Previous Article in Journal
Orthogonal Families of Bicircular Quartics, Quadratic Differentials, and Edwards Normal Form

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

An Extended Hilbert-Type Inequality with Two Internal Variables Involving One Partial Sums

by
Aizhen Wang
and
Bicheng Yang
*
School of Mathematics, Guangdong University of Education, Guangzhou 510303, China
*
Author to whom correspondence should be addressed.
Axioms 2023, 12(9), 871; https://doi.org/10.3390/axioms12090871
Submission received: 26 July 2023 / Revised: 17 August 2023 / Accepted: 22 August 2023 / Published: 10 September 2023

Abstract

:
By the use of the techniques of analysis and some useful formulas, we give a new extension of Hilbert-type inequality with two internal variables involving one partial sums, which is a refinement of a published inequality. We provide a few equivalent conditions of the best possible constant related to multi parameters. We obtain the equivalent inequalities, the operator expressions as well as a few inequalities with the particular parameters as applications.
MSC:
26D15; 47A05

1. Introduction

Assuming that and $0 < ∑ n = 1 ∞ b n q < ∞ ,$ we have the well known Hardy-Hilbert’s inequality as follows (ref. [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 the best possible.
In 2006, by putting Krnic et al. [2] gave a generalization of Inequality (1) as follows:
with the best possible constant factor  ($B ( u , v ) : = ∫ 0 ∞ t u − 1 ( 1 + t ) u + v d t ( u , v > 0 )$ is the Beta function). For $p = q = 2 ,$ $λ 1 = λ 2 = λ 2$ in Inequality (2), it deduces to the inequality in Yang’s paper [3].
Inequalities (1) and (2) with their integral analogues played an important role in analysis and its applications (ref. [4,5,6,7,8,9,10,11,12,13,14]). By using the weight functions in 2016, Hong et al. [15] obtained a few equivalent conditions of the extension of Inequality (1) with the best constant factor related to multi parameters. Some further results were provided by [16,17,18,19,20].
In 2019, by using Inequality (2), Adiyasuren et. al. [21] gave an extension of Inequality (2) involving partial sums: if $λ i ∈ ( 0 , 1 ] ∩ ( 0 , λ )$ $( λ ∈ ( 0 , 2 ] ;$ $i = 1 , 2 ) ,$ $λ 1 + λ 2 = λ ,$  then it follows that
with the best constant , involving two partial sums $A m : = ∑ i = 1 m a i$ and $B n : = ∑ k = 1 n b k$ $( m , n ∈ { 1 , 2 , ⋯ } )$, $A m = o ( e t m )$, $B n = o ( e t n )$  such that
In 2021, Liao et al. [22] gave an extension of Inequality (3) with the kernel as $1 ( m α + n β ) λ ( α , β ∈ ( 0 , 1 ] )$ involving one partial sums. But the constant factor in this inequality seems not to be the best possible unless at $α = β = 1$. In 2023, by using the mid-value theorem, Hong et al. [23,24] gave a new inequality as well as the reverse with the same kernel as [22] involving two partial sums, and proved that the constant factor is the best possible in some conditions.
In this article, by applying the methods in [23], and using the techniques of introduced parameters, some useful formulas and the mid-value theorem, we give a new extension of Hardy-Hilbert’s inequality with the internal variables involving one partial sums, which is a refinement of the inequality in [22]. We also provide a few equivalent statements of the best possible constant factor related to several parameters. As applications, we obtain the equivalent inequalities, the operator expressions as well as a few inequalities with the particular parameters. The lemmas and theorems provide an extensive account of this type of inequalities.

2. Some Lemmas

In what follows of this article, we assume that $p > 1 ( q > 1 ) , 1 p + 1 q = 1 ,$ $λ ∈ ( 0 , 5 ] ,$ $α , β ∈ ( 0 , 1 ] ,$  $λ ^ 1 : = λ − λ 2 p + λ 1 q ,$ $λ ^ 2 : = λ − λ 1 q + λ 2 p .$ We still suppose that  $A m : = ∑ j = 1 m a j ,$ $( m , n ∈ N$ $= { 1 , 2 , ⋯ } ) ,$ such that $A m = o ( e t m α )$ $( t > 0 ; m → ∞ ) ,$ and
For showing the main results, we give the following key lemma by using the mid-value theorem:
Lemma 1.
If  $t > 0$then the following inequality holds:
$∑ m = 1 ∞ e − t m α a m ≤ t α ∑ m = 1 ∞ e − t m α m α − 1 A m .$
Proof.
For $A m e − t m α = o ( 1 ) ( m → ∞ )$, in view of Abel’s summation formula, we obtain
$∑ m = 1 ∞ e − t m α a m = lim m → ∞ A m e − t m α + ∑ m = 1 ∞ A m [ e − t m α − e − t ( m + 1 ) α ] = ∑ m = 1 ∞ A m [ e − t m α − e − t ( m + 1 ) α ] .$
We set function $g ( x ) = e − t x α , x ∈ [ m , m + 1 ] .$ Then we find that $g ′ ( x ) = − t α x α − 1 e − t x α ,$ and for $x ∈ [ m , m + 1 ] ,$ $α ∈ ( 0 , 1 ] ,$ the function $h ( x ) : = x α − 1 e − t x α$ is decreasing. In view of the mid-value theorem, it follows that
$∑ m = 1 ∞ e − t m α a m = − ∑ m = 1 ∞ A m ( g ( m + 1 ) − g ( m ) ) = − ∑ m = 1 ∞ A m g ′ ( m + θ ) = t α ∑ m = 1 ∞ h ( m + θ ) A m ≤ t α ∑ m = 1 ∞ h ( m ) A m = t α ∑ m = 1 ∞ m α − 1 e − t m α A m ( θ ∈ ( 0 , 1 ) ) ,$
namely, Equation (6) follows.
The lemma is proved. □
For showing the inequalities in Lemma 3, we need the following lemma:
Lemma 2.
(ref. [4], (2.2.3)). (i) Assuming that $( − 1 ) k d k d t k g ( t ) > 0 ,$ $t ∈ [ m , ∞ )$, $g ( k ) ( ∞ ) = 0$ $( k = 0 , 1 , 2 , 3 )$,  are Bernoulli functions and Bernoulli numbers of j-order, it follows that
$∫ m ∞ P 2 q − 1 ( t ) g ( t ) d t = − ε q B 2 q 2 q g ( m ) ( 0 < ε q < 1 ; q = 1 , 2 , ⋯ ) .$
In particular, for  $q = 1 ,$  $B 2 = 1 6$, we have
$− 1 12 g ( m ) < ∫ m ∞ P 1 ( t ) g ( t ) d t < 0 ;$
for  $q = 2 ,$  $B 4 = − 1 30$, we have
$0 < ∫ m ∞ P 3 ( t ) g ( t ) d t < 1 120 g ( m ) .$
(ii) (ref. [4], (2.3.2)) If $f ( t ) ( > 0 ) ∈ C 3 [ m , ∞ ) , f ( k ) ( ∞ ) = 0 ( k = 0 , 1 , 2 , 3 )$, then the following Euler-Maclaurin summation formula holds:
$∑ k = m ∞ f ( k ) = ∫ m ∞ f ( t ) d t + 1 2 f ( m ) + ∫ m ∞ P 1 ( t ) f ′ ( t ) d t ,$
$∫ m ∞ P 1 ( t ) f ′ ( t ) d t = − 1 12 f ′ ( m ) + 1 6 ∫ m ∞ P 3 ( t ) f ‴ ( t ) d t .$
Lemma 3.
For $s ∈ ( 0 , 6 ] ,$ $s 2 ∈ ( 0 , 2 β ] ∩ ( 0 , s ) ,$ $k s ( s 2 ) : = B ( s 2 , s − s 2 )$, indicate the weight coefficient as follows:
$ϖ s ( s 2 , m ) : = m α ( s − s 2 ) ∑ n = 1 ∞ β n β s 2 − 1 ( m α + n β ) s ( m ∈ N ) .$
We have
$0 < k s ( s 2 ) ( 1 − O ( 1 m α s 2 ) ) < ϖ s ( s 2 , m ) < k s ( s 2 ) ( m ∈ N ) .$
where, we indicate $O ( 1 m α s 2 ) : = 1 k s ( s 2 ) ∫ 0 1 m α u s 2 − 1 ( 1 + u ) s d u > 0 .$
Proof.
For fixed $m ∈ N$, we define $g ( m , t )$ as follows:
$g ( m , t ) : = β t β s 2 − 1 ( m α + t β ) s ( t > 0 ) .$
In view of Inequality (10), it follows that
$∑ n = 1 ∞ g ( m , n ) = ∫ 1 ∞ g ( m , t ) d t + 1 2 g ( m , 1 ) + ∫ 1 ∞ P ( t ) 1 g ′ ( m , t ) d t = ∫ 0 ∞ g ( m , t ) d t − h ( m ) , h ( m ) : = ∫ 0 1 g ( m , t ) d t − 1 2 g ( m , 1 ) − ∫ 1 ∞ P 1 ( t ) g ′ ( m , t ) d t .$
We obtain that $− 1 2 g ( m , 1 ) = − β 2 ( m α + 1 ) s$ and find
$∫ 0 1 g ( m , t ) d t = β ∫ 0 1 t β s 2 − 1 ( m α + t β ) s d t = u = t β ∫ 0 1 u s 2 − 1 ( m α + u ) s d u = 1 s 2 ∫ 0 1 d u s 2 ( m α + u ) s = 1 s 2 u s 2 ( m α + u ) s | 0 1 + s s 2 ∫ 0 1 u s 2 ( m α + u ) s + 1 d u = 1 s 2 1 ( m α + 1 ) s + s s 2 ( s 2 + 1 ) ∫ 0 1 d u s 2 + 1 ( m α + u ) s + 1 > 1 s 2 1 ( m α + 1 ) s + s s 2 ( s 2 + 1 ) [ u s 2 + 1 ( m α + u ) s + 1 ] 0 1 + s ( s + 1 ) s 2 ( s 2 + 1 ) ( m α + 1 ) s + 2 ∫ 0 1 u s 2 + 1 d u = 1 s 2 1 ( m α + 1 ) s + λ s 2 ( s 2 + 1 ) 1 ( m α + 1 ) s + 1 + s ( s + 1 ) s 2 ( s 2 + 1 ) ( s 2 + 2 ) 1 ( m α + 1 ) s + 2 , − g ′ ( m , t ) = − β ( β s 2 − 1 ) t β s 2 − 2 ( m α + t β ) s + β 2 s t β + β s 2 − 2 ( m α + t β ) s + 1 = − β ( β s 2 − 1 ) t β s 2 − 2 ( m α + t β ) s + β 2 s ( m α + t β − m α ) t β s 2 − 2 ( m α + t β ) s + 1 = β ( β s − β s 2 + 1 ) t β s 2 − 2 ( m α + t β ) s − β 2 s m α t β s 2 − 2 ( m α + t β ) s + 1 .$
For , it follows that
$( − 1 ) i d i d t i [ t β s 2 − 2 ( m α + t β ) s ] > 0 , ( − 1 ) i d i d t i [ t β s 2 − 2 ( m α + t β ) s + 1 ] > 0 ( i = 0 , 1 , 2 , 3 ) .$
By Inequalities (8)–(11), we obtain
$β ( β s − β s 2 + 1 ) ∫ 1 ∞ P 1 ( t ) t β s 2 − 2 ( m α + t β ) s d t > − β ( β s − β s 2 + 1 ) 12 ( m α + 1 ) s , − β 2 m α s ∫ 1 ∞ P 1 ( t ) t β s 2 − 2 ( m α + t β ) s + 1 d t = β 2 m α s 12 ( m α + 1 ) s + 1 − β 2 m α s 6 ∫ 1 ∞ P 3 ( t ) [ t β s 2 − 2 ( m α + t β ) s + 1 ] ″ d t > β 2 m α s 12 ( m α + 1 ) s + 1 − β 2 m α s 720 [ t β s 2 − 2 ( m α + t β ) s + 1 ] t = 1 ″ > β 2 ( m α + 1 − 1 ) s 12 ( m α + 1 ) s + 1 − β 2 ( m α + 1 ) s 720 [ ( s + 1 ) ( s + 2 ) β 2 ( m α + 1 ) s + 3 + β ( s + 1 ) ( 5 − β − 2 β s 2 ) ( m α + 1 ) s + 2 + ( 2 − β s 2 ) ( 3 − β s 2 ) ( m α + 1 ) s + 1 ] = β 2 s 12 ( m α + 1 ) s − β 2 s 12 ( m α + 1 ) s + 1 − β 2 s 720 [ ( s + 1 ) ( s + 2 ) β 2 ( m α + 1 ) s + 2 + β ( s + 1 ) ( 5 − β − 2 β s 2 ) ( m α + 1 ) s + 1 + ( 2 − β s 2 ) ( 3 − β s 2 ) ( m α + 1 ) s ] .$
Then it follows that $h ( m ) > 1 ( m α + 1 ) s h 1 + λ ( m α + 1 ) s + 1 h 2 + s ( s + 1 ) ( m α + 1 ) s + 2 h 3 ,$ where, we set
$h 1 : = 1 s 2 − β 2 − β − β 2 s 2 12 − β 2 s ( 2 − β s 2 ) ( 3 − β s 2 ) 720 ,$
$h 2 : = 1 s 2 ( s 2 + 1 ) − β 2 12 − β 3 ( s + 1 ) ( 5 − β − 2 β s 2 ) 720 ,$
and $h 3 : = 1 s 2 ( s 2 + 1 ) ( s 2 + 2 ) − β 4 ( s + 2 ) 720 .$ We 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, we indicate the function $g ( σ ) ( σ ∈ ( 0 , 2 β ] )$ as follows:
$g ( σ ) : = 720 − ( 420 β + 6 s β 2 ) σ + ( 60 β 2 + 5 s β 3 ) σ 2 − s β 4 σ 3 .$
We obtain that for $β ∈ ( 0 , 1 ] , s ∈ ( 0 , 6 ]$,
$g ′ ( σ ) = − ( 420 β + 6 s β 2 ) + 2 ( 60 β 2 + 5 s β 3 ) σ − 3 β 4 σ 2 ≤ − 420 β − 6 s β 2 + 2 ( 60 β 2 + 5 s β 3 ) 2 β = ( 14 s β − 180 ) β < 0 ,$
and then it follows that $h 1 ≥ g ( s 2 ) 720 s 2 ≥ g ( 2 / β ) 720 s 2 = 1 6 s 2 > 0 .$ We find that for $s 2 ∈ ( 0 , 2 β ]$,
$h 2 > β 2 6 − β 2 12 − 5 ( s + 1 ) β 2 720 = ( 1 12 − s + 1 140 ) β 2 > 0 ,$
and $h 3 ≥ ( 1 24 − s + 2 720 ) β 3 > 0 ( 0 < s ≤ 6 ) .$
Therefore, we have $h ( m ) > 0 ,$ and then setting $u = m − α t β ,$ it follows that
$ϖ s ( s 2 , m ) = m α ( s − s 2 ) ∑ n = 1 ∞ g ( m , n ) < m α ( s − s 2 ) ∫ 0 ∞ g ( m , t ) d t = β m α ( s − s 2 ) ∫ 0 ∞ t β s 2 − 1 d t ( m α + t β ) s = ∫ 0 ∞ u s 2 − 1 d u ( 1 + u ) s = B ( s 2 , s − s 2 ) = k s ( s 2 ) .$
By Inequality (10), it follows that
$∑ n = 1 ∞ g ( m , n ) = ∫ 1 ∞ g ( m , t ) d t + 1 2 g ( m , 1 ) + ∫ 1 ∞ P 1 ( t ) g ′ ( m , t ) d t = ∫ 1 ∞ g ( m , t ) d t + H ( m ) , H ( m ) : = 1 2 g ( m , 1 ) + ∫ 1 ∞ P 1 ( t ) g ′ ( m , t ) d t .$
We have fond that $1 2 g ( m , 1 ) = β 2 ( m α + 1 ) s$, 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 $s 2 ∈ ( 0 , 2 β ] ∩ ( 0 , s ) , 0 < s ≤ 6 ,$ in view of Inequality (7), it follows that
$− β ( β s − β s 2 + 1 ) ∫ 1 ∞ P 1 ( t ) t β s 2 − 2 ( m α + t β ) s d t > 0 ,$
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, we have
$H ( m ) > β 2 ( m α + 1 ) s − β 2 s 12 ( m α + 1 ) s ≥ β 2 ( m α + 1 ) s − 6 β 12 ( m α + 1 ) s = 0 ,$
and then it follows that
$ϖ s ( s 2 , m ) = m α ( s − s 2 ) ∑ n = 1 ∞ g ( m , n ) > m α ( s − s 2 ) ∫ 1 ∞ g ( m , t ) d t = m α ( s − s 2 ) ∫ 0 ∞ g ( m , t ) d t − m α ( s − s 2 ) ∫ 0 1 g ( m , t ) d t = k s ( s 2 ) [ 1 − 1 k s ( s 2 ) ∫ 0 1 m α u s 2 − 1 ( 1 + u ) s d u ] > 0 ,$
where, $O ( 1 m α s 2 ) = 1 k s ( s 2 ) ∫ 0 1 m α u s 2 − 1 ( 1 + u ) s d u ,$ satisfying
$0 < ∫ 0 1 m α u s 2 − 1 ( 1 + u ) s d u < ∫ 0 1 m α u s 2 − 1 d u = 1 s 2 m α s 2 .$
Hence, Inequalities (13) follow.
This proves the lemma. □
By applying the above lemma, we obtain an extended Hardy-Hilbert’s inequality as follows.
Lemma 4.
The following inequality is valid:
$I λ + 1 : = ∑ n = 1 ∞ ∑ m = 1 ∞ m α − 1 ( m α + n β ) λ + 1 A m b n < ( 1 β k λ + 1 ( λ 2 ) ) 1 p ( 1 α k λ + 1 ( λ 1 + 1 ) ) 1 q × ( ∑ m = 1 ∞ m − p α λ ^ 1 − 1 A m p ) 1 p ( ∑ n = 1 ∞ n q ( 1 − β λ ^ 2 ) − 1 b n q ) 1 q .$
Proof.
By the symmetry, for $s 1 ∈ ( 0 , 2 α ] ∩ ( 0 , s ) , s ∈ ( 0 , 6 ] ,$ the inequalities for the next weight coefficient is obtained as follows
$0 < k s ( s 1 ) ( 1 − O ( 1 n β s 1 ) ) < ω s ( s 1 , n ) : = n β ( s − s 1 ) ∑ m = 1 ∞ α m α s 1 − 1 ( m α + n β ) s < k s ( s 1 ) : = B ( s 1 , s − s 1 ) ( n ∈ N ) ,$
where, we set $O ( 1 n β s 1 ) : = 1 k s ( s 1 ) ∫ 0 1 n β u s 1 − 1 ( 1 + u ) s d u > 0 ( n ∈ N ) .$
By using H$o ¨$ lder’s inequality (ref. [25]), we find
$I λ + 1 = ∑ n = 1 ∞ ∑ m = 1 ∞ 1 ( m α + n β ) λ + 2 [ m α ( − λ ) 1 / q ( β n β − 1 ) 1 / p m α − 1 n β ( 1 − λ 2 ) / p ( α m α − 1 ) 1 / q A m ] [ n β ( 1 − λ 2 ) / p ( α m α − 1 ) 1 / q m α ( − λ ) 1 / q ( β n β − 1 ) 1 / p b n ] ≤ [ ∑ m = 1 ∞ ∑ n = 1 ∞ β m p ( α − 1 ) ( m α + n β ) λ + 1 m α ( − λ ) 1 ( p − 1 ) n β − 1 A m p n β ( 1 − λ 2 ) ( α m α − 1 ) p − 1 ] 1 p × [ ∑ n = 1 ∞ ∑ m = 1 ∞ α ( m α + n β ) 1 n β ( 1 − λ 2 ) ( q − 1 ) m α − 1 b n q m α ( − λ ) 1 ( β n β − 1 ) q − 1 ] 1 q = ( 1 β ∑ m = 1 ∞ ϖ λ + 1 ( λ 2 , m ) m − p α λ ^ 1 − 1 A m p ) 1 p × ( 1 α ∑ n = 1 ∞ ω λ + 1 ( λ + 1 1 , n ) n q ( 1 − β λ ^ 2 ) − 1 b n q ) 1 q .$
Then by Inequalities (13) and (15) (for ), in view of Inequality (6), we have Inequality (14).
This proves the lemma. □

3. Main Results

In view of Lemma 1 and Lemma 3, we obtain the following main results:
Theorem 1.
We have an inequality as follows:
$I : = ∑ m = 1 ∞ ∑ n = 1 ∞ a m b n ( m α + n β ) λ < ( α β ) 1 p λ ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q × ( ∑ m = 1 ∞ m − p α λ ^ 1 − 1 A m p ) 1 p [ ∑ n = 1 ∞ n q ( 1 − β λ ^ 2 ) − 1 b n q ] 1 q .$
In particular, for $λ 1 + λ 2 = λ$, we have
and the following inequality:
$∑ m = 1 ∞ ∑ n = 1 ∞ a m b n ( m α + n β ) λ < ( α β ) 1 p λ B ( λ 1 + 1 , λ 2 ) × ( ∑ m = 1 ∞ m − p α λ 1 − 1 A m p ) 1 p [ ∑ n = 1 ∞ n q ( 1 − β λ 2 ) − 1 b n q ] 1 q .$
Proof.
In view of the following equality relate to the Gamma function:
$1 ( m α + n β ) λ = 1 Γ ( λ ) ∫ 0 ∞ t λ − 1 e − ( m α + n β ) t d t ,$
and Inequality (6), we obtain
$I = 1 Γ ( λ ) ∑ m = 1 ∞ ∑ n = 1 ∞ a m b n ∫ 0 ∞ t λ − 1 e − ( m α + n β ) t d t = 1 Γ ( λ ) ∫ 0 ∞ t λ − 1 ( ∑ m = 1 ∞ e − m α t a m ) ( ∑ n = 1 ∞ e − n β t b n ) d t ≤ 1 Γ ( λ ) ∫ 0 ∞ t λ − 1 ( t α ∑ m = 1 ∞ e − m α t m α − 1 A m ) ( ∑ n = 1 ∞ e − n β t b n ) d t = α Γ ( λ ) ∑ m = 1 ∞ ∑ n = 1 ∞ m α − 1 A m b n ∫ 0 ∞ t λ e − ( m α + n β ) t d t = α Γ ( λ + 1 ) Γ ( λ ) ∑ m = 1 ∞ ∑ n = 1 ∞ m α − 1 ( m α + n β ) λ + 1 A m b n .$
Hence by Inequality (14), it follows that Inequality (16) is valid.
This proves the theorem. □
In the following two theorems, we give a few equivalent conditions of the best value related to multi parameters in Inequality (16).
Theorem 2.
Suppose that  If $λ 1 + λ 2 = λ ,$ then the constant factor $( α β ) 1 p λ$ $( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q$ is the best possible in Inequality (16).
Proof.
Now, we prove that
$( α β ) 1 p λ B ( λ 1 + 1 , λ 2 ) ( = ( α β ) 1 p λ 1 B ( λ 1 , λ 2 ) )$
is the best possible constant in Inequality (18).
For any $ε ∈ ( 0 , min { p λ 1 , q λ 2 } )$, we put
Since $0 < λ 1 − ε p ≤ 2 α − 1 , 0 < α ( λ 1 − ε p ) ≤ 2 − α < 2 ,$ in view of (2.2.24) (ref. [5]), we find
($c 1$ is a constant). We observe that $A ˜ m = o ( e t m α ) ( t > 0 ; m → ∞ )$.
If there exists a positive constant , such that Inequality (18) is value as we replace by $M$. Then, substitution of $a m = a ˜ m ,$ $b n = b ˜ n$ and $A m = A ˜ m$ in Inequality (18), we find
$I ˜ : = ∑ n = 1 ∞ ∑ m = 1 ∞ a ˜ m b ˜ n ( m α + n β ) λ < M ( ∑ m = 1 ∞ m − p α λ 1 − 1 A ˜ m p ) 1 p [ ∑ n = 1 ∞ n q ( 1 − β λ 2 ) − 1 b ˜ n q ] 1 q .$
Setting $a ( x ) → 0 ( x → ∞ ) ,$ we find
$lim x → ∞ ( 1 + a ( x ) ) p − 1 a ( x ) = lim x → ∞ p ( 1 + a ( x ) ) p − 1 a ′ ( x ) a ′ ( x ) = lim x → ∞ p ( 1 + a ( x ) ) p − 1 = p ,$
namely, $( 1 + a ( x ) ) p = 1 + O ( a ( x ) ) ( x → ∞ ) .$ Then we obtain
Hence, we have
$∑ m = 1 ∞ m − p α λ 1 − 1 A ˜ m p ≤ [ 1 α ( λ 1 − ε p ) ] p ∑ m = 1 ∞ m − α ε − 1 ( 1 + | c 1 | m − α ( λ 1 − ε p ) + | O ( m − 1 ) 1 | ) p = [ 1 α ( λ 1 − ε p ) ] p ∑ m = 1 ∞ m − α ε − 1 ( 1 + O ( m − α ( λ 1 − ε p ) ) + O ˜ ( m − 1 ) ) = [ 1 α ( λ 1 − ε p ) ] p [ ∑ m = 1 ∞ m − α ε − 1 + ∑ m = 1 ∞ ( O ( m − α ( λ 1 + ε q ) − 1 + O ˜ ( m − α ε − 2 ) ) ] = [ 1 α ( λ 1 − ε p ) ] p ( ∑ m = 2 ∞ m − α ε − 1 + O 1 ( 1 ) ) < [ 1 α ( λ 1 − ε p ) ] p ( ∫ 1 ∞ x − α ε − 1 d x + O 1 ( 1 ) ) = [ 1 α ( λ 1 − ε p ) ] p ( 1 α ε + O 1 ( 1 ) ) .$
We still find that
$∑ n = 1 ∞ n q ( 1 − β λ 2 ) − 1 b ˜ n q = 1 + ∑ n = 2 ∞ n − β ε − 1 < 1 + ∫ 1 ∞ y − β ε − 1 d y = 1 + 1 β ε .$
Then, we obtain $I ˜ < M ε 1 α ( λ 1 − ε p ) ( 1 α + ε O 1 ( 1 ) ) 1 p ( 1 β + ε ) 1 q .$
In view of Inequality (15) (for ), we find
$I ˜ = ∑ n = 1 ∞ ∑ m = 1 ∞ m α ( λ 1 − ε p ) − 1 ( m α + n β ) λ n β ( λ 2 − ε q ) − 1 = 1 α ∑ n = 1 ∞ n − β ε − 1 [ n β ( λ 2 + ε q ) ∑ m = 1 ∞ α m α ( λ 1 − ε p ) − 1 ( m α + n β ) λ ] ≥ 1 α k λ ( λ 1 − ε p ) ∑ n = 1 ∞ n − β ε − 1 ( 1 − O ( 1 n β ( λ 1 − ε p ) ) ) = 1 α k λ ( λ 1 − ε p ) ( ∑ n = 1 ∞ n − β ε − 1 − ∑ n = 1 ∞ O ( 1 n β ( λ 1 + ε q ) + 1 ) ) > 1 α k λ ( λ 1 − ε p ) ( ∫ 1 ∞ y − β ε − 1 d y − O 2 ( 1 ) ) = 1 α ε B ( λ 1 − ε p , λ 2 + ε p ) ( 1 β − ε O 2 ( 1 ) ) .$
Based on the above results, we have
$1 α B ( λ 1 − ε p , λ 2 + ε p ) ( 1 β − ε O 2 ( 1 ) ) < ε I ˜ < M 1 α ( λ 1 − ε p ) ( 1 α + ε O 1 ( 1 ) ) 1 p ( 1 β + ε O 2 ( 1 ) ) 1 q .$
Setting $ε → 0 +$, based on the continuity of the Beta function, we find
$1 α β B ( λ 1 , λ 2 ) ≤ M 1 α λ 1 ( 1 α ) 1 p ( 1 β ) 1 q ,$
namely,
Hence, the constant factor in Inequality (18) is the best possible.
The theorem is proved. □
Theorem 3.
Assume that  If the constant factor $( α β ) 1 p λ$ $( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q$ in Inequality (16) is the best value, then for
$λ − λ 1 − λ 2 ≤ min { p ( 2 α − 1 − λ 1 ) , q ( 2 β − λ 2 ) } ,$
we have $λ 1 + λ 2 = λ .$
Proof.
For , we find $λ ^ 1 + λ ^ 2 = λ ,$ and For $λ − λ 1 − λ 2 ≤ p ( 2 α − 1 − λ 1 ) ,$ we have $λ ^ 1 ≤ 2 α − 1$; for $λ − λ 1 − λ 2$ $≤ q ( 2 β − λ 2 ) ,$ we find $λ ^ 2 ≤ 2 β$. For $λ i = λ ^ i ( i = 1 , 2 )$ in Inequality (18), we obtain
$∑ m = 1 ∞ ∑ n = 1 ∞ a m b n ( m α + n β ) λ < ( α β ) 1 p λ B ( λ ^ 1 + 1 , λ ^ 2 ) × ( ∑ m = 1 ∞ m − p α λ ^ 1 − 1 A m p ) 1 p [ ∑ n = 1 ∞ n q ( 1 − β λ ^ 2 ) − 1 b n q ] 1 q .$
In view of H$o ¨$ lder’s inequality (ref. [25]), we obtain
$B ( λ ^ 1 + 1 , λ ^ 2 ) = k λ + 1 ( λ − λ 2 p + λ 1 q + 1 ) = ∫ 0 ∞ 1 ( 1 + u ) λ + 1 u λ − λ 2 p + λ 1 q d u = ∫ 0 ∞ 1 ( 1 + u ) λ + 1 ( u λ − λ 2 p ) ( u λ 1 q ) d u ≥ [ ∫ 0 ∞ 1 ( 1 + u ) λ + 1 u λ − λ 2 d u ] 1 p [ ∫ 0 ∞ 1 ( 1 + u ) λ + 1 u λ 1 d u ] 1 q = [ ∫ 0 ∞ 1 ( 1 + v ) λ + 1 v λ 2 − 1 d v ] 1 p [ ∫ 0 ∞ 1 ( 1 + u ) λ + 1 u λ 1 d u ] 1 q = ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q .$
If constant factor $( α β ) 1 p λ$ $( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q$ in Inequality (16) is the best possible, then by Inequalities (16) and (20), it follows that
$( α β ) 1 p λ ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q ≤ ( α β ) 1 p λ B ( λ ^ 1 + 1 , λ ^ 2 ) ( ∈ R + ) ,$
which means that
$B ( λ ^ 1 + 1 , λ ^ 2 ) ≥ ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q .$
Then by Inequality (21), we have
$B ( λ ^ 1 + 1 , λ ^ 2 ) = ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q ;$
Hence, Inequality (21) keeps the form of equality.
Since Inequality (21) keeps the form of equality, there exist constants $A$ and $B$, such that they are not both zero and (ref.) $A u λ − λ 2 = B u λ 1 a . e .$ in $R + .$ Assuming that $A ≠ 0$, we have $u λ − λ 2 − λ 1 = B A a . e .$ in $R + ,$ and $λ − λ 2 − λ 1 = 0$. Hence, we have $λ 1 + λ 2 = λ$.
This proves the theorem. □
Remark 1.
Since the constant in Inequality (18) is the best value, inequality Inequality (18) is a refinement of the inequalities in [16].

4. Equivalent Inequalities and Operator Expressions

Theorem 4.
The following inequality is valid equivalent to Inequality (16):
$J : = { ∑ n = 1 ∞ n p β λ ^ 2 − 1 [ ∑ m = 1 ∞ a m ( m α + n β ) λ ] p } 1 p < ( α β ) 1 p λ ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q ( ∑ m = 1 ∞ m − p α λ ^ 1 − 1 A m p ) 1 p .$
For $λ 1 + λ 2 = λ$, the following inequality is valid, which is equivalent to Inequality (18):
${ ∑ n = 1 ∞ n p β λ 2 − 1 [ ∑ m = 1 ∞ a m ( m α + n β ) λ ] p } 1 p < ( α β ) 1 p λ B ( λ 1 + 1 , λ 2 ) ( ∑ m = 1 ∞ m − p α λ 1 − 1 A m p ) 1 p .$
Proof.
Assuming that Inequality (23) is value, by H$o ¨$ lder’s inequality, we find
$I = ∑ n = 1 ∞ [ n − 1 p + β λ ^ 2 ∑ m = 1 ∞ a m ( m α + n β ) λ ] ( n 1 p − β λ ^ 2 b n ) ≤ J [ ∑ n = 1 ∞ n q ( 1 − β λ ^ 2 ) − 1 b n q ] 1 q .$
By Inequalities (16) and (23) follows. Assuming that Inequality (16) is valid, setting
$b n : = n p β λ ^ 2 − 1 [ ∑ m = 1 ∞ a m ( m α + n β ) λ ] p − 1 , n ∈ N ,$
it follows that
$∑ n = 1 ∞ n q ( 1 − β λ ^ 2 ) − 1 b n q = J p = I .$
If $J = ∞$, then it is impossible that makes Inequality (23) value, namely, $J < ∞$; if $J = 0$, then Inequality (23) is naturally valid. Assume that $0 < J < ∞ .$ In view of Inequality (16), we find
$J p = I < ( α β ) 1 p λ ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q ( ∑ m = 1 ∞ m − p α λ ^ 1 − 1 A m p ) 1 p J p − 1 ,$
$J < ( α β ) 1 p λ ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q ( ∑ m = 1 ∞ m − p α λ ^ 1 − 1 A m p ) 1 p .$
Hence, we have Inequality (23), which is equivalent to Inequality (16).
This proves the theorem. □
By the equivalency of Inequalities (16) and (23), we have
Theorem 5.
Assume that
If $λ 1 + λ 2 = λ$, then $( α β ) 1 p λ$ $( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q$ is the best possible constant factor in Inequality (23). If the same constant factor in Inequality (23) is the best possible, then for $λ − λ 1 − λ 2 ≤ min { p ( 2 α − 1 − λ 1 ) , q ( 2 β − λ 2 ) }$, we have $λ 1 + λ 2 = λ$.
Proof.
We prove that the following constant factor
$( α β ) 1 p λ B ( λ 1 + 1 , λ 21 ) ( = ( α β ) 1 p λ 1 B ( λ 1 , λ 21 ) )$
in Inequality (24) is the best possible. Otherwise, we would reach a contradiction that the same constant in Inequality (18) is not the best value by using Inequality (25) (for $λ 1 + λ 2 = λ$).
If the constant in Inequality (23) is the best value, then the same constant in Inequality (16) is still the best value. Otherwise, by Inequality (26) (for $λ 1 + λ 2 = λ$), we would reach a contradiction that the same constant factor in Inequality (24) is not the best value.
This proves theorem. □
We indicate the functions as follows: $ϕ ( m ) : = m − p α λ ^ 1 − 1 ,$ $ψ ( n ) : = n q ( 1 − β λ ^ 2 ) − 1 ,$ wherefrom,
$ψ 1 − p ( n ) : = n p β λ ^ 2 − 1 ( m , n ∈ N ) .$
We still indicate some linear spaces as follows:
$l q , ψ : = { b = { b n } n = 1 ∞ ; | | b | | q , ψ = ( ∑ n = 1 ∞ ψ ( n ) | b n | q ) 1 q < ∞ } , l p , ψ 1 − p : = { c = { c n } n = 1 ∞ ; | | b | | p , ψ 1 − p = ( ∑ n = 1 ∞ ψ 1 − p ( n ) | c n | p ) 1 p < ∞ } , l p , ϕ : = { d = { d m } m = 1 ∞ ; | | d | | p , ϕ = ( ∑ m = 1 ∞ ϕ ( m ) | d m | p ) 1 p < ∞ } , B : = { a = { a m } m = 1 ∞ ; A = { A m } m = 1 ∞ ∈ l p , ϕ } ( A m = ∑ i = 1 m a i ) .$
For $a = { a m } m = 1 ∞ ( > 0 ) ∈ B$, setting $c = { c n } n = 1 ∞ :$ $c n : = ∑ m = 1 ∞ a m ( m α + n β ) λ$, inequality (23) can be rewritten as:
$| | c | | p , ψ 1 − p < ( α β ) 1 p λ ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q | | A | | p , ϕ < ∞ ,$
namely, $c ∈ l p , ψ 1 − p$.
Definition 1.
Define an operator $T : B → l p , ψ 1 − p$ as follows: For any $a ∈ B ,$ there exists a unique representation $c = T a ∈ l p , ψ 1 − p$, such that for any $n ∈ N ,$ we have $T a ( n ) = c n$. Define the formal inner product of $T a$ and $b ∈ l q , ψ$ and the norm of $T$ as follows:
$( T a , b ) : = ∑ n = 1 ∞ b n ∑ m = 1 ∞ a m n ( m α + n β ) λ ,$
$| | T | | : = sup a ( ≠ 0 ) ∈ B | | T a | | p , ψ 1 − p | | A | | p , ϕ .$
By using Theorems 2–4, we have
Theorem 6.
Assume that $a ( > 0 )$ $∈ B , b ( > 0 ) ∈ l q , ψ , | | b | | q , ψ > 0 .$ We have the following equivalent inequalities:
$( T a , b ) < ( α β ) 1 p λ ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q | | A | | p , ϕ | | b | | q , ψ ,$
$| | T a | | p , ψ 1 − p < ( α β ) 1 p λ ( k λ + 1 ( λ 2 ) ) 1 p ( k λ + 1 ( λ 1 + 1 ) ) 1 q | | A | | p , ϕ .$
Moreover, assume that If $λ 1 + λ 2 = λ$, then the constant factor $( α β ) 1 p$ $λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q$ in Inequality (27) and (28) is the best possible, namely, $| | T | | =$ $( α β ) 1 p λ 1$ . On the other hand, if the constant factor in Inequality (26) (or Inequality (27)) is the best possible, then for
$λ − λ 1 − λ 2 ≤ min { p ( 2 α − 1 − λ 1 ) , q ( 2 β − λ 2 ) } ,$
we have $λ 1 + λ 2 = λ$.
Remark 2.
(i) For  in Inequalities (18) and (24), we have the equivalent inequalities as follows:
$∑ m = 1 ∞ ∑ n = 1 ∞ a m b n m α + n β < ( α β ) 1 p π q sin ( π / p ) × [ ∑ m = 1 ∞ m ( 1 − p ) α − 1 A m p ] 1 p [ ∑ n = 1 ∞ n ( q − 1 ) ( 1 − β ) b n q ] 1 q ,$
$[ ∑ n = 1 ∞ n β − 1 ( ∑ m = 1 ∞ a m m α + n β ) p ] 1 p < ( α β ) 1 p π q sin ( π / p ) [ ∑ m = 1 ∞ m ( 1 − p ) α − 1 A m p ] 1 p .$
(ii) For  $λ = 1 , λ 1 = 1 p , λ 2 = 1 q$  in Inequalities (18) and (24), the equivalent inequalities are valid as follows:
$∑ m = 1 ∞ ∑ n = 1 ∞ a m b n m α + n β < ( α β ) 1 p π p sin ( π / p ) × ( ∑ m = 1 ∞ m − α − 1 A m p ) 1 p ( ∑ n = 1 ∞ n q − β − 1 b n q ) 1 q ,$
$[ ∑ n = 1 ∞ n ( p − 1 ) β − 1 ( ∑ m = 1 ∞ a m m α + n β ) p ] 1 p < ( α β ) 1 p π p sin ( π / p ) ( ∑ m = 1 ∞ m − α − 1 A m p ) 1 p .$
(iii) when  $p = q = 2 ,$  both Inequalities (29) and (31) deduce to
$∑ m = 1 ∞ ∑ n = 1 ∞ a m b n m α + n β < ( α β ) 1 2 π 2 ( ∑ m = 1 ∞ m − α − 1 A m 2 ∑ n = 1 ∞ n 1 − β b n 2 ) 1 2 ,$
and both Inequalities (30) and (32) deduce to the following inequality equivalent to Inequality (33):
$[ ∑ n = 1 ∞ n β − 1 ( ∑ m = 1 ∞ a m m α + n β ) 2 ] 1 2 < ( α β ) 1 2 π 2 ( ∑ m = 1 ∞ m − α − 1 A m 2 ) 1 2$
We observe that the above constants are all the best value.

5. Conclusions

In this article, by means of the idea of introduced parameters and the techniques of real analysis, using Euler-Maclaurin summation formula as well as the mid-value theorem, we give a new extended Hardy-Hilbert’s inequality with the power functions as the internal variables involving one partial sums in Theorem 1, which is a refinement of a published inequality. We provide a few equivalent conditions of the best possible constant related to multi parameters in Theorem. 2 and 3. As applications, the equivalent inequalities are fond in Theorem. 4 and 5, and the operator expressions as well as a few inequalities with the particular parameters are considered in Theorem 5 and Remark 2.

Author Contributions

B.Y. carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. A.W. 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 Science Foundation of China (No. 61772140), and the 2022 Guangdong Provincial Education Science Planning Project (Higher Education Project) (2022GXJK290).

Data Availability Statement

We declare that the data and material in this paper can be used publicly.

Acknowledgments

The authors thank the referee for his useful proposal to revise the paper.

Conflicts of Interest

The authors declare that they 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]
3. Yang, B.C. On a generalization of Hilbert double series theorem. J. Nanjing Univ. Math. Biquarterly 2001, 18, 145–152. [Google Scholar]
4. Yang, B.C. The Norm of Operator and Hilbert-Type Inequalities; Science Press: Beijing, China, 2009. [Google Scholar]
5. Krnić, M.; Pečarić, J. General Hilbert’s and Hardy’s inequalities. Math. Inequal. Appl. 2005, 8, 29–51. [Google Scholar] [CrossRef]
6. Perić, I.; Vuković, P. Multiple Hilbert’s type inequalities with a homogeneous kernel. Banach J. Math. Anal. 2011, 5, 33–43. [Google Scholar] [CrossRef]
7. Huang, Q.L. A new extension of Hardy-Hilbert-type inequality. J. Inequal. Appl. 2015, 2015, 397. [Google Scholar] [CrossRef]
8. He, B. A multiple Hilbert-type discrete inequality with a new kernel and best possible constant factor. J. Math. Anal. Appl. 2015, 431, 902–990. [Google Scholar] [CrossRef]
9. Xu, J.S. Hardy-Hilbert’s inequalities with two parameters. Adv. Math. 2007, 36, 63–76. [Google Scholar]
10. 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]
11. Zeng, Z.; Gandhi, K.R.R.; 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]
12. Xin, D.M. A Hilbert-type integral inequality with the homogeneous kernel of zero degree. Math. Theory Appl. 2010, 30, 70–74. [Google Scholar]
13. Azar, L.E. The connection between Hilbert and Hardy inequalities. J. Inequal. Appl. 2013, 2013, 452. [Google Scholar] [CrossRef]
14. 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]
15. 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]
16. 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]
17. Hong, Y.; Huang, Q.L.; Yang, B.C.; Liao, J.L. 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. Inequal. Appl. 2017, 2017, 316. [Google Scholar] [CrossRef]
18. Xin, D.M.; Yang, B.C.; Wang, A.Z. Equivalent property of a Hilbert-type integral inequality related to the beta function in the whole plane. J. Funct. Spaces 2018, 2018, 2691816. [Google Scholar] [CrossRef]
19. 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]
20. Hong, Y.; Chen, Q. Equivalent parameter conditions for the construction of Hilbert-type integral inequalities with a class of non-homogeneous kernels. J. South China Norm. Univ. (Nat. Sci. Ed.) 2020, 52, 124–128. [Google Scholar]
21. Adiyasuren, V.; Batbold, T.; Azar, L.E. A new discrete Hilbert-type inequality involving partial sums. J. Inequal. Appl. 2019, 2019, 127. [Google Scholar] [CrossRef]
22. Liao, J.Q.; Wu, S.H.; Yang, B.C. A multi parameter Hardy–Hilbert-type inequality containing partial sums as the terms of series. J. Math. 2021, 2021, 5264623. [Google Scholar] [CrossRef]
23. Yang, B.C.; Wu, S.H. A weighted Generalization of Hardy–Hilbert-type inequality involving two partial sums. Mathematics 2023, 11, 3212. [Google Scholar] [CrossRef]
24. Liao, J.Q.; Yang, B.C. A New Reverse Extended Hardy-Hilbert’s Inequality with Two Partial Sums and Parameters. Axioms 2023, 12, 678. [Google Scholar] [CrossRef]
25. 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

Wang, A.; Yang, B. An Extended Hilbert-Type Inequality with Two Internal Variables Involving One Partial Sums. Axioms 2023, 12, 871. https://doi.org/10.3390/axioms12090871

AMA Style

Wang A, Yang B. An Extended Hilbert-Type Inequality with Two Internal Variables Involving One Partial Sums. Axioms. 2023; 12(9):871. https://doi.org/10.3390/axioms12090871

Chicago/Turabian Style

Wang, Aizhen, and Bicheng Yang. 2023. "An Extended Hilbert-Type Inequality with Two Internal Variables Involving One Partial Sums" Axioms 12, no. 9: 871. https://doi.org/10.3390/axioms12090871

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