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On a more accurate multidimensional Mulholland-type inequality

Abstract

In this paper, by using the way of weight coefficients and technique of real analysis, a more accurate multidimensional discrete Mulholland-type inequality with the best possible constant factor is given, which is an extension of the Mulholland inequality. The equivalent form, the operator expression with the norm as well as a few particular cases are also considered.

MSC:26D15, 47A07.

1 Introduction

Suppose that p>1, 1 p + 1 q =1, f(x),g(y)0, f L p ( R + ), g L q ( R + ), f p = { 0 f p ( x ) d x } 1 p >0, g q >0. We have the following Hardy-Hilbert integral inequality (cf. [1]):

0 0 f ( x ) g ( y ) x + y dxdy< π sin ( π / p ) f p g q ,
(1)

where the constant factor π sin ( π / p ) is the best possible. Assuming that a m , b n 0, a= { a m } m = 1 l p , b= { b n } n = 1 l q , a p = { m = 1 a m p } 1 p >0, b q >0, we have the following Hardy-Hilbert inequality with the same best constant π sin ( π / p ) (cf. [1]):

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

Inequalities (1) and (2) are important in analysis and its applications (cf. [16]). Also we have the following Mulholland inequality (cf. [1]):

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 .
(3)

In 1998, by introducing an independent parameter λ(0,1], Yang [7] gave an extension of (1) for p=q=2. Yang [5] gave some extensions of (1) and (2) as follows: If λ 1 , λ 2 ,λR, λ 1 + λ 2 =λ, k λ (x,y) is a non-negative homogeneous function of degree −λ, with

k( λ 1 )= 0 k λ (t,1) t λ 1 1 dt R + ,

ϕ(x)= x p ( 1 λ 1 ) 1 , ψ(x)= x q ( 1 λ 2 ) 1 , f(x),g(y)0,

f L p , ϕ ( R + )= { f ; f p , ϕ : = { 0 ϕ ( x ) | f ( x ) | p d x } 1 p < } ,

g L q , ψ ( R + ), f p , ϕ , g q , ψ >0, then

0 0 k λ (x,y)f(x)g(y)dxdy<k( λ 1 ) f p , ϕ g q , ψ ,
(4)

where the constant factor k( λ 1 ) is 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 with respect to x>0 (y>0), then for a m , b n 0,

a 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, it follows that

m = 1 n = 1 k λ (m,n) a m b n <k( λ 1 ) a p , ϕ b q , ψ ,
(5)

where the constant factor k( λ 1 ) is still the best possible.

Clearly, for λ=1, k 1 (x,y)= 1 x + y , λ 1 = 1 q , λ 2 = 1 p , inequality (3) reduces to (1), while (5) reduces to (2). Some other results including the multidimensional Hilbert-type integral inequalities are provided by [824].

About half-discrete Hilbert-type inequalities with the non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors are the best possible. However, Yang [25] gave a result with the kernel 1 ( 1 + n x ) λ (0<λ2) by introducing a variable and proved that the constant factor is the best possible. In 2011 Yang [26] gave a half-discrete Hardy-Hilbert inequality with the best possible constant factor. Zhong et al. [2733] investigated several half-discrete Hilbert-type inequalities with particular kernels. Using the way of weight functions and the techniques of discrete and integral Hilbert-type inequalities with some additional conditions on the kernel, a half-discrete Hilbert-type inequality with a general homogeneous kernel of degree λR and a best constant factor k( λ 1 ) is obtained as follows:

0 f(x) n = 1 k λ (x,n) a n dx<k( λ 1 ) f p , ϕ a q , ψ
(6)

(see Yang and Chen [34]). At the same time, a half-discrete Hilbert-type inequality with a general non-homogeneous kernel and the best constant factor is given by Yang [35].

In this paper, by using the way of weight coefficients and technique of real analysis, a more accurate multidimensional discrete Mulholland-type inequality with the best possible constant factor is given, which is an extension of (3). The equivalent form, the operator expression with the norm as well as a few particular cases are also considered.

2 Some lemmas

Lemma 1 If ( 1 ) i h ( i ) (t)>0 (t>0; i=1,2), then for b>0, 0<α1,

( 1 ) i d i d x i h ( ( b + ln α x ) 1 α ) >0(x>1;i=1,2).
(7)

Proof We find

d d x h ( ( b + ln α x ) 1 α ) = 1 x h ( ( b + ln α x ) 1 α ) ( b + ln α x ) 1 α 1 ln α 1 x < 0 , d 2 d x 2 h ( ( b + ln α x ) 1 α ) = d d x [ 1 x h ( ( b + ln α x ) 1 α ) ( b + ln α x ) 1 α 1 ln α 1 x ] d 2 d x 2 h ( ( b + ln α x ) 1 α ) = 1 x 2 h ( ( b + ln α x ) 1 α ) ( b + ln α x ) 1 α 1 ln α 1 x d 2 d x 2 h ( ( b + ln α x ) 1 α ) = + 1 x 2 h ( ( b + ln α x ) 1 α ) ( b + ln α x ) 2 α 2 ln 2 α 2 x d 2 d x 2 h ( ( b + ln α x ) 1 α ) = + α ( 1 α 1 ) 1 x 2 h ( ( b + ln α x ) 1 α ) ( b + ln α x ) 1 α 2 ln 2 α 2 x d 2 d x 2 h ( ( b + ln α x ) 1 α ) = + ( α 1 ) 1 x 2 h ( ( b + ln α x ) 1 α ) ( b + ln α x ) 1 α 1 ln α 2 x d 2 d x 2 h ( ( b + ln α x ) 1 α ) = [ h ( ( b + ln α x ) 1 α ) ( b + ln α x ) ln x d 2 d x 2 h ( ( b + ln α x ) 1 α ) = + h ( ( b + ln α x ) 1 α ) ( b + ln α x ) 1 α ln α x d 2 d x 2 h ( ( b + ln α x ) 1 α ) = + b ( α 1 ) h ( ( b + ln α x ) 1 α ) ] 1 x 2 ( b + ln α x ) 1 α 2 ln α 2 x > 0 .

Then we have (7). □

If i 0 , j 0 N (N is the set of positive integers), α,β>0, we set

x α := ( k = 1 i 0 | x k | α ) 1 α ( x = ( x 1 , , x i 0 ) R i 0 ) ,
(8)
y β := ( k = 1 j 0 | y k | β ) 1 β ( y = ( y 1 , , y j 0 ) R j 0 ) .
(9)

Lemma 2 If sN, γ,M>0, Ψ(u) is a non-negative measurable function in (0,1], and

D M := { x R + s ; i = 1 s x i γ M γ } = { x ; i = 1 s ( x i M ) γ 1 } ,

then we have (cf. [36])

D M Ψ ( i = 1 s ( x i M ) γ ) d x 1 d x s = M s Γ s ( 1 γ ) γ s Γ ( s γ ) 0 1 Ψ ( u ) u s γ 1 d u .
(10)

Lemma 3 If sN, γ>0, ε>0, d=( d 1 ,, d s ) [ 1 2 , 1 ] s , then

A s ( ε ) : = m ln ( m + d ) γ s ε 1 i = 1 s ( m i + d i ) = Γ s ( 1 γ ) ε s ε / γ γ s 1 Γ ( s γ ) + O ( 1 ) ( ε 0 + ) .
(11)

Proof For M> s 1 / γ , we set

Ψ(u)= { 0 , 0 < u < s M γ , ( M u 1 / γ ) s ε , s M γ u 1 .

Then by the decreasing property and (10), it follows that

A s ( ε ) { x R + s ; x i e d i } ln ( x + d ) γ s ε d x i = 1 s ( x i + d i ) = u i = ln ( x i + d i ) { u R + s ; u i 1 } u γ s ε d u = lim M D M Ψ ( i = 1 s ( x i M ) γ ) d x 1 d x s = lim M M s Γ s ( 1 γ ) γ s Γ ( s γ ) s / M γ 1 ( M u 1 / γ ) s ε u s γ 1 d u = Γ s ( 1 γ ) ε s ε / γ γ s 1 Γ ( s γ ) .

In the following, by mathematical induction we prove that, for any sN,

A s (ε) O s (1)+ Γ s ( 1 γ ) ε s ε / γ γ s 1 Γ ( s γ ) ( ε 0 + ) .
(12)

For s=1, by the Hermite-Hadamard inequality (cf. [37]), it follows that

A 1 ( ε ) = m 1 = 1 2 ln 1 ε ( m 1 + d 1 ) m 1 + d 1 + m 1 = 3 ln 1 ε ( m 1 + d 1 ) m 1 + d 1 O 1 ( 1 ) + 5 2 ln 1 ε ( x + d 1 ) d x x + d 1 O 1 ( 1 ) + e d 1 ln 1 ε ( x + d 1 ) d x x + d 1 = u = ln ( x + d 1 ) O 1 ( 1 ) + 1 u 1 ε d u = O 1 ( 1 ) + 1 ε ,

and then (12) is valid. Assuming that (12) is valid for s1N, then for s, we set

A s ( ε ) = { m N s ; i 0 , m i 0 = 1 , 2 } ln ( m + d ) γ s ε 1 i = 1 s ( m i + d i ) + { m N s ; m i 3 } ln ( m + d ) γ s ε 1 i = 1 s ( m i + d i ) .

There exist constants a,b R + , such that

{ m N s ; i 0 , m i 0 = 1 , 2 } ln ( m + d ) γ s ε 1 i = 1 s ( m i + d i ) a + b { m N s 1 ; m i 1 } ln ( m + d ) γ ( s 1 ) ( 1 + ε ) 1 i = 1 s 1 ( m i + d i ) .

By the assumption of mathematical induction for s1, we find

{ m N s 1 ; m i 1 } ln ( m + d ) γ ( s 1 ) ( 1 + ε ) 1 i = 1 s 1 ( m i + d i ) O s 1 ( 1 ) + Γ s 1 ( 1 γ ) ( 1 + ε ) ( s 1 ) ( 1 + ε ) / γ γ s 2 Γ ( s 1 γ ) ,

and then

{ m N s ; i 0 , m i 0 = 1 , 2 } ln ( m + d ) γ s ε 1 i = 1 s ( m i + d i ) O s (1).

By Lemma 1 and the Hermite-Hadamard inequality (cf. [37]), we obtain

{ m N s ; m i 3 } ln ( m + d ) γ s ε 1 i = 1 s ( m i + d i ) { x R + s ; x i 5 2 } ln ( x + d ) γ s ε 1 i = 1 s ( x i + d i ) d x { x R + s ; x i e d i } ln ( x + d ) γ s ε 1 i = 1 s ( x i + d i ) d x = u i = ln ( x i + d i ) { u R + s ; u i 1 } u γ s ε d u = Γ s ( 1 γ ) ε s ε / γ γ s 1 Γ ( s γ ) .

Hence we prove that (12) is valid for sN. Therefore, we have (11). □

Lemma 4 If C is the set of complex numbers and C =C{}, z k C{z|Rez0,Imz=0} (k=1,2,,n) are different points, the function f(z) is analytic in C except for z i (i=1,2,,n), and z= is a zero point of f(z) whose order is not less than 1, then for αR, we have

0 f(x) x α 1 dx= 2 π i 1 e 2 π α i k = 1 n Res [ f ( z ) z α 1 , z k ] ,
(13)

where 0<Imlnz=argz<2π. In particular, if z k (k=1,,n) are all poles of order 1, setting φ k (z)=(z z k )f(z) ( φ k ( z k )0), then

0 f(x) x α 1 dx= π sin π α k = 1 n ( z k ) α 1 φ k ( z k ).
(14)

Proof By [[38], p.118], we have (13). We find

1 e 2 π α i = 1 cos 2 π α i sin 2 π α = 2 i sin π α ( cos π α + i sin π α ) = 2 i e i π α sin π α .

In particular, since f(z) z α 1 = 1 z z k ( φ k (z) z α 1 ), it is obvious that

Res [ f ( z ) z α 1 , a k ] = z k α 1 φ k ( z k )= e i π α ( z k ) α 1 φ k ( z k ).

Then by (13), we obtain (14). □

Example 1 For sN, we set

k λ (x,y)= k = 1 s 1 ( x λ / s + c k y λ / s ) (0< c 1 << c s ,0<λs).

For 0< λ 1 i 0 , 0< λ 2 j 0 , λ 1 + λ 2 =λ, by (14), we find

k s ( λ 1 ) : = 0 k = 1 s 1 t λ / s + c k t λ 1 1 d t = u = t λ / s s λ 0 k = 1 s 1 u + c k u s λ 1 λ 1 d u = π s λ sin ( π s λ 1 λ ) k = 1 s c k s λ 1 λ 1 j = 1 ( j k ) s 1 c j c k R + .
(15)

In particular, for s=1, we obtain

k 1 ( λ 1 )= 1 λ 0 u ( λ 1 / λ ) 1 u + c 1 du= π λ sin ( π λ 1 λ ) c 1 λ 1 λ 1 .

Definition 1 For sN, 0<α,β1, 0< c 1 << c s , 0<λs, 0< λ 1 i 0 , 0< λ 2 j 0 , λ 1 + λ 2 =λ, τ=( τ 1 ,, τ i 0 ) [ 1 2 , 1 ] i 0 , σ=( σ 1 ,, σ j 0 ) [ 1 2 , 1 ] j 0 , ln(m+τ)=(ln( m 1 + τ 1 ),,ln( m i 0 + τ i 0 )) R + i 0 , ln(n+σ)=(ln( n 1 + σ 1 ),,ln( n j 0 + σ j 0 )) R + j 0 , we define two weight coefficients w λ ( λ 2 ,n) and W λ ( λ 1 ,m) as follows:

w λ ( λ 2 , n ) : = m ln ( n + σ ) β λ 2 ln ( m + τ ) α λ 1 i 0 k = 1 s [ ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ] i = 1 i 0 ( m i + τ i ) , W λ ( λ 1 , m ) : = n ln ( m + τ ) α λ 1 ln ( n + σ ) β λ 2 j 0 k = 1 s [ ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ] j = 1 j 0 ( n j + σ j ) ,
(16)

where m = m i 0 = 1 m 1 = 1 and n = n j 0 = 1 n 1 = 1 .

Lemma 5 Let the assumptions as in Definition  1 be fulfilled. Then:

  1. (i)

    we have

    w λ ( λ 2 ,n)< K 2 ( n N j 0 ) ,
    (17)
W λ ( λ 1 ,m)< K 1 ( m N i 0 ) ,
(18)

where

K 1 := Γ j 0 ( 1 β ) β j 0 1 Γ ( j 0 β ) k s ( λ 1 ), K 2 := Γ i 0 ( 1 α ) α i 0 1 Γ ( i 0 α ) k s ( λ 1 ),
(19)

and k s ( λ 1 ) is indicated by (15);

  1. (ii)

    for p>1, 0<ε<pmin{ λ 1 ,1 λ 2 }, setting λ ˜ 1 = λ 1 ε p , λ ˜ 2 = λ 2 + ε p , we have

    0< K ˜ 2 ( 1 θ ˜ λ ( n ) ) < w λ ( λ ˜ 2 ,n),
    (20)

where

θ ˜ λ ( n ) : = 1 k s ( λ ˜ 1 ) 0 i 0 λ / ( α s ) / ln ( n + σ ) β λ / s v s λ 1 λ 1 k = 1 s ( v + c k ) d v θ ˜ λ ( n ) = O ( 1 ln ( n + σ ) β λ ˜ 1 ) ,
(21)
K ˜ 2 = Γ i 0 ( 1 α ) k s ( λ ˜ 1 ) α i 0 1 Γ ( i 0 α ) R + .
(22)

Proof By Lemma 1, the Hermite-Hadamard inequality (cf. [37]), (10), and (15), it follows that

w λ ( λ 2 , n ) < ( 1 2 , ) i 0 ln ( n + σ ) β λ 2 ln ( x + τ ) α λ 1 i 0 d x k = 1 s [ ln ( x + τ ) α λ / s + c k ln ( n + σ ) β λ / s ] i = 1 i 0 ( x i + τ i ) = u i = ln ( x i + τ i ) { u R + i 0 ; u i > ln ( 1 2 + τ i ) } ln ( n + σ ) β λ 2 u α λ 1 i 0 k = 1 s [ u α λ / s + c k ln ( n + σ ) β λ / s ] d u R + i 0 n σ β λ 2 u α λ 1 i 0 k = 1 s [ u α λ / s + c k ln ( n + σ ) β λ / s ] d u = lim M D M ln ( n + σ ) β λ 2 M λ 1 i 0 [ i = 1 j 0 ( u i M ) α ] ( λ 1 i 0 ) / α k = 1 s { M λ s [ i = 1 i 0 ( u i M ) α ] λ α s + c k ln ( n + σ ) β λ s } d u = lim M M i 0 Γ i 0 ( 1 α ) α i 0 Γ ( i 0 α ) 0 1 ln ( n + σ ) β λ 2 M λ 1 i 0 t ( λ 1 i 0 ) / α t i 0 α 1 k = 1 s ( M λ s t λ α s + c k ln ( n + σ ) β λ s ) d t = lim M M λ 1 Γ i 0 ( 1 α ) α i 0 Γ ( i 0 α ) 0 1 ln ( n + σ ) β λ 2 t λ 1 α 1 k = 1 s ( M λ s t λ α s + c k ln ( n + σ ) β λ s ) d t = t = ln ( n + σ ) β α M α v α s / λ s Γ i 0 ( 1 α ) λ α i 0 1 Γ ( i 0 α ) 0 v s λ 1 λ 1 k = 1 s ( v + c k ) d v = Γ i 0 ( 1 α ) α i 0 1 Γ ( i 0 α ) k s ( λ 1 ) = K 2 .

Hence, we have (17). In the same way, we have (18).

By the decreasing property and (10), similarly to the proof of (11), we find

w λ ( λ ˜ 2 , n ) > { x R + i 0 ; x i e τ i } ln ( n + σ ) β λ ˜ 2 ln ( x + τ ) α λ ˜ 1 i 0 d x k = 1 s [ ln ( x + τ ) α λ / s + c k ln ( n + σ ) β λ / s ] i = 1 i 0 ( x i + τ i ) = u i = ln ( x i + τ i ) ln ( n + σ ) β λ ˜ 2 { u R + i 0 ; u i 1 } u α λ ˜ 1 i 0 d u k = 1 s [ u α λ / s + c k ln ( n + σ ) β λ / s ] = ln ( n + σ ) β λ ˜ 2 lim M D M { i = 1 i 0 ( u i M ) α } λ ˜ 1 i 0 α M λ ˜ 1 i 0 d u 1 d u i 0 k = 1 s [ { i = 1 i 0 ( u i M ) α } λ α s M λ s + c k ln ( n + σ ) β λ / s ] = M i 0 Γ i 0 ( 1 α ) α i 0 Γ ( i 0 α ) ln ( n + σ ) β λ ˜ 2 lim M i 0 M α 1 t λ ˜ 1 i 0 α M λ ˜ 1 i 0 t i 0 α 1 k = 1 s [ t λ α s M λ s + c k ln ( n + σ ) β λ / s ] d t = s Γ i 0 ( 1 α ) λ α i 0 1 Γ ( i 0 α ) i 0 λ / ( α s ) ln ( n + σ ) β λ / s v s λ ˜ 1 λ 1 d v k = 1 s ( v + c k ) = K ˜ 2 ( 1 θ ˜ λ ( n ) ) > 0 , 0 < θ ˜ λ ( n ) = s λ k s ( λ ˜ 1 ) 0 i 0 λ / ( α s ) / ln ( n + σ ) β λ / s v s λ 1 λ 1 k = 1 s ( v + c k ) d v 0 s λ k s ( λ ˜ 1 ) k = 1 s c k 0 i 0 λ / ( α s ) / ln ( n + σ ) β λ / s v s λ ˜ 1 λ 1 d v 0 = 1 λ ˜ 1 k s ( λ ˜ 1 ) k = 1 s c k i 0 λ ˜ 1 / α ln ( n + σ ) β λ ˜ 1 .

Hence, we have (20) and (21). □

3 Main results and operator expressions

Setting Φ(m):= i = 1 i 0 ( m i + τ i ) p 1 ln ( m + τ ) α p ( i 0 λ 1 ) i 0 (m N i 0 ) and Ψ(n):= j = 1 j 0 ( n j + σ j ) q 1 ln ( n + σ ) β q ( j 0 λ 2 ) j 0 (n N j 0 ), wherefrom

[ Ψ ( n ) ] 1 p = j = 1 j 0 ( n j + σ j ) 1 ln ( n + σ ) β p λ 2 j 0 ,

we have the following.

Theorem 1 If sN, 0<α,β1, 0< c 1 << c s , 0<λs, 0< λ 1 i 0 , 0< λ 2 j 0 , λ 1 + λ 2 =λ, τ [ 1 2 , 1 ] i 0 , σ [ 1 2 , 1 ] j 0 , then for p>1, 1 p + 1 q =1, a m , b n 0, 0< a p , Φ , b q , Ψ <, we have

I : = n m a m b n k = 1 s [ ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ] < K 1 1 p K 2 1 q a p , Φ b q , Ψ ,
(23)

where the constant factor

K 1 1 p K 2 1 q = [ Γ j 0 ( 1 β ) β j 0 1 Γ ( j 0 β ) ] 1 p [ Γ i 0 ( 1 α ) β i 0 1 Γ ( i 0 α ) ] 1 q k s ( λ 1 )
(24)

is the best possible ( k s ( λ 1 ) is indicated by (15)).

Proof By the Hölder inequality (cf. [37]), we have

I = n m 1 k = 1 s [ ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ] × [ ln ( m + τ ) α ( i 0 λ 1 ) / q ln ( n + σ ) β ( j 0 λ 2 ) / p i = 1 i 0 ( m i + τ i ) 1 / q j = 1 j 0 ( n j + σ j ) 1 / p a m ] × [ ln ( n + σ ) β ( j 0 λ 2 ) / p ln ( m + τ ) α ( i 0 λ 1 ) / q j = 1 j 0 ( n j + σ j ) 1 / p i = 1 i 0 ( m i + τ i ) 1 / q b n ] { m W λ ( λ 1 , m ) i = 1 i 0 ( m i + τ i ) p 1 ln ( m + τ ) α p ( i 0 λ 1 ) i 0 a m p } 1 p × { n w λ ( λ 2 , n ) j = 1 j 0 ( n j + σ j ) q 1 ln ( n + σ ) β q ( j 0 λ 2 ) j 0 b n q } 1 q .

Then by (17) and (18), we have (23).

For 0<ε<pmin{ λ 1 ,1 λ 2 }, λ ˜ 1 = λ 1 ε p , λ ˜ 2 = λ 2 + ε p , we set

a ˜ m = ln ( m + τ ) α i 0 + λ 1 ε p 1 i = 1 i 0 ( m i + τ i ) , b ˜ n = ln ( n + σ ) β j 0 + λ 2 ε q 1 j = 1 j 0 ( n j + σ j ) ( m N i 0 , n N j 0 ) .

Then by (11) and (20)-(22), we obtain

a ˜ p , Φ b ˜ q , Ψ = { m i = 1 i 0 ( m i + τ i ) p 1 ln ( m + τ ) α p ( i 0 λ 1 ) i 0 a ˜ m p } 1 p a ˜ p , Φ b ˜ q , Ψ = × { n j = 1 j 0 ( n j + σ j ) q 1 ln ( n + σ ) β q ( j 0 λ 2 ) j 0 b ˜ n q } 1 q a ˜ p , Φ b ˜ q , Ψ = { m ln ( m + τ ) α i 0 ε 1 i = 1 i 0 ( m i + τ i ) } 1 p a ˜ p , Φ b ˜ q , Ψ = × { n ln ( n + σ ) β j 0 ε 1 j = 1 j 0 ( n j + σ j ) } 1 q a ˜ p , Φ b ˜ q , Ψ = 1 ε [ Γ i 0 ( 1 α ) i 0 ε / α α i 0 1 Γ ( i 0 α ) + ε O ( 1 ) ] 1 p [ Γ j 0 ( 1 β ) j 0 ε / β β j 0 1 Γ ( j 0 β ) + ε O ˜ ( 1 ) ] 1 q ,
(25)
I ˜ : = n [ m a ˜ m k = 1 s ( ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ) ] b ˜ n I ˜ = n w λ ( λ ˜ 2 , n ) ln ( n + σ ) β j 0 ε 1 j = 1 j 0 ( n j + σ j ) I ˜ > K ˜ 2 n ( 1 O ( 1 ln ( n + σ ) β λ ˜ 1 ) ) ln ( n + σ ) β j 0 ε 1 j = 1 j 0 ( n j + σ j ) I ˜ = K ˜ 2 [ Γ j 0 ( 1 β ) ε j 0 ε / β β j 0 1 Γ ( j 0 β ) + O ˜ ( 1 ) O ( 1 ) ] .
(26)

If there exists a constant K K 1 1 p K 2 1 q , such that (23) is valid when replacing K 1 1 p K 2 1 q by K, then we have

( K 2 + o ( 1 ) ) [ Γ j 0 ( 1 β ) j 0 ε / β β j 0 1 Γ ( j 0 β ) + ε O ˜ ( 1 ) ε O ( 1 ) ] < ε I ˜ < ε K a ˜ p , φ b ˜ q , ψ = K [ Γ i 0 ( 1 α ) i 0 ε / α α i 0 1 Γ ( i 0 α ) + ε O ( 1 ) ] 1 p [ Γ j 0 ( 1 β ) j 0 ε / β β j 0 1 Γ ( j 0 β ) + ε O ˜ ( 1 ) ] 1 q .

For ε 0 + , we find

Γ j 0 ( 1 β ) Γ i 0 ( 1 α ) k s ( λ 1 ) β j 0 1 Γ ( j 0 β ) α i 0 1 Γ ( i 0 α ) K [ Γ i 0 ( 1 α ) α i 0 1 Γ ( i 0 α ) ] 1 p [ Γ j 0 ( 1 β ) β j 0 1 Γ ( j 0 β ) ] 1 q ,

and then K 1 1 p K 2 1 q K. Hence, K= K 1 1 p K 2 1 q is the best possible constant factor of (23). □

Theorem 2 With the assumptions of Theorem  1, for 0< a p , Φ <, we have the following inequality with the best constant factor K 1 1 p K 2 1 q :

J : = { n [ Ψ ( n ) ] 1 p ( m a m k = 1 s [ ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ] ) p } 1 p < K 1 1 p K 2 1 q a p , Φ ,
(27)

which is equivalent to (23).

Proof We set b n as follows:

b n := [ Ψ ( n ) ] 1 p ( m a m k = 1 s ( ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ) ) p 1 .

Then it follows that J p = b q , Ψ q . If J=0, then (27) is trivially valid, since 0< a p , Φ <; if J=, then it is a contradiction since the right hand side of (27) is finite. Suppose that 0<J<. Then by (23), we find

b q , Ψ q = J p =I< K 1 1 p K 2 1 q a p , Φ b q , Ψ ,

namely, b q , Ψ q 1 =J< K 1 1 p K 2 1 q a p , Φ , and then (27) follows.

On the other hand, assuming that (27) is valid, by the Hölder inequality, we have

I = n ( Ψ ( n ) ) 1 q [ m a m k = 1 s ( ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ) ] × [ ( Ψ ( n ) ) 1 q b n ] J b q , Ψ .
(28)

Then by (27), we have (23). Hence (27) and (23) are equivalent.

By the equivalency, the constant factor K 1 1 p K 2 1 q in (27) is the best possible. Otherwise, we would reach a contradiction by (28) that the constant factor K 1 1 p K 2 1 q in (23) is not the best possible. □

For p>1, we define two real weight normal discrete spaces l p , φ and l q , ψ as follows:

l p , φ : = { a = { a m } ; a p , Φ = { m Φ ( m ) a m p } 1 p < } , l q , ψ : = { b = { b n } ; b q , Ψ = { n Ψ ( n ) b n q } 1 q < } .

With the assumptions of Theorem 2, in view of J< K 1 1 p K 2 1 q a p , Φ , we have the following definition.

Definition 2 Define a multidimensional Hilbert-type operator T: l p , Φ l p , Ψ 1 p as follows: For a l p , Φ , there exists an unique representation Ta l p , Ψ 1 p , satisfying for n N j 0 ,

(Ta)(n):= m a m k = 1 s [ ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ] .
(29)

For b l q , Ψ , we define the following formal inner product of Ta and b as follows:

(Ta,b):= n m a m b n k = 1 s [ ln ( m + τ ) α λ / s + c k ln ( n + σ ) β λ / s ] .
(30)

Then by Theorem 1 and Theorem 2, for 0< a p , φ , b q , ψ <, we have the following equivalent inequalities:

(Ta,b)< K 1 1 p K 2 1 q a p , Φ b q , Ψ ,
(31)
T a p , Ψ 1 p < K 1 1 p K 2 1 q a p , Φ .
(32)

It follows that T is bounded since

T:= sup a ( θ ) l p , Φ T a p , Ψ 1 p a p , Φ K 1 1 p K 2 1 q .
(33)

Since the constant factor K 1 1 p K 2 1 q in (32) is the best possible, we have:

Corollary 1 With the assumptions of Theorem  2, T is defined by Definition  2, it follows that

T= K 1 1 p K 2 1 q = [ Γ j 0 ( 1 β ) β j 0 1 Γ ( j 0 β ) ] 1 p [ Γ i 0 ( 1 α ) α i 0 1 Γ ( i 0 α ) ] 1 q k s ( λ 1 ).
(34)

Remark 1 (i) Setting Φ 1 (m):= i = 1 i 0 ( m i + 1 ) p 1 ln ( m + 1 ) α p ( i 0 λ 1 ) i 0 (m N i 0 ) and Ψ 1 (n):= j = 1 j 0 ( n j + 1 ) q 1 ln ( n + 1 ) β q ( j 0 λ 2 ) j 0 (n N j 0 ), then putting τ=σ=1 in (23) and (27), we have the following equivalent inequalities with the best constant factor K 1 1 p K 2 1 q :

n m a m b n k = 1 s [ ln ( m + 1 ) α λ / s + c k ln ( n + 1 ) β λ / s ] < K 1 1 p K 2 1 q a p , Φ 1 b q , Ψ 1 ,
(35)
{ n [ Ψ 1 ( n ) ] 1 p ( m a m k = 1 s [ ln ( m + 1 ) α λ / s + c k ln ( n + 1 ) β λ / s ] ) p } 1 p < K 1 1 p K 2 1 q a p , Φ 1 .
(36)

Hence, (23) and (27) are more accurate inequalities than (35) and (36).

  1. (ii)

    Putting i 0 = j 0 =1, λ=s, ϕ 1 (m):= ( m + 1 ) p 1 ln p ( 1 λ 1 ) 1 (m+1) (mN) and ψ 1 (n):= ( n + 1 ) q 1 ln q ( 1 λ 2 ) 1 (n+1) (nN), in (32), we have the following new inequality:

    m = 1 n = 1 a m b n k = 1 s ln ( m + 1 ) ( n + 1 ) c k < π sin ( π λ 1 ) k = 1 s j = 1 ( j k ) s c k λ 1 1 c j c k a p , ϕ 1 b q , ψ 1 .
    (37)

In particular, for s= c k =1, λ 1 = 1 q , λ 2 = 1 p in (37), we can deduce (4). Hence, (23) is an extension of (4).

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Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 61370186), and 2013 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2013KJCX0140).

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Correspondence to Bicheng Yang.

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

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Chen, Q., Yang, B. On a more accurate multidimensional Mulholland-type inequality. J Inequal Appl 2014, 322 (2014). https://doi.org/10.1186/1029-242X-2014-322

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Keywords

  • Mulholland-type inequality
  • weight coefficient
  • equivalent form
  • operator
  • norm