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A new extended Mulholland's inequality involving one partial sum

  • Ling Peng and Bicheng Yang EMAIL logo
Published/Copyright: September 25, 2024

Abstract

By using the weight coefficients and the techniques of real analysis, a new extended Mulholland’s inequality with multi-parameters involving one partial sum is given. The equivalent statements of the best value related to several parameters are provided. The equivalent forms, some inequalities in particular parameters, and the operator expressions are obtained.

MSC 2010: 26D15; 26D10; 47A05

1 Introduction

Suppose that p > 1 , 1 p + 1 q = 1 , a m , b n 0 , 0 < m = 1 a m p < , and 0 < n = 1 b n q < . The following Hardy-Hilbert’s inequality with the best value π sin ( π p ) was provided (cf. [1], Theorem 315):

(1) 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 .

With regard to a similar assumption, the following Mulholland’s inequality with the same best value was obtained (cf. [1], Theorem 343):

(2) m = 2 n = 2 a m b n ln m n < π sin ( π p ) m = 2 a m p m 1 p 1 p n = 2 b n q n 1 q 1 q .

In 2006, an extension of (1) was given by Krnić and Pečarić [2] as follows:

(3) 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 ,

where λ i ( 0 , 2 ] ( i = 1 , 2 ) , λ 1 + λ 2 = λ ( 0 , 4 ] , the constant B ( λ 1 , λ 2 ) is the best value, and

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 , (3) deduces to Yang’s result in [3], which was published in 2001. In 2019, following the idea of (3), Adiyasuren et al. [4] gave an extension of (3) involving two partial sums as follows: For λ ( 0 , 2 ] , λ i ( 0 , 1 ] ( 0 , λ ) ( i = 1 , 2 ) , λ 1 + λ 2 = λ , a m , b n 0 ,

(4) 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 value, and A m = i = 1 m a i and B n = k = 1 n b k ( m , n { 1 , 2 , } ) , satisfy A m = o ( e t m ) , B n = o ( e t n ) ( t > 0 ; m , n ) ,

0 < m = 1 m p λ 1 1 A m p < and 0 < n = 1 n q λ 2 1 B n q < .

We can see some new extensions and applications of (1) and (2) with their integral analogs in [514]. In 2016, Hong and Wen [15] considered some equivalent conditions of the best value related to some parameters in the general form of (1). The further results were provided by [1621]. In 2021, Gu and Yang [22] gave an extension of (4) with the kernel as 1 ( m α + n β ) λ , where α , β ( 0 , 1 ] . But the constant factor is not the best possible unless at α = β = 1 . In 2023, Yang et al. [23] gave an improvement of [22] and proved that the constant factor in the new inequality is the best possible for α , β ( 0 , 1 ] . Recently, Huang et al. [24,25] provided the same kinds of applications in Hilbert-type inequalities.

In this article, following the way of [15], by the use of the weight coefficients and the techniques of real analysis, a new extended Mulholland’s inequality with multi-parameters involving one partial sum is given. The equivalent conditions of the best value related to several parameters are provided. The equivalent forms, a few inequalities for particular parameters, and the operator expressions are obtained.

2 Some lemmas

In this section, it is necessary to introduce the following conditions on the parameters that we use in the following results.

Assumption 1

Assume that p > 1 ( q > 1 ) , 1 p + 1 q = 1 , λ > 0 , λ 1 ( 0 , 2 ] ( 0 , λ + 1 ) , λ 2 ( 0 , 1 ] ( 0 , λ ) , λ ˆ 1 λ λ 2 p + λ 1 q , λ ˆ 2 λ λ 1 q + λ 2 p , N = { 1 , 2 , } , a m , b n 0 , B n k = 2 n b k ( m , n N 2 N \ { 1 } ) , B n = o ( e t ln n ) ( t > 0 ; n ) , and

(5) 0 < m = 2 ln p ( 1 λ ˆ 1 ) 1 m m 1 p a m p < and 0 < n = 2 B n q n ln q λ ˆ 2 + 1 n < .

Lemma 1

(cf. [5], (2.2.3)) (i) Assuming that ( 1 ) d i d t i g ( t ) > 0 , t [ k , ) ( k N ) , g ( i ) ( ) = 0 ( i = 0 , 1 , 2 , 3 ) , P i ( t ) , B i ( i N ) are the Bernoulli functions and Bernoulli numbers of i-order, we have

(6) k P 2 q 1 ( t ) g ( t ) d t = ε q B 2 q 2 q g ( k ) ( 0 < ε q < 1 ; q = 1 , 2 , ) .

In particular, for q = 1 , B 2 = 1 6 , it follows that

(7) 1 12 g ( k ) < k P 1 ( t ) g ( t ) d t < 0 .

(ii) If ( 1 ) i d i d t i g ( t ) > 0 , t [ k , ) ( k N ) , g ( i ) ( ) = 0 ( i = 0 , 1 ), then we obtain (cf. [5], (2.2.13))

(8) 1 8 g ( k ) < k P 1 ( t ) g ( t ) d t < 0 .

(iii) (cf. [5], (2.3.2)) Assuming that f ( t ) ( 0 ) C [ k , ) ( k N ) , f ( i ) ( ) = 0 ( i = 0 , 1 ) , the following expression is valid:

(9) i = k f ( i ) = k f ( t ) d t + 1 2 f ( k ) + k P 1 ( t ) f ( t ) d t .

Lemma 2

Let s > 0 , s 2 ( 0 , 2 ] ( 0 , s ) , k s ( s 2 ) = B ( s 2 , s s 2 ) . Define the weight coefficient as follows:

(10) ϖ s ( s 2 , m ) ( ln m ) s s 2 n = 2 ( ln n ) s 2 1 ( ln m n ) s n .

Then the following inequalities are valid:

(11) 0 < k s ( s 2 ) 1 O 1 ln s 2 m < ϖ s ( s 2 , m ) < k s ( s 2 ) ( m N 2 ) ,

where we indicate that O 1 ln s 2 m 1 k s ( s 2 ) 0 ln 2 ln m u s 2 1 ( 1 + u ) s d u ( > 0 ) .

Proof

For fixed m N 2 , we define a positive function g m ( t ) as follows:

(12) g m ( t ) ln s 2 1 t ( ln m + ln t ) s t ( t > 1 ) .

(i) For s 2 ( 0 , 1 ] ( 0 , s ) , we find

g m ( t ) = ( 1 s 2 ) ln s 2 2 t ( ln m + ln t ) s t 2 s ln s 2 1 t ( ln m + ln t ) s + 1 t 2 ln s 2 1 t ( ln m + ln t ) s t 2 < 0 .

By means of the decreasing property of series, it follows that

(13) 2 g m ( t ) d t < n = 2 g m ( n ) < 1 g m ( t ) d t .

(ii) For s 2 ( 1 , 2 ] ( 0 , s ) , by (9), we have

n = 2 g m ( n ) = 2 g m ( t ) d t + 1 2 g m ( 2 ) + 2 P 1 ( t ) g m ( t ) d t = 1 g m ( t ) d t h ( m ) , h ( m ) 1 2 g m ( t ) d t 1 2 g m ( 2 ) 2 P 1 ( t ) g m ( t ) d t .

We obtain 1 2 g m ( 2 ) = ln s 2 1 2 4 ( ln m + ln 2 ) s . Setting u = ln t , by integration by parts, we find

1 2 g m ( t ) d t = 0 ln 2 u s 2 1 ( ln m + u ) s d u = 0 ln 2 d u s 2 s 2 ( ln m + u ) s = u s 2 s 2 ( ln m + u ) s 0 ln 2 0 ln 2 u s 2 s 2 d 1 ( ln m + u ) s = ln s 2 2 s 2 ( ln m + ln 2 ) s + s s 2 ( s 2 + 1 ) 0 ln 2 d u s 2 + 1 ( ln m + u ) s + 1 > ln s 2 2 s 2 ( ln m + ln 2 ) s + s s 2 ( s 2 + 1 ) ln s 2 + 1 2 ( ln m + ln 2 ) s + 1 .

Since ln t t 2 > 0 , ln t t 2 = 1 2 ln t t 3 < 0 ( t > 2 ) , by (8), we have

g m ( t ) = ( s 2 1 ) ln s 2 2 t ( ln m + ln t ) s t 2 s ln s 2 t ( ln m + ln t ) s + 1 ln 1 t t 2 ln s 2 2 t ( ln m + ln t ) s ln t t 2 ,

2 P 1 ( t ) ( s 2 1 ) ln s 2 2 t ( ln m + ln t ) s t 2 d t = ( 1 s 2 ) 2 P 1 ( t ) ln s 2 2 t ( ln m + ln t ) s t 2 d t > 0 ,

2 P 1 ( t ) s ln s 2 t ( ln m + ln t ) s + 1 ln 1 t t 2 + ln s 2 2 t ( ln m + ln t ) s ln t t 2 d t > 1 8 s ln s 2 2 2 ( ln m + ln 2 ) s + 1 ln 1 2 2 2 + ln s 2 2 2 ( ln m + ln 2 ) s ln 2 2 2 ,

2 P 1 ( t ) g m ( t ) d t > s ln s 2 1 2 8 ( ln m + ln 2 ) s + 1 2 2 ln s 2 1 2 8 ( ln m + ln 2 ) s 2 2 .

Then we obtain

h ( m ) > ln s 2 2 s 2 ( ln m + ln 2 ) s + s s 2 ( s 2 + 1 ) ln s 2 + 1 2 ( ln m + ln 2 ) s + 1 ln s 2 1 2 4 ( ln m + ln 2 ) s s ln s 2 2 8 ( ln m + ln 2 ) s + 1 2 2 ln s 2 1 2 8 ( ln m + ln 2 ) s 2 2 = ln s 2 1 2 ( ln m + ln 2 ) s ln 2 s 2 1 4 1 32 + s ln s 2 1 2 ( ln m + ln 2 ) s + 1 ln 2 2 s 2 ( s 2 + 1 ) 1 32 ln s 2 1 2 ( ln m + ln 2 ) s ln 2 2 9 32 + s ln s 2 1 2 ( ln m + ln 2 ) s + 1 ln 2 2 6 1 32 > 0 .

By (9), we still have

n = 2 g m ( n ) = 2 g m ( t ) d t + 1 2 g m ( 2 ) + 2 P 1 ( t ) g m ( t ) d t = 2 g m ( t ) d t + h 1 ( m ) , h 1 ( m ) 1 2 g m ( 2 ) + 2 P 1 ( t ) g m ( t ) d t .

For s 2 ( 1 , 2 ] ( 0 , s ) , in view of (8), we find

2 P 1 ( t ) ( s 2 1 ) s 2 2 t ( ln m + ln t ) s t 2 d t = ( s 2 1 ) 2 P 1 ( t ) ln s 2 2 t ( ln m + ln t ) s t 2 d t > ( s 2 1 ) ln s 2 2 2 32 ( ln m + ln 2 ) s , 2 P 1 ( t ) s ln s 2 2 t ( ln m + ln t ) s + 1 ln t t 2 + ln s 2 2 t ( ln m + ln t ) s ln t t 2 d t > 0 , 2 P 1 ( t ) g m ( t ) d t > ( s 2 1 ) ln s 2 2 2 32 ( ln m + ln 2 ) s , h 1 ( m ) > ln s 2 1 2 4 ( ln m + ln 2 ) s ( s 2 1 ) ln s 2 2 2 32 ( ln m + ln 2 ) s ln s 2 2 2 4 ( ln m + ln 2 ) s ln 2 1 8 > 0 .

Hence, we also have (13).

In a word, for s 2 ( 0 , 2 ] ( 0 , s ) , by (13), setting u = ln t ln m , we find

ϖ s ( s 2 , m ) = ln s s 2 m n = 2 g m ( n ) < ln s s 2 m 1 g m ( t ) d t = 0 u s 2 1 d u ( 1 + u ) s = B ( s 2 , s s 2 ) = k s ( s 2 ) ,

ϖ s ( s 2 , m ) > ln s s 2 m 2 g m ( t ) d t = ln 2 ln m u s 2 1 d u ( 1 + u ) s = k s ( s 2 ) 1 O 1 ln s 2 m > 0 ,

where we indicate that O 1 ln s 2 m = 1 k s ( s 2 ) 0 ln 2 ln m u s 2 1 ( 1 + u ) s d u , satisfying

0 < 0 ln 2 ln m u s 2 1 ( 1 + u ) s d u 0 ln 2 ln m u s 2 1 d u = 1 s 2 ln 2 ln m s 2 .

Therefore, inequalities (11) follow.

This proves the lemma.□

The following extended Mulholland’s inequality is valid.

Lemma 3

Let Assumption 1 hold. Let us define

k λ + 1 ( λ 2 + 1 ) = B ( λ 2 + 1 , λ λ 2 ) ,

k λ + 1 ( λ 1 ) = B ( λ 1 , λ + 1 λ 1 ) . Then the following inequality is valid:

(14) n = 2 m = 2 a m B n ( ln m n ) λ + 1 n < ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q m = 2 ln p ( 1 λ ^ 1 ) 1 m m 1 p a m p 1 p n = 2 B n q n ln q λ ^ 2 + 1 n 1 q .

Proof

By the symmetry, for s > 0 , s 1 ( 0 , 2 ] ( 0 , s ) , the following estimations for the next weight coefficient are valid:

(15) 0 < k s ( s 1 ) 1 O 1 ln s 1 n < ω s ( s 1 , n ) ( ln n ) s s 1 m = 2 ( ln n ) s 1 1 ( ln m n ) s m < k s ( s 1 ) B ( s 1 , s s 1 ) ( n N 2 ) ,

where O 1 ln s 1 n 1 k s ( s 1 ) 0 ln 2 ln n u s 1 1 ( 1 + u ) s d u > 0 .

Let us denote I λ + 1 n = 2 m = 2 a m B n ( ln m n ) λ + 1 n . By Hölder’s inequality (cf. [26]),

I λ + 1 n = 2 m = 2 1 ( ln m n ) λ + 1 m 1 q ln λ 2 p n n 1 p ln ( λ 1 1 ) q m a m × ln ( λ 1 1 ) q m m 1 q ln λ 2 p n 1 n 1 q B n n = 2 m = 2 1 ( ln m n ) λ + 1 m p 1 ln λ 2 n n ln ( λ 1 1 ) ( p 1 ) m a m p 1 p × n = 2 m = 2 1 ( ln m n ) λ + 1 ln λ 1 1 m m ln λ 2 ( q 1 ) n B n q n 1 q = m = 2 ϖ λ + 1 ( λ 2 + 1 , m ) ln p ( 1 λ ^ 1 ) 1 m 1 p a m p 1 p × n = 2 ω λ + 1 ( λ 1 , n ) B n q n ln q λ ^ 2 + 1 n 1 q .

Then by (11) and (15), for s = λ + 1 > 1 , s 1 = λ 1 ( 0 , 2 ] ( 0 , λ + 1 ) ,

s 2 = λ 2 + 1 ( 0 , 2 ] ( 0 , λ + 1 ) ,

in view of (5), we obtain (14).

This proves the lemma.□

Lemma 4

Let b n , B n , n N 2 be as in Assumption 1. If t > 0 , then the following inequality is valid:

(16) n = 2 e t ln n b n t n = 2 e t ln n 1 n B n .

Proof

Since B n e t ln n = o ( 1 ) ( n ) , by Abel’s summation by parts formula, it follows that

n = 2 e t ln n b n = lim n B n e t ln n + n = 2 B n [ e t ln n e t ln ( n + 1 ) ] = n = 2 B n [ e t ln n e t ln ( n + 1 ) ] .

We set a function f ( x ) = e t ln x , x [ n , n + 1 ] . Then we find f ( x ) = t x e t ln x = h ( x ) , where h ( x ) 1 x e ln x is decreasing in [ n , n + 1 ] . In view of the differentiation mid-value theorem, we obtain

n = 2 e t ln n b n = n = 2 B n ( f ( n + 1 ) f ( n ) ) = n = 2 B n f ( n + θ n ) = n = 2 h ( n + θ n ) B n n = 2 h ( n ) B n = t n = 2 e t ln n 1 n B n ( θ n ( 0 , 1 ) ) ,

namely, (16) follows.

This proves the lemma.□

3 Main results

We have the following extended Mulholland’s inequality involving one partial sum.

Theorem 1

Let Assumption 1 hold. Let k λ + 1 ( λ 2 + 1 ) and k λ + 1 ( λ 1 ) be defined as in Lemma 3. Then the following inequality is valid:

(17) m = 2 n = 2 a m b n ( ln m n ) λ < λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q × m = 2 ln p ( 1 λ ˆ 1 ) 1 m m 1 p a m p 1 p n = 2 B n q n ln q λ ˆ 2 + 1 n 1 q .

In particular, for λ 1 + λ 2 = λ , we have

0 < m = 2 ln p ( 1 λ 1 ) 1 m m 1 p a m p < , 0 < n = 2 B n q n ln q λ 2 + 1 n < ,

and the following inequality:

(18) m = 2 n = 2 a m b n ( ln m n ) λ < λ B ( λ 1 , λ 2 + 1 ) m = 2 ln p ( 1 λ 1 ) 1 m m 1 p a m p 1 p n = 2 B n q n ln q λ 2 + 1 n 1 q .

Proof

Use the following expression:

1 ( ln m n ) λ = 1 ( ln m + ln n ) λ = 1 Γ ( λ ) 0 t λ 1 e ( ln m + ln n ) t d t ,

where Γ is the gamma function. Let us denote I m = 2 n = 2 a m b n ( ln m n ) λ . By (16), it follows that

I = 1 Γ ( λ ) m = 2 n = 2 a m b n 0 t λ 1 e ( ln m + ln n ) t d t = 1 Γ ( λ ) 0 t λ 1 m = 2 e t ln m a m n = 2 e t ln n b n d t 1 Γ ( λ ) 0 t λ 1 m = 2 e t ln m a m t n = 2 e t ln n 1 n B n d t = 1 Γ ( λ ) m = 2 n = 2 a m 1 n B n 0 t ( λ + 1 ) 1 e ( ln m + ln n ) t d t = Γ ( λ + 1 ) Γ ( λ ) m = 2 n = 2 1 ( ln m + ln n ) λ + 1 n a m B n = Γ ( λ + 1 ) Γ ( λ ) I λ + 1 ,

where I λ + 1 is defined as in (14). Then by (14), in view of Γ ( λ + 1 ) = λ Γ ( λ ) , we have (17).

This proves the theorem.□

Theorem 2

Suppose that λ > 0 , λ 1 ( 0 , 2 ] ( 0 , λ ) , λ 2 ( 0 , 1 ] ( 0 , λ ) . If λ 1 + λ 2 = λ , then the constant λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q in (17) is the best value.

Proof

We now prove that the constant λ B ( λ 1 , λ 2 + 1 ) ( = λ 2 B ( λ 1 , λ 2 ) ) in (18) is the best value following the condition. For any number ε , 0 < ε < min { p λ 1 , q λ 2 } , we set

a ˜ m 1 m ln λ 1 ε p 1 m , b ˜ n 1 n ln λ 2 ε q 1 n ( m , n N 2 ) .

Since 0 < λ 2 ε q < 1 , by means of the decreasingness property of series (cf. [26]), we obtain

B ˜ n k = 2 n b ˜ k < 1 n 1 y ln λ 2 ε q 1 y d y = 1 λ 2 ε q ln λ 2 ε q n ( n N 2 ) ,

and observe that B ˜ n = o ( e t ln n ) ( t > 0 ; n ) .

If there exists a positive constant M λ 2 B ( λ 1 , λ 2 ) ( = λ B ( λ 1 , λ 2 + 1 ) ) , such that (18) is valid when we replace λ B ( λ 1 , λ 2 + 1 ) by M , then in particular, substitution of a m = a ˜ m , b n = b ˜ n and B n = B ˜ n in (18), we have

(19) I ˜ n = 2 m = 2 a ˜ m b ˜ n ( ln m n ) 2 < M m = 2 ln p ( 1 λ 1 ) 1 m 1 p a ˜ m p 1 p n = 2 B ˜ n q n ln q λ 2 + 1 n 1 q .

By (19) (cf. [26]), we find

I ˜ < M λ 2 ε q m = 2 ln ε 1 m m 1 p n = 2 ln ε 1 n n 1 q = M λ 2 ε q ln ε 1 2 2 + m = 3 ln ε 1 m m < M λ 2 ε q ln ε 1 2 2 + 2 ln ε 1 x x d x = M ε λ 2 ε q ε ln ε 1 2 2 + ln ε 2 .

In view of (15) (for s = λ > 0 , s 1 = λ 1 ε p ( 0 , 2 ) ( 0 , λ ) ), we have

I ˜ = n = 2 m = 2 ln λ 1 ε p 1 m ( ln m n ) λ m n ln λ 2 ε q 1 n = n = 2 1 n ln ε 1 n ln λ 2 + ε p n m = 2 ln λ 1 ε p 1 m ( ln m n ) λ m k λ λ 1 ε p n = 2 1 n ln ε 1 n 1 O 1 ln λ 1 ε p n = k λ λ 1 ε p n = 2 1 n ln ε 1 n n = 2 1 n O 1 ln λ 1 + ε q + 1 n > k λ λ 1 ε p 2 1 y ln ε 1 y d y O ( 1 ) = 1 ε B λ 1 ε p , λ 2 + ε p ( ln ε 2 ε O ( 1 ) ) .

Then we find

B λ 1 ε p , λ 2 + ε p ( ln ε 2 ε O ( 1 ) ) < ε I ˜ < M λ 2 ε q ε ln ε 1 2 2 + ln ε 2 .

For ε 0 + , we obtain B ( λ 1 , λ 2 ) M λ 2 , namely, λ 2 B ( λ 1 , λ 2 ) M .

Hence, M = λ 2 B ( λ 1 , λ 2 ) ( = λ B ( λ 1 , λ 2 + 1 ) ) is the best value in (18).

This proves the theorem.□

Theorem 3

Assume that λ 1 ( 0 , 2 ] ( 0 , λ ) , λ 2 ( 0 , 1 ] ( 0 , λ ) ( λ > 0 ) . If the constant λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 4 in (17) is the best value, then for

λ λ 1 λ 2 min { p ( 2 λ 1 ) , q ( 1 λ 2 ) } ,

we have λ 1 + λ 2 = λ .

Proof

For λ ˆ 1 = λ λ 2 p + λ 1 q = λ λ 1 λ 2 p + λ 1 , λ ˆ 2 = λ λ 1 q + λ 2 p = λ λ 1 λ 2 q + λ 2 , we find λ ˆ 1 + λ ˆ 2 = λ , 0 < λ ˆ 1 , λ ˆ 2 < λ p + λ q = λ , and B ( λ ˆ 1 , λ ˆ 2 + 1 ) ( 0 , ) . For λ λ 1 λ 2 p ( 2 λ 1 ) , we have λ ˆ 1 2 ; for λ λ 1 λ 2 q ( 1 λ 2 ) , we have λ ˆ 2 1 . Substitution of λ i = λ ˆ i ( i = 1 , 2 ) in (18), it follows that

(20) m = 2 n = 2 a m b n ( ln m n ) λ < λ B ( λ ˆ 1 , λ ˆ 2 + 1 ) m = 2 ln p ( 1 λ ˆ 1 ) 1 m m 1 p a m p 1 p n = 2 B n q n ln n q λ ˆ 2 + 1 n 1 q .

By Hölder’s inequality with weight (cf. [26]), we obtain

(21) B ( λ ˆ 1 , λ ˆ 2 + 1 ) = k λ + 1 ( λ ˆ 1 ) = k λ + 1 λ λ 2 p + λ 1 q = 0 u λ λ 2 p + λ 1 q 1 ( 1 + u ) λ + 1 d u = 0 1 ( 1 + u ) λ + 1 u λ λ 2 1 p u λ 1 1 q d u 0 1 ( 1 + u ) λ + 1 u λ λ 2 1 d u 1 p 0 1 ( 1 + u ) λ + 1 u λ 1 1 d u 1 q = 0 1 ( 1 + v ) λ + 1 v ( λ 2 + 1 ) 1 d v 1 p 0 1 ( 1 + u ) λ + 1 u λ 1 1 d u 1 q = ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q .

If the constant λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q in (17) is the best value, then compared with the constant factors in (17) and (20), we have

λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q λ B ( λ ˆ 1 , λ ˆ 2 + 1 ) ( R + ) ,

which follows that

(22) B ( λ ˆ 1 , λ ˆ 2 + 1 ) ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q .

Hence, (21) keep the form of equality. We observe that (21) keep the form of equality if and only if there exist constants A and B , such that they are not both zero and (cf. [26]) A u λ λ 2 1 = B u λ 1 1 a.e. in R + . Suppose that A 0 . We have 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 .

This proves the theorem.□

4 Equivalent forms, some particular inequalities, and the operator expressions

The following inequalities equivalent to (17) and (18), respectively, are valid.

Theorem 4

Let Assumption 1 hold. Let k λ + 1 ( λ 2 + 1 ) and k λ + 1 ( λ 1 ) be defined as in Lemma 3. Then inequality (23) is equivalent to (17):

(23) m = 2 ln q λ ˆ 1 1 m m n = 2 b n ( ln m n ) λ q 1 q < λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q n = 2 B n q ln q λ ˆ 2 + 1 n 1 q .

In particular, for λ 1 + λ 2 = λ , inequality (24) is equivalent to (18):

(24) m = 2 ln q λ 1 1 m m n = 2 b n ( ln m n ) λ q 1 q < λ B ( λ 1 , λ 2 + 1 ) n = 2 B n q n ln q λ 2 + 1 n 1 q .

Proof

Assume that inequality (23) holds. Let us denote

J m = 2 ln q λ ˆ 1 1 m m n = 2 b n ( ln m n ) λ q 1 q .

Using notation I = m = 2 n = 2 a m b n ( ln m n ) λ from the proof of Theorem 1, with Hölder’s inequality, it follows that

(25) I = m = 2 ln λ ˆ 1 1 q m m 1 q a m ln λ ˆ 1 1 q m m 1 q n = 2 b n ( ln m m ) λ m = 2 ln p 1 λ ˆ 1 1 m m 1 p a m p 1 p J .

Then by (23), we have (17).

On the other hand, suppose that (17) is valid. We set

a m ln q λ 1 1 m m n = 2 b n ( ln m n ) λ q 1 , m N 2 .

We find J q = I . If J = 0 , then (23) is naturally valid; if J = , then (23) is impossible valid, namely, J < . Suppose that 0 < J < . By (17), we have

0 < m = 2 ln p ( 1 λ ˆ 1 ) 1 m m 1 p a m p = J q = I < λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q J q 1 n = 2 B n q n ln q λ ˆ 2 + 1 n 1 q < , J < λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q n = 2 B n q n ln q λ ˆ 2 + 1 n 1 q .

Hence, we have (23), which is equivalent to (17).

This proves the theorem.□

Theorem 5

Assume that λ > 0 , λ 1 ( 0 , 2 ] ( 0 , λ ) , λ 2 ( 0 , 1 ] ( 0 , λ ) . If λ 1 + λ 2 = λ , then the constant λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q in (23) is the best value. On the other hand, if the constant λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 9 in (23) is the best value, then for

λ λ 1 λ 2 min { p ( 2 λ 1 ) , q ( 1 λ 2 ) } ,

we have λ 1 + λ 2 = λ .

Proof

If λ 1 + λ 2 = λ , then the constant in (17) is the best value. We can show that the constant in (23) is the best value. Otherwise, by using (25), we can prove that the constant in (17) is not the best value. On the other hand, if the constant in (23) is the best value, then for the equivalency of (23) and (17), in view of J q = I (see the proof of Theorem 4.1), we can prove that the constant in (17) is the best value. Then by Theorem 3.3, for

λ λ 1 λ 2 min { p ( 2 λ 1 ) , q ( 1 λ 2 ) } ,

we have λ 1 + λ 2 = λ .

This proves the theorem.□

Remark 1

(i) For λ = 1 , λ 1 = 1 q , λ 2 = 1 p in (18) and (24), the following equivalent forms are valid:

(26) m = 2 n = 2 a m b n ln m n < π p sin ( π p ) m = 2 a m p m 1 p 1 p n = 2 B n q n ln q n 1 q ,

(27) m = 2 1 m n = 2 b n ln m q 1 q < π p sin ( π p ) n = 2 B n q n ln q n 1 q .

(ii) For λ = 1 , λ 1 = 1 p , λ 2 = 1 q in (18) and (24), the following equivalent dual forms of (26) and (27) are valid:

(28) m = 2 n = 2 a m b n ln m n < π q sin ( π p ) m = 2 ln p 2 m m 1 p a m p 1 p n = 2 B n q n ln 2 n 1 q ,

(29) m = 2 ln q 2 m m n = 2 b n ln m n q 1 q < π q sin ( π p ) n = 2 B n q n ln 2 n 1 q .

In particular, for p = q = 2 , both (26) and (28) deduce to

(30) m = 2 n = 2 a m b n ln m n < π 2 m = 2 m a m 2 n = 2 B n 2 n ln 2 n 1 2 ,

and both (27) and (29) deduce to the following equivalent form of (30):

(31) m = 2 1 m n = 2 b n ln m n 2 1 2 < π 2 n = 2 B n 2 n ln 2 n 1 2 .

The constants in the above inequalities are all the best values.

We set the following functions: φ ( m ) ln p ( 1 λ ^ 1 ) 1 m m 1 p , ψ ( n ) ln q ( 1 λ ^ 2 ) 1 n n 1 q , and Ψ ( n ) 1 n ln q λ ^ 2 + 1 n , wherefrom,

ψ 1 q ( m ) ln q λ ^ 1 1 m m ( m , n N 2 ) .

Define the normed linear spaces as follows:

l p , φ a = { a m } m = 2 ; a p , φ = m = 2 φ ( m ) a m p 1 p < , l q , ψ b = { b n } n = 2 ; b q , ψ = n = 2 ψ ( n ) b n q 1 q < , l q , Ψ B = { B n } n = 2 ; B q , Ψ = n = 2 Ψ ( n ) B n q 1 q < , l q , φ 1 q c = { c m } m = 2 ; c q , φ 1 q = m = 2 φ 1 q ( m ) c m q 1 q < , l ˜ q , ψ b = { b n } n = 2 l q , ψ ; B n = i = 2 n b i , B = { B n } n = 2 l q , Ψ .

For b = { b n } n = 2 ( 0 ) l ˜ q , ψ , setting c = { c m } m = 2 , c m n = 2 b n ( ln m n ) λ .

Inequality (23) can be rewritten as follows:

c q , φ 1 q λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q B q , Ψ < ,

namely, c l q , φ 1 q .

Definition 1

Define an extended Mulholland’s operator T : l ˜ q , ψ l q , φ 1 q as follows: For any b l ˜ q , ψ , there exists a unique representation c = T b l q , φ 1 q , such that for any m N 2 , T b ( m ) = c m . Define the formal inner product of a l p , φ and T b , and the norm of T as follows:

( a , T b ) m = 2 a m n = 2 b n ( ln m n ) λ , T sup b ( 0 ) l ˜ q , ψ T b q , φ 1 q B q , ψ .

Theorem 6

Assuming that b ( 0 ) l ˜ q , ψ , B q , Ψ > 0 , a ( 0 ) l p , φ , a p , φ > 0 , the following equivalent operator inequalities are valid:

(32) ( a , T b ) < λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 q a p , q B q , Ψ ,

(33) T b q , φ 1 q < λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 4 B q , Ψ .

Moreover, assume that λ 1 ( 0 , 2 ] ( 0 , λ ) , λ 2 ( 0 , 1 ] ( 0 , λ ) ( λ > 0 ) . If λ 1 + λ 2 = λ , then the constant λ ( k λ + 1 ( λ 2 + 1 ) ) 1 p ( k λ + 1 ( λ 1 ) ) 1 4 in (32) and (33) is the best value, namely,

T = λ 2 B ( λ 1 , λ 2 ) .

On the other hand, if the same constant in (32) (or (33)) is the best value, then for

λ λ 1 λ 2 min { p ( 2 λ 1 ) , q ( 1 λ 2 ) } ,

we have λ 1 + λ 2 = λ .

5 Conclusions

In this article, by using the techniques of real analysis, a new extended Mulholland’s inequality with multi-parameters involving one partial sum is given in Theorem 3.1. The equivalent conditions of the best possible value related to several parameters are provided by Theorems 3.2 and 3.3. The equivalent forms and some inequalities in particular parameters are obtained in Theorems 4.1, 4.2, and Remark 4.1. The operator expressions are considered in Theorem 4.3. The idea and the methods provide an extensive account of discrete Hilbert-type inequalities.

Acknowledgement

The authors thank the referees for their useful proposal to reform this article.

  1. Funding information: This work was supported by the National Natural Science Foundation (No. 61772140).

  2. Author contributions: BY carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. LP participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.

  3. Conflict of interest: The authors declare that they have no conflict of interest.

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2024-03-03
Revised: 2024-06-18
Accepted: 2024-07-22
Published Online: 2024-09-25

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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