J. Math. Inequalities 2021, 15, 1375–1389. [Google Scholar] [CrossRef]
18. Hong, Y.; Wen, Y. A necessary and sufficient condition of that Hilbert type series inequality with homogeneous kernel has the best constant factor. Ann. Math. 2016, 37A, 329–336. [Google Scholar]
19. Hong, Y. On the structure character of Hilbert’s type integral inequality with homogeneous kernel and application. J. Jilin Univ. (Sci. Ed.) 2017, 55, 189–194. [Google Scholar]
20. Hong, Y.; Huang, Q.L.; Yang, B.C.; Liao, J.Q. The necessary and sufficient conditions for the existence of a kind of Hilbert-type multiple integral inequality with the non-homogeneous kernel and its applications. J. Inequalities Appl. 2017, 2017, 316. [Google Scholar] [CrossRef]
21. Chen, Q.; He, B.; Hong, Y.; Li, Z. Equivalent parameter conditions for the validity of half-discrete Hilbert-type multiple integral inequality with generalized homogeneous kernel. J. Funct. Spaces 2020, 2020, 7414861. [Google Scholar] [CrossRef]
22. He, B.; Hong, Y.; Chen, Q. The equivalent parameter conditions for constructing multiple integral half-discrete Hilbert-type inequalities with a class of non-homogeneous kernels and their applications. Open Math. 2021, 19, 400–411. [Google Scholar] [CrossRef]
23. Hong, Y.; Huang, Q.L.; Chen, Q. The parameter conditions for the existence of the Hilbert-type multiple integral inequality and its best constant factor. Ann. Funct. Anal. 2020, 12, 7. [Google Scholar] [CrossRef]
24. Hong, Y.; Chen, Q. Equivalent parameter conditions for the construction of Hilbert-type integral inequalities with a class of non-homogeneous kernels. J. S. China Norm. Univ. (Nat. Sci. Ed.) 2020, 52, 124–128. [Google Scholar]
25. Hong, Y.; Chen, Q. Research Progress and Applications of Hilbert-Type Series Inequalitie. J. Jilin Univ. (Sci. Ed.) 2021, 59, 1131–1140. [Google Scholar]
26. Hong, H.; Chen, Q.; Wu, C.Y. The best matching parameters for semi-discrete Hilbert-type inequality with quasi-homogeneous kernel. Math. Appl. 2021, 34, 779–785. [Google Scholar]
27. Hong, Y.; He, B. The optimal matching parameter of half-discrete Hilbert-type multiple integral inequalities with non-homogeneous kernels and applications. Chin. Quart. J. Math. 2021, 36, 252–262. [Google Scholar]
28. Luo, R.; Yang, B.C.; Huang, X.S. A reverse extended Hardy–Hilbert’s inequality with parameters. J. Inequalities Appl. 2023, 2023, 58, Springer Science and Business Media LLC. [Google Scholar] [CrossRef]
29. Nave, O. Modification of Semi-Analytical Method Applied System of ODE. Mod. Appl. Sci. 2020, 14, 75, Canadian Center of Science and Education. [Google Scholar] [CrossRef]
30. Yang, B.C.; Debnath, L. Discrete Hilbert-Type Inequalities; Bentham Science Publishers Ltd.: Dubai, United Arab Emirate, 2011. [Google Scholar]
31. Kuang, J.C. Applied Inequalities; Shangdong Science and Technology Press: Jinan, China, 2004. [Google Scholar]
 Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

Share and Cite

MDPI and ACS Style

Liao, J.; Yang, B. A New Reverse Extended Hardy–Hilbert’s Inequality with Two Partial Sums and Parameters. Axioms 2023, 12, 678. https://doi.org/10.3390/axioms12070678

AMA Style

Liao J, Yang B. A New Reverse Extended Hardy–Hilbert’s Inequality with Two Partial Sums and Parameters. Axioms. 2023; 12(7):678. https://doi.org/10.3390/axioms12070678

Chicago/Turabian Style

Liao, Jianquan, and Bicheng Yang. 2023. "A New Reverse Extended Hardy–Hilbert’s Inequality with Two Partial Sums and Parameters" Axioms 12, no. 7: 678. https://doi.org/10.3390/axioms12070678

Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.